insert not working

Question

niths -1 Posting Whiz in Training

13 Years Ago

hi all,
i had a table with image details.so wen i click on submit button that image details should insert into assignwork table.but that is not working.

<html>

<body>
<link rel="stylesheet" type="text/css" href="css/table.css"/>
<form action="useajax.php" method="post">
<table align="center" id="customers" style="width: 90%">
<tr>
  <th>S.no</th>
  <th>Imageid</th>
  <th>Name</th>
  <th>Size</th>
  <th>Type</th>
  <th>Assigned to</th>
  </tr>
  <?php
    include 'connection.php';
    $q=$_GET["q"];
    $query= mysql_query("SELECT id,imageid,name,size,type from images where uploadeddate='".$q."'");
    mysql_error();
    $num=mysql_num_rows($query);
$i=0;
while ($i < $num){
$f1=mysql_result($query,$i,"id");
$f2=mysql_result($query,$i,"imageid");
$f3=mysql_result($query,$i,"name");
$f4=mysql_result($query,$i,"size");
$f5=mysql_result($query,$i,"type");
if($i%2==0)
{
?>
<tr class="">
    <td><?php echo $f1;?></td>
    <td><?php echo $f2; ?></td>
    <td><?php echo $f3; ?></td>
    <td><?php echo $f4; ?></td>
    <td><?php echo $f5; ?></td>
    <td>
        <select name="assignedto">
            <option><b>SELECT</b></option>
            <option>sss</option>
            <option>jack</option>
            <option>john</option>
            <option>rick</option>
        </select>
    </td>
</tr> 
<?php
}
else
{
?>
<tr class="alt">
    <td><?php echo $f1;?></td>
    <td><?php echo $f2; ?></td>
    <td><?php echo $f3; ?></td>
    <td><?php echo $f4; ?></td>
    <td><?php echo $f5; ?></td>
    <td>
        <select>
            <option><b>SELECT</b></option>
            <option>sss</option>
            <option>jack</option>
            <option>john</option>
            <option>rick</option>
        </select>
    </td>
</tr> 
<?php
}
    $i++;
}
?>
<br></table>
<table align="center">
<input type="submit" name="btn" value="submit">   
</table>
</form>
</body>
</html>
<?php
if(isset($_POST['btn']))
{
     $sql= mysql_query("INSERT INTO assignwork (imageid,name,size,type,assignedto,assignedby) VALUES ($imageid,$name,$size,$type,$assignedto,$_session[username])");
mysql_error();
}
?>

Actually in query for inserting we need to cal the names of that text or anything.but here how can i call them. so please can anyone help me..
Thank u.

php

4 Contributors
8 Replies
86 Views
7 Hours Discussion Span
Latest Post 13 Years Ago Latest Post by HemantPHP

All 8 Replies

muralibobby2015 17 Posting Pro

13 Years Ago

try this

$sql= mysql_query("INSERT INTO assignwork (imageid,name,size,type,assignedto,assignedby) VALUES ('$imageid','$name','$size','$type','$assignedto','$_session[username]')");

rajarajan2017

13 Years Ago

// create query 
    $query = "INSERT INTO symbols (country, animal) VALUES ('$country', '$animal')"; 
     
    // execute query 
    $result = mysql_query($query) or die ("Error in query: $query. ".mysql_error());
	     
    // print message with ID of inserted record 
    echo "New record inserted with ID ".mysql_insert_id();

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niths -1 Posting Whiz in Training · Answer 1 · 2010-07-26T12:13:19+00:00

try this

$sql= mysql_query("INSERT INTO assignwork (imageid,name,size,type,assignedto,assignedby) VALUES ('$imageid','$name','$size','$type','$assignedto','$_session[username]')");

i tried this i am getting the values inserted as $imageid,$name,$size,$type,$assignedto like this.

niths -1 Posting Whiz in Training · Answer 2 · 2010-07-26T12:16:59+00:00

niths -1 Posting Whiz in Training

13 Years Ago

rajarajan i tried as u said i got the error as query was empty.

muralibobby2015 17 Posting Pro · Answer 3 · 2010-07-26T12:20:55+00:00

muralibobby2015 17 Posting Pro

13 Years Ago

put echo . see now all values are coming or not?

