Member Avatar for muneer_alam
muneer_alam 0 Newbie Poster
Re: select value from drop down list of one table and insert that value into another tabl 
<?php include("connect.php");?>
<?php include("cityname.php");?>
<?php // select from city
$query= "select city_id from city where(city_name='$city_name')";
//exe query
$result=mysql_query($query) or die();
$row=mysql_fetch_array($rsult);
$city_id=$row['city_id'];
// then insert into location
if($city_id){
$l_id=$_POST['l_id'];
$location=$_POST['location'];
$q = "INSERT INTO location (loc_id,city_id,loc_city)VALUES('','$city_id','$location')";
$re=mysql_query($q) or die (mysql_error());
if($re){
echo "roger that";
}
}
?>

when i am using this code following i find this code

Warning: mysql_query() [function.mysql-query]: Access denied for user 'ODBC'@'localhost' (using password: NO) in C:\xampp\htdocs\newzstreet\addloc1.php on line 29

Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in C:\xampp\htdocs\newzstreet\addloc1.php on line 29


currently i am using Apache server

Web Development php
Member Avatar for muneer_alam
muneer_alam 0 Newbie Poster
Re: select value from drop down list of one table and insert that value into another tabl

Error SummaryHTTP Error 500.0 - Internal Server ErrorThe page cannot be displayed because an internal server error has occurred.Detailed Error InformationModule FastCgiModule
Notification ExecuteRequestHandler
Handler PHP_via_FastCGI
Error Code 0x00000000
Requested URL http://localhost:80/muneer/newzstreet/addloc1.php
Physical Path C:\inetpub\wwwroot\muneer\newzstreet\addloc1.php
Logon Method Anonymous
Logon User Anonymous

HOw can i remove this error?......I am using IIS server not apache.....

Web Development php
Member Avatar for muneer_alam
muneer_alam 0 Newbie Poster
Role based Access control in php

How can i apply Role based Access control in php ..........

Web Development php
Member Avatar for muneer_alam
muneer_alam 0 Newbie Poster
Re: select value from drop down list of one table and insert that value into another tabl

My problem is not resolved yet when i am running this code a problem is displaying

HTTP Error 500.0 - Internal Server ErrorThe page cannot be displayed because an internal server error has occurred.

what is this error????/

Web Development php
Member Avatar for muneer_alam
muneer_alam 0 Newbie Poster
Re: select value from drop down list of one table and insert that value into another tabl

in my city table there are two entry
city_id 1,2
city_name noida ,lucknow
city_status true,true

in my location table this is empty.....because i am inserting this...

Web Development php
Member Avatar for muneer_alam
muneer_alam 0 Newbie Poster
Re: select value from drop down list of one table and insert that value into another tabl

i am using mysql and iis for running application.....all are current version...

(if no, make sure you have data in the database for your query.) what is the meaning of this...........

Web Development php
Member Avatar for muneer_alam
muneer_alam 0 Newbie Poster
Re: Select the radio button and send value into table

this is working very fine....which output i want

Web Development php
Member Avatar for muneer_alam
muneer_alam 0 Newbie Poster
Re: Select the radio button and send value into table

A very very thanks for this code this is very helpfull.....

Web Development php
Member Avatar for muneer_alam
muneer_alam 0 Newbie Poster
Re: Select the radio button and send value into table

Thanks for your reply

is this correct or wrong method........if wrong then try to correct this....

Web Development php
Member Avatar for muneer_alam
muneer_alam 0 Newbie Poster
Re: Select the radio button and send value into table 
<?PHP
if (isset($_POST['Submit']))
{
$selected_radio = $_POST['page'];

if ($selected_radio = = 'admin') 
{
echo "include("adminfetch.php")";
}
else if ($selected_radio = = 'customer') 
{
echo "include("customerfetch.php")";
}
}

?>

this is my code..

Web Development php
Member Avatar for muneer_alam
muneer_alam 0 Newbie Poster
Select the radio button and send value into table

Hello all of you,,,,,,,,

Currently i am facing a problem.....problem is that.i am using two table admin or customer.I have one form.....i want when i select admin button then it will insert the values into admin table and when i will select customer button then it will insert into the customer table....here is my form.....

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<table width="300" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#CCCCCC">
<tr>
<form name="form1" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<td>
<table width="100%" border="0" cellpadding="3" cellspacing="1" bgcolor="#FFFFFF">
<tr>
<td colspan="3"><strong> Login Page </strong></td>
</tr>
<tr>
<td width="78">Username</td>
<td width="6">:</td>
<td width="294"><input name="username" type="text" id="myusername"></td>
</tr>
<tr>
<td>Password</td>
<td>:</td>
<td><input name="password" type="text" id="mypassword"></td>
</tr>
<tr>
<td><INPUT TYPE="radio" NAME="page" VALUE="admin" <?PHP print $admin_status; ?> >Admin</td><td>&nbsp;&nbsp;&nbsp;&nbsp;</td>
<td><INPUT TYPE="radio" NAME="page" VALUE="customer" <?PHP print $customer_status; ?> >Customer</td>

</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td><input type="submit" name="Submit" value="Login"></td>
</tr>
</table>
</td>
</form>
</tr>
</table>
</body>
</html>

i do not want to insert the value of radio button......

plz reply me.....

Web Development php
Member Avatar for muneer_alam
muneer_alam 0 Newbie Poster
Re: select value from drop down list of one table and insert that value into another tabl

Thanks for your reply but i have tried this one also there was no result.is there any problem on my code or query......

plz reply me....

