[RESOLVED] passing variables through the URL

Sooz719

I have a page which updates a mysql database from the input of a form, if the update is successfull i have told it to redirect to another page using

     [code=php]print("<SCRIPT LANGUAGE=\"JavaScript1.1\">\n");
 print("location = \"jobedited.php?ReferenceCode= echo $id \";\n");
 print("</SCRIