php

Question

niths -1 Posting Whiz in Training

14 Years Ago

hi all,
i am not getting inserted the dropdown value to the back end except that everything is inserting. so can u please help me.

<?php
ob_start();
@session_start();
require_once ("check.php");
createsessions($username,$password,$userid,$projectid);
    $con=mysql_connect("10.70.1.50","invensis","invensis");
    if(!$con)
        {
            die('Could not connect: ' . mysql_error());
        }
    mysql_select_db("database_myproject",$con);
    $sql="SELECT projectname from projects where createdby='$_SESSION[username]'";
    mysql_error();
    $result=mysql_query($sql);
    $options=""; 
    while ($row=mysql_fetch_array($result)) 
    {
        $id=$row["projectid"];
        $projectname=$row["projectname"];
        $options.="<option value=\"$id\">".$projectname.'</option>';
    }
    $sql2="SELECT * FROM users where reportingto='$_SESSION[username]' and role='2'";
    mysql_error();
    $result2=mysql_query($sql2);
    $num=mysql_num_rows($result2);
    
    ?> 
<head>
<body>
<form action="projectassign.php" method="post">
<table width="100%">
<tr><td> <img src="Logofinalcopy.gif">
</td></tr>
    <tr>
        <td bgcolor="aqua"><h2>Project assign</h2></td>
    </tr></table>
    <br>
    <table align="center">
    <tr>
        <td>projectassign</td>
        <td><select name="id">
        <option value="<?echo '$projectassign';?>" >----select project----<?=$options?>
        </select>
        </td>
    </tr>
    </table>  
    <br><br><br>  
    <table border="1" cellspacing="0" cellpadding="0" align="center" style="width: 500px">
    <tr>
        <th></th>
        <th>userid</th>
        <th>username</th>
           
    </tr>
    <?php
    $sno=1;
    while($row=mysql_fetch_array($result2, MYSQL_ASSOC)){
    ?>
     <tr>
<td><input name="checkbox[]" type="checkbox" id="checkbox[]" value="<? echo $row['userid']; ?>"></td>
<td><? echo $row['userid']; ?></td>
<td><? echo $row['username'];?></td>
</tr>
<?php
    $sno=$sno+1;
}
?>
    </table>
    
    <?php
if(isset($_POST['Submit']))
    
    {
        $checkbox=$_POST['checkbox'];
        for($i=0;$i<count($checkbox);$i++)
        {
        $del_id = $checkbox[$i];        
        $sql = "Insert into projectassign(projectassign,userid,assigneddate)values('$_POST[projectassign]','$del_id','')";
         mysql_error();
        $result = mysql_query($sql);
        
        }   
    }      
    ?>
    <input type="submit" name="Submit"  value="Submit">
    </form>
    </body>
    </head>

php

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Latest Post 14 Years Ago Latest Post by niths

All 27 Replies

cwarn23 387 Occupation: Genius

14 Years Ago

I received your pm and your code should be as follows.

<?php
ob_start();
@session_start();
require_once ("check.php");
createsessions($username,$password,$userid,$projectid);
    $con=mysql_connect("10.70.1.50","invensis","invensis");
    if(!$con)
        {
            die('Could not connect: ' . mysql_error());
        }
    mysql_select_db("database_myproject",$con);
    $sql="SELECT projectname from projects where createdby='$_SESSION[username]'";
    mysql_error();
    $result=mysql_query($sql);
    $options=""; 
    while ($row=mysql_fetch_array($result)) 
    {
        $id=$row["projectid"];
        $projectname=$row["projectname"];
        $options.="<option value=\"$id\">".$projectname.'</option>';
    }
    $sql2="SELECT * FROM users where reportingto='$_SESSION[username]' and role='2'";
    mysql_error();
    $result2=mysql_query($sql2);
    $num=mysql_num_rows($result2);
    
    ?> 
<head>
<body>
<form action="projectassign.php" method="post">
<table width="100%">
<tr><td> <img src="Logofinalcopy.gif">
</td></tr>
    <tr>
        <td bgcolor="aqua"><h2>Project assign</h2></td>
    </tr></table>
    <br>
    <table align="center">
    <tr>
        <td>projectassign</td>
        <td><select name="id">
        <option value="<?php echo '$projectassign';?>" >----select project----</option><?php echo $options ?>
        </select>
        </td>
    </tr>
    </table>  
    <br><br><br>  
    <table border="1" cellspacing="0" cellpadding="0" align="center" style="width: 500px">
    <tr>
        <th></th>
        <th>userid</th>
        <th>username</th>
           
    </tr>
    <?php
    $sno=1;
    while($row=mysql_fetch_array($result2, MYSQL_ASSOC)){
    ?>
     <tr>
<td><input name="checkbox[]" type="checkbox" id="checkbox[]" value="<? echo $row['userid']; ?>"></td>
<td><? echo $row['userid']; ?></td>
<td><? echo $row['username'];?></td>
</tr>
<?php
    $sno=$sno+1;
}
?>
    </table>
    
    <?php
if(isset($_POST['Submit']))
    
    {
        $checkbox=$_POST['checkbox'];
        for($i=0;$i<count($checkbox);$i++)
        {
        $del_id = $checkbox[$i];        
        $sql = "Insert into projectassign(projectassign,userid,assigneddate)values('".$_POST['projectassign']."','$del_id','')";
         mysql_error();
        $result = mysql_query($sql);
        
        }   
    }      
    ?>
    <input type="submit" name="Submit"  value="Submit">
    </form>
    </body>
    </head>

cwarn23 387 Occupation: Genius

14 Years Ago

Could you make your problem more clearer. I will try debugging again which is as follows but if that doesn't solve it you will need to be a bit more descriptive.

<?php
ob_start();
@session_start();
require_once ("check.php");
createsessions($username,$password,$userid,$projectid);
    $con=mysql_connect("10.70.1.50","invensis","invensis");
    if(!$con)
        {
            die('Could not connect: ' . mysql_error());
        }
    mysql_select_db("database_myproject",$con);
    $sql="SELECT projectname from projects where createdby='".mysql_real_escape_string($_SESSION['username'])."'";
    mysql_error();
    $result=mysql_query($sql);
    $options=""; 
    while ($row=mysql_fetch_array($result)) 
    {
        $id=$row["projectid"];
        $projectname=$row["projectname"];
        $options.="<option value=\"$id\">".$projectname.'</option>';
    }
    $sql2="SELECT * FROM users where reportingto='".mysql_real_escape_string($_SESSION['username'])."' and role='2'";
    mysql_error();
    $result2=mysql_query($sql2);
    $num=mysql_num_rows($result2);
    
    ?> 
<head>
<body>
<form action="projectassign.php" method="post">
<table width="100%">
<tr><td> <img src="Logofinalcopy.gif">
</td></tr>
    <tr>
        <td bgcolor="aqua"><h2>Project assign</h2></td>
    </tr></table>
    <br>
    <table align="center">
    <tr>
        <td>projectassign</td>
        <td><select name="id">
        <option value="<?php echo '$projectassign';?>" >----select project----</option><?php echo $options ?>
        </select>
        </td>
    </tr>
    </table>  
    <br><br><br>  
    <table border="1" cellspacing="0" cellpadding="0" align="center" style="width: 500px">
    <tr>
        <th></th>
        <th>userid</th>
        <th>username</th>
           
    </tr>
    <?php
    $sno=1;
    while($row=mysql_fetch_array($result2, MYSQL_ASSOC)){
    ?>
     <tr>
<td><input name="checkbox[]" type="checkbox" id="checkbox[]" value="<? echo $row['userid']; ?>"></td>
<td><? echo $row['userid']; ?></td>
<td><? echo $row['username'];?></td>
</tr>
<?php
    $sno=$sno+1;
}
?>
    </table>
    