Web Development php
Member Avatar for muneer_alam
muneer_alam 0 Newbie Poster
Re: select value from drop down list of one table and insert that value into another tabl

Query failed: Access denied for user 'ODBC'@'localhost' (using password: NO)


when i am accessing file addloc1.php

plz find as soon as possible....

Web Development php
Member Avatar for muneer_alam
muneer_alam 0 Newbie Poster
Re: select value from drop down list of one table and insert that value into another tabl

Here is my code which i am using here...........
droplistcity.php

<html>
<head>
<script language="JavaScript">
function onChange() {
  var Current=document.form1.select1.selectedIndex;
  document.form1.currentText3.value=document.form1.select1.options[Current].text;
}
</script>
</head>
<body>
<form name="form1" method="POST">
<?php include("connect.php")?>
<table border="1" align="center">
<tr><td>List of City</td>
<td><select name="select1" onChange="onChange()">
    <option value="">Select City</option>
<?php
 $reza=mysql_query("select city_id,city_name from city order by city_id");
                    while ($row=mysql_fetch_array($reza)){
					   $id1=$row["city_id"];
                      $id2=$row["city_name"];
                      
				 echo "<option value=\"$id2\">$id2</option>";
                    }
mysql_close($con);
?>
</select></td>
</tr>
<tr><td>Selected City</td>
<td><input
  name="currentText3"
  type="text"
  value=""
></td>
</tr></form>
</body>
</html>

cityname.php

<?php include("connect.php");?>
<?php 
$city_name = $_POST['currentText3'];
$result=mysql_query("select * from city where city_name='$city_name'");
while($row = mysql_fetch_array($result))
  {
  echo $row['city_name'] ;
  echo "<br />";
  }
mysql_close($con);
?>

addloc.php

<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<form name="form1"action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<table border="1" align="center">
<input type="hidden"  name="l_id" />
<tr>
<?php include("droplistcity.php");?>
</tr>
<tr>
<td>Location Name</td>
<td><input type="text" name="location" /></td></tr>
<tr>
<td></td><td><input type="submit" value="added!">
</td>
</tr>
</table>
</form>
</body>
</html>
<?php include("connect.php");?>
<?php include("cityname.php");?>

<?php
$l_id = $_POST['l_id'];
$city_id = $_POST['city_name'];
$location = $_POST['location'];
$query = "INSERT INTO location (loc_id, city_id,loc_city) VALUES('$l_id',(SELECT city_id FROM city WHERE city_id = '$city_id'),'$location')";
$result = mysql_query($query) or die("Query failed: ".mysql_error());
mysql_close($con);
?>



here is my all code........plz reply as soon as possible.......
Web Development php
Member Avatar for muneer_alam
muneer_alam 0 Newbie Poster
Re: select value from drop down list of one table and insert that value into another tabl

Here is my code which i am using here...........
droplistcity.php

<html>
<head>
<script language="JavaScript">
function onChange() {
  var Current=document.form1.select1.selectedIndex;
  document.form1.currentText3.value=document.form1.select1.options[Current].text;
}
</script>
</head>
<body>
<form name="form1" method="POST">
<?php include("connect.php")?>
<table border="1" align="center">
<tr><td>List of City</td>
<td><select name="select1" onChange="onChange()">
    <option value="">Select City</option>
<?php
 $reza=mysql_query("select city_id,city_name from city order by city_id");
                    while ($row=mysql_fetch_array($reza)){
					   $id1=$row["city_id"];
                      $id2=$row["city_name"];
                      
				 echo "<option value=\"$id2\">$id2</option>";
                    }
mysql_close($con);
?>
</select></td>
</tr>
<tr><td>Selected City</td>
<td><input
  name="currentText3"
  type="text"
  value=""
></td>
</tr></form>
</body>
</html>

cityname.php

<?php include("connect.php");?>
<?php 
$city_name = $_POST['currentText3'];
$result=mysql_query("select * from city where city_name='$city_name'");
while($row = mysql_fetch_array($result))
  {
  echo $row['city_name'] ;
  echo "<br />";
  }
mysql_close($con);
?>

addloc.php

<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<form name="form1"action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<table border="1" align="center">
<input type="hidden"  name="l_id" />
<tr>
<?php include("droplistcity.php");?>
</tr>
<tr>
<td>Location Name</td>
<td><input type="text" name="location" /></td></tr>
<tr>
<td></td><td><input type="submit" value="added!">
</td>
</tr>
</table>
</form>
</body>
</html>
<?php include("connect.php");?>
<?php include("cityname.php");?>

<?php
$l_id = $_POST['l_id'];
$city_id = $_POST['city_name'];
$location = $_POST['location'];
$query = "INSERT INTO location (loc_id, city_id,loc_city) VALUES('$l_id',(SELECT city_id FROM city WHERE city_id = '$city_id'),'$location')";
$result = mysql_query($query) or die("Query failed: ".mysql_error());
mysql_close($con);
?>



here is my all code........plz reply as soon as possible.......
Web Development php
Member Avatar for muneer_alam
muneer_alam 0 Newbie Poster
select value from drop down list of one table and insert that value into another tabl

Hi,,
My problem is that i have two table city and location.In city table i have 3 field city_id(Primary key),city_name,city_status where city_status is true or false.In location table i have 3 field also loc_id(Primary key),city_id,loc_city.i done a code in which city_name comes into drop down list from city table.my problem starts here when i select one of this value and want to insert into location table...then this show error..is there any code available related to my problem plz send my id......

Web Development php