$imageid,$name,$size,$type,$assigned

i thing you didnot extrated $_POST any where

Edited 13 Years Ago by muralibobby2015 because: n/a

niths -1 Posting Whiz in Training · Answer 4 · 2010-07-26T12:25:57+00:00

i think values are not cuming because i am getting an error as 'query was empty'.

INSERT INTO assignwork (imageid,name,size,type,assignedto,assignedby) VALUES ($imageid,$name,$size,$type,$assignedto,$_session[username])

In the above code '$imageid' should be the name and similarly $name,$type,$assignedto should be the names but i didnt mention any names for them.

rajarajan2017 · Answer 5 · 2010-07-26T12:26:38+00:00

you came closer,

$query = "INSERT INTO symbols (country, animal) VALUES ('$country', '$animal')";

form your query as above then put the echo $query after that, now you get the result of the $query in the browser, then copy and paste the result in your sql editor of phpadmin and check whether it is running or not.

Paste your echo $query ouput here?

HemantPHP 0 Light Poster · Answer 6 · 2010-07-26T12:53:05+00:00

hi all,
i had a table with image details.so wen i click on submit button that image details should insert into assignwork table.but that is not working.

<html>

<body>
<link rel="stylesheet" type="text/css" href="css/table.css"/>
<form action="useajax.php" method="post">
<table align="center" id="customers" style="width: 90%">
<tr>
  <th>S.no</th>
  <th>Imageid</th>
  <th>Name</th>
  <th>Size</th>
  <th>Type</th>
  <th>Assigned to</th>
  </tr>
  <?php
    include 'connection.php';
    $q=$_GET["q"];
    $query= mysql_query("SELECT id,imageid,name,size,type from images where uploadeddate='".$q."'");
    mysql_error();
    $num=mysql_num_rows($query);
$i=0;
while ($i < $num){
$f1=mysql_result($query,$i,"id");
$f2=mysql_result($query,$i,"imageid");
$f3=mysql_result($query,$i,"name");
$f4=mysql_result($query,$i,"size");
$f5=mysql_result($query,$i,"type");
if($i%2==0)
{
?>
<tr class="">
    <td><?php echo $f1;?></td>
    <td><?php echo $f2; ?></td>
    <td><?php echo $f3; ?></td>
    <td><?php echo $f4; ?></td>
    <td><?php echo $f5; ?></td>
    <td>
        <select name="assignedto">
            <option><b>SELECT</b></option>
            <option>sss</option>
            <option>jack</option>
            <option>john</option>
            <option>rick</option>
        </select>
    </td>
</tr> 
<?php
}
else
{
?>
<tr class="alt">
    <td><?php echo $f1;?></td>
    <td><?php echo $f2; ?></td>
    <td><?php echo $f3; ?></td>
    <td><?php echo $f4; ?></td>
    <td><?php echo $f5; ?></td>
    <td>
        <select>
            <option><b>SELECT</b></option>
            <option>sss</option>
            <option>jack</option>
            <option>john</option>
            <option>rick</option>
        </select>
    </td>
</tr> 
<?php
}
    $i++;
}
?>
<br></table>
<table align="center">
<input type="submit" name="btn" value="submit">   
</table>
</form>
</body>
</html>
<?php
if(isset($_POST['btn']))
{
     $sql= mysql_query("INSERT INTO assignwork (imageid,name,size,type,assignedto,assignedby) VALUES ($imageid,$name,$size,$type,$assignedto,$_session[username])");
mysql_error();
}
?>

Actually in query for inserting we need to cal the names of that text or anything.but here how can i call them. so please can anyone help me..
Thank u.

Hello,

Just echo your query after insert query execution with using die(), And than provide that query string to us for more understanding your problem :-)

insert not working

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