    <?php
if(isset($_POST['Submit']))
    
    {
        $checkbox=$_POST['checkbox'];
        for($i=0;$i<count($checkbox);$i++)
        {
        $del_id = mysql_real_escape_string($checkbox[$i]);        
        $sql = "INSERT INTO projectassign(projectassign,userid,assigneddate)values('".mysql_real_escape_string($_POST['projectassign'])."','$del_id','')";
         mysql_error();
        $result = mysql_query($sql);
        
        }   
    }      
    ?>
    <input type="submit" name="Submit"  value="Submit">
    </form>
    </body>
    </head>

cwarn23 387 Occupation: Genius

14 Years Ago

If you mean the mysql query on line 78 is not inserting one of the four columns then that is because you never specified the fourth column. Perhaps you should have a second look at the query on line 78.

cwarn23 387 Occupation: Genius

14 Years Ago

Then on line 41 replace with the following.

<td><select name="projectassign">

cwarn23 387 Occupation: Genius

14 Years Ago

try adding print_r($_POST); at the bottom of the page and see what php displays. It should show the key used for your project assign value.

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niths -1 Posting Whiz in Training · Answer 1 · 2010-03-30T11:03:03+00:00

I received your pm and your code should be as follows.

<?php
ob_start();
@session_start();
require_once ("check.php");
createsessions($username,$password,$userid,$projectid);
    $con=mysql_connect("10.70.1.50","invensis","invensis");
    if(!$con)
        {
            die('Could not connect: ' . mysql_error());
        }
    mysql_select_db("database_myproject",$con);
    $sql="SELECT projectname from projects where createdby='$_SESSION[username]'";
    mysql_error();
    $result=mysql_query($sql);
    $options=""; 
    while ($row=mysql_fetch_array($result)) 
    {
        $id=$row["projectid"];
        $projectname=$row["projectname"];
        $options.="<option value=\"$id\">".$projectname.'</option>';
    }
    $sql2="SELECT * FROM users where reportingto='$_SESSION[username]' and role='2'";
    mysql_error();
    $result2=mysql_query($sql2);
    $num=mysql_num_rows($result2);
    
    ?> 
<head>
<body>
<form action="projectassign.php" method="post">
<table width="100%">
<tr><td> <img src="Logofinalcopy.gif">
</td></tr>
    <tr>
        <td bgcolor="aqua"><h2>Project assign</h2></td>
    </tr></table>
    <br>
    <table align="center">
    <tr>
        <td>projectassign</td>
        <td><select name="id">
        <option value="<?php echo '$projectassign';?>" >----select project----</option><?php echo $options ?>
        </select>
        </td>
    </tr>
    </table>  
    <br><br><br>  
    <table border="1" cellspacing="0" cellpadding="0" align="center" style="width: 500px">
    <tr>
        <th></th>
        <th>userid</th>
        <th>username</th>
           
    </tr>
    <?php
    $sno=1;
    while($row=mysql_fetch_array($result2, MYSQL_ASSOC)){
    ?>
     <tr>
<td><input name="checkbox[]" type="checkbox" id="checkbox[]" value="<? echo $row['userid']; ?>"></td>
<td><? echo $row['userid']; ?></td>
<td><? echo $row['username'];?></td>
</tr>
<?php
    $sno=$sno+1;
}
?>
    </table>
    
    <?php
if(isset($_POST['Submit']))
    
    {
        $checkbox=$_POST['checkbox'];
        for($i=0;$i<count($checkbox);$i++)
        {
        $del_id = $checkbox[$i];        
        $sql = "Insert into projectassign(projectassign,userid,assigneddate)values('".$_POST['projectassign']."','$del_id','')";
         mysql_error();
        $result = mysql_query($sql);
        
        }   
    }      
    ?>
    <input type="submit" name="Submit"  value="Submit">
    </form>
    </body>
    </head>

i am still not getting inserted the value selected in dropdown box, but the check box value it is inserting in data base.

niths -1 Posting Whiz in Training · Answer 2 · 2010-03-30T11:25:48+00:00

Could you make your problem more clearer. I will try debugging again which is as follows but if that doesn't solve it you will need to be a bit more descriptive.

<?php
ob_start();
@session_start();
require_once ("check.php");
createsessions($username,$password,$userid,$projectid);
    $con=mysql_connect("10.70.1.50","invensis","invensis");
    if(!$con)
        {
            die('Could not connect: ' . mysql_error());
        }
    mysql_select_db("database_myproject",$con);
    $sql="SELECT projectname from projects where createdby='".mysql_real_escape_string($_SESSION['username'])."'";
    mysql_error();
    $result=mysql_query($sql);
    $options=""; 
    while ($row=mysql_fetch_array($result)) 
    {
        $id=$row["projectid"];
        $projectname=$row["projectname"];
        $options.="<option value=\"$id\">".$projectname.'</option>';
    }
    $sql2="SELECT * FROM users where reportingto='".mysql_real_escape_string($_SESSION['username'])."' and role='2'";
    mysql_error();
    $result2=mysql_query($sql2);
    $num=mysql_num_rows($result2);
    
    ?> 
<head>
<body>
<form action="projectassign.php" method="post">
<table width="100%">
<tr><td> <img src="Logofinalcopy.gif">
</td></tr>
    <tr>
        <td bgcolor="aqua"><h2>Project assign</h2></td>
    </tr></table>
    <br>
    <table align="center">
    <tr>
        <td>projectassign</td>
        <td><select name="id">
        <option value="<?php echo '$projectassign';?>" >----select project----</option><?php echo $options ?>
        </select>
        </td>
    </tr>
    </table>  
    <br><br><br>  
    <table border="1" cellspacing="0" cellpadding="0" align="center" style="width: 500px">
    <tr>
        <th></th>
        <th>userid</th>
        <th>username</th>
           
    </tr>
    <?php
    $sno=1;
    while($row=mysql_fetch_array($result2, MYSQL_ASSOC)){
    ?>
     <tr>
<td><input name="checkbox[]" type="checkbox" id="checkbox[]" value="<? echo $row['userid']; ?>"></td>
<td><? echo $row['userid']; ?></td>
<td><? echo $row['username'];?></td>
</tr>
<?php
    $sno=$sno+1;
}
?>
    </table>
    
    <?php
if(isset($_POST['Submit']))
    
    {
        $checkbox=$_POST['checkbox'];
        for($i=0;$i<count($checkbox);$i++)
        {
        $del_id = mysql_real_escape_string($checkbox[$i]);        
        $sql = "INSERT INTO projectassign(projectassign,userid,assigneddate)values('".mysql_real_escape_string($_POST['projectassign'])."','$del_id','')";
         mysql_error();
        $result = mysql_query($sql);
        
        }   
    }      
    ?>
    <input type="submit" name="Submit"  value="Submit">
    </form>
    </body>
    </head>

i am doing project management in which an admin can asign a project to his user. so for that i need to have a dropdown of list of projects present under him and the users with checkbox userid,username are displaying. so now admin need to assign a project to his users. so he selected a project from dropdown and assigned that to one user by clicking on check box. in the data base i am having a table of project assign with fields assignid,projectname,userid,asigned date. so when i selected a user and a project and click submit button iam getting userid value and assigned date but in the projectname nothing is displaying. actually the selected value in the dropdown box should get displayed. so i think u understood my problem.

niths -1 Posting Whiz in Training · Answer 3 · 2010-03-30T11:48:12+00:00

If you mean the mysql query on line 78 is not inserting one of the four columns then that is because you never specified the fourth column. Perhaps you should have a second look at the query on line 78.

Actually assignid is nothing but sl.no. so it will increment. and project name(which is not assigning) and userid(getting inserted) and assigned date(get inserted)

niths -1 Posting Whiz in Training · Answer 4 · 2010-03-30T12:06:07+00:00

Then on line 41 replace with the following.
<td><select name="projectassign">

i have made changes but it is not inserting

niths -1 Posting Whiz in Training · Answer 5 · 2010-03-30T12:24:14+00:00

try adding print_r($_POST); at the bottom of the page and see what php displays. It should show the key used for your project assign value.

i am getting Array() at the end .

cwarn23 387 Occupation: Genius Team Colleague Featured Poster · Answer 6 · 2010-03-30T12:26:47+00:00

i am getting Array() at the end .

Did you post the form? if it only displays Array() then that means no $_POST data has been sent.

niths -1 Posting Whiz in Training · Answer 7 · 2010-03-30T12:31:42+00:00

Did you post the form? if it only displays Array() then that means no $_POST data has been sent.

i think i posted it. please verify my code. where should i post it.

cwarn23 387 Occupation: Genius Team Colleague Featured Poster · Answer 8 · 2010-03-30T12:35:23+00:00

On line 83 just after the last bracket. That is if the filename is projectassign.php as it is being posted to projectassign.php

niths -1 Posting Whiz in Training · Answer 9 · 2010-03-30T12:42:32+00:00

On line 83 just after the last bracket. That is if the filename is projectassign.php as it is being posted to projectassign.php

ok. But wat should i write code there. please tell me. i am already posting this in the form. again we need to post it??

cwarn23 387 Occupation: Genius Team Colleague Featured Poster · Answer 10 · 2010-03-30T12:45:47+00:00

To make things easier make the following file projectassign.php and in it is as below. Now what does the print_r at the bottom tell you in the webpage?

<?php
ob_start();
@session_start();
require_once ("check.php");
createsessions($username,$password,$userid,$projectid);
    $con=mysql_connect("10.70.1.50","invensis","invensis");
    if(!$con)
        {
            die('Could not connect: ' . mysql_error());
        }
    mysql_select_db("database_myproject",$con);
    $sql="SELECT projectname from projects where createdby='".mysql_real_escape_string($_SESSION['username'])."'";
    mysql_error();
    $result=mysql_query($sql);
    $options=""; 
    while ($row=mysql_fetch_array($result)) 
    {
        $id=$row["projectid"];
        $projectname=$row["projectname"];
        $options.="<option value=\"$id\">".$projectname.'</option>';
    }
    $sql2="SELECT * FROM users where reportingto='".mysql_real_escape_string($_SESSION['username'])."' and role='2'";
    mysql_error();
    $result2=mysql_query($sql2);
    $num=mysql_num_rows($result2);
    
    ?> 
<head>
<body>
<form action="projectassign.php" method="post">
<table width="100%">
<tr><td> <img src="Logofinalcopy.gif">
</td></tr>
    <tr>
        <td bgcolor="aqua"><h2>Project assign</h2></td>
    </tr></table>
    <br>
    <table align="center">
    <tr>
        <td>projectassign</td>
        <td><select name="projectassign">
        <option value="<?php echo '$projectassign';?>" >----select project----</option><?php echo $options ?>
        </select>
        </td>
    </tr>
    </table>  
    <br><br><br>  
    <table border="1" cellspacing="0" cellpadding="0" align="center" style="width: 500px">
    <tr>
        <th></th>
        <th>userid</th>
        <th>username</th>
           
    </tr>
    <?php
    $sno=1;
    while($row=mysql_fetch_array($result2, MYSQL_ASSOC)){
    ?>
     <tr>
<td><input name="checkbox[]" type="checkbox" id="checkbox[]" value="<? echo $row['userid']; ?>"></td>
<td><? echo $row['userid']; ?></td>
<td><? echo $row['username'];?></td>
</tr>
<?php
    $sno=$sno+1;
}
?>
    </table>
    
    <?php
if(isset($_POST['Submit']))
    
    {
        $checkbox=$_POST['checkbox'];
        for($i=0;$i<count($checkbox);$i++)
        {
        $del_id = mysql_real_escape_string($checkbox[$i]);        
        $sql = "INSERT INTO projectassign(projectassign,userid,assigneddate)values('".mysql_real_escape_string($_POST['projectassign'])."','$del_id','')";
         mysql_error();
        $result = mysql_query($sql);
        
        }   
    }      print_r($_POST);
    ?>
    <input type="submit" name="Submit"  value="Submit">
    </form>
    </body>
    </head>

niths -1 Posting Whiz in Training · Answer 11 · 2010-03-30T12:51:19+00:00

To make things easier make the following file projectassign.php and in it is as below. Now what does the print_r at the bottom tell you in the webpage?

<?php
ob_start();
@session_start();
require_once ("check.php");
createsessions($username,$password,$userid,$projectid);
    $con=mysql_connect("10.70.1.50","invensis","invensis");
    if(!$con)
        {
            die('Could not connect: ' . mysql_error());
        }
    mysql_select_db("database_myproject",$con);
    $sql="SELECT projectname from projects where createdby='".mysql_real_escape_string($_SESSION['username'])."'";
    mysql_error();
    $result=mysql_query($sql);
    $options=""; 
    while ($row=mysql_fetch_array($result)) 
    {
        $id=$row["projectid"];
        $projectname=$row["projectname"];
        $options.="<option value=\"$id\">".$projectname.'</option>';
    }
    $sql2="SELECT * FROM users where reportingto='".mysql_real_escape_string($_SESSION['username'])."' and role='2'";
    mysql_error();
    $result2=mysql_query($sql2);
    $num=mysql_num_rows($result2);
    
    ?> 
<head>
<body>
<form action="projectassign.php" method="post">
<table width="100%">
<tr><td> <img src="Logofinalcopy.gif">
</td></tr>
    <tr>
        <td bgcolor="aqua"><h2>Project assign</h2></td>
    </tr></table>
    <br>
    <table align="center">
    <tr>
        <td>projectassign</td>
        <td><select name="projectassign">
        <option value="<?php echo '$projectassign';?>" >----select project----</option><?php echo $options ?>
        </select>
        </td>
    </tr>
    </table>  
    <br><br><br>  
    <table border="1" cellspacing="0" cellpadding="0" align="center" style="width: 500px">
    <tr>
        <th></th>
        <th>userid</th>
        <th>username</th>
           
    </tr>
    <?php
    $sno=1;
    while($row=mysql_fetch_array($result2, MYSQL_ASSOC)){
    ?>
     <tr>
<td><input name="checkbox[]" type="checkbox" id="checkbox[]" value="<? echo $row['userid']; ?>"></td>
<td><? echo $row['userid']; ?></td>
<td><? echo $row['username'];?></td>
</tr>
<?php
    $sno=$sno+1;
}
?>
    </table>
    
    <?php
if(isset($_POST['Submit']))
    
    {
        $checkbox=$_POST['checkbox'];
        for($i=0;$i<count($checkbox);$i++)
        {
        $del_id = mysql_real_escape_string($checkbox[$i]);        
        $sql = "INSERT INTO projectassign(projectassign,userid,assigneddate)values('".mysql_real_escape_string($_POST['projectassign'])."','$del_id','')";
         mysql_error();
        $result = mysql_query($sql);
        
        }   
    }      print_r($_POST);
    ?>
    <input type="submit" name="Submit"  value="Submit">
    </form>
    </body>
    </head>

i had inserted the code as it is. now in the web page beside submit button it is displaying
Array()

cwarn23 387 Occupation: Genius Team Colleague Featured Poster · Answer 12 · 2010-03-30T12:59:24+00:00

Then there are two things to consider. One is did you post the $_POST in the first place and the second is have you got globals disabled in your php.ini file as they should. So your problem currently is your post data is not being sent to $_POST. If you want to test your globals in the php.ini file use the following.

<?php
if (!ini_get('register_globals')) {
die('$_POST should work!');
} else {
die ('$_POST won't work due to the php.ini settings.');
}
?>

And let me know which message pops up.

niths -1 Posting Whiz in Training · Answer 13 · 2010-03-30T13:10:17+00:00

Then there are two things to consider. One is did you post the $_POST in the first place and the second is have you got globals disabled in your php.ini file as they should. So your problem currently is your post data is not being sent to $_POST. If you want to test your globals in the php.ini file use the following.
<?php
if (!ini_get('register_globals')) {
die('$_POST should work!');
} else {
die ('$_POST won't work due to the php.ini settings.');
}
?>
And let me know which message pops up.

i am getting this in the browser
Array won't work due to the php.ini settings.

cwarn23 387 Occupation: Genius Team Colleague Featured Poster · Answer 14 · 2010-03-30T13:14:14+00:00

i am getting this in the browser
Array won't work due to the php.ini settings.

It means you will need to go into your php.ini settings and change the register globals to disabled. So if you like post your php.ini file in code tags and I will try and fix.

niths -1 Posting Whiz in Training · Answer 15 · 2010-03-30T13:38:14+00:00

It means you will need to go into your php.ini settings and change the register globals to disabled. So if you like post your php.ini file in code tags and I will try and fix.

i had send u to ur private message

cwarn23 387 Occupation: Genius Team Colleague Featured Poster · Answer 16 · 2010-03-30T13:38:36+00:00

I received your pm and register_globals = Off appears in the php.ini file which is what it should have been. But could you add at the very end of your file the following to see if it displays anything after the post. <?php echo $projectassign.$id; ?> That should confirm weather globals is on or off as we are getting mixed results. So tell me if that displays anything.

niths -1 Posting Whiz in Training · Answer 17 · 2010-03-30T13:43:31+00:00

I received your pm and register_globals = Off appears in the php.ini file which is what it should have been. But could you add at the very end of your file the following to see if it displays anything after the post. <?php echo $projectassign.$id; ?> That should confirm weather globals is on or off as we are getting mixed results. So tell me if that displays anything.

nothing it is displaying extra.

cwarn23 387 Occupation: Genius Team Colleague Featured Poster · Answer 18 · 2010-03-30T13:45:38+00:00

Then could you check your url bar after posting to see that the variables haven't gone to the url.

niths -1 Posting Whiz in Training · Answer 19 · 2010-03-30T14:10:52+00:00

Then could you check your url bar after posting to see that the variables haven't gone to the url.

means where should i check

cwarn23 387 Occupation: Genius Team Colleague Featured Poster · Answer 20 · 2010-03-30T14:13:32+00:00

means where should i check

check inside the url address bar after you have posted. Make sure something like the following hasn't happened. http://localhost/projectassign.php?formvar1=test&var2=another&tickbox=1&missingthing=here

niths -1 Posting Whiz in Training · Answer 21 · 2010-03-30T14:19:42+00:00

check inside the url address bar after you have posted. Make sure something like the following hasn't happened. http://localhost/projectassign.php?formvar1=test&var2=another&tickbox=1&missingthing=here

no nothing like that
it is http://localhost/Project/projectassign.php as it is even after clicking submit button .

niths -1 Posting Whiz in Training · Answer 22 · 2010-03-30T14:25:46+00:00

check inside the url address bar after you have posted. Make sure something like the following hasn't happened. http://localhost/projectassign.php?formvar1=test&var2=another&tickbox=1&missingthing=here

after selecting projectassign and a check box i click on submit button it again gives the same page. and in the data base after refreshing except projectname everything is displaying fine..

php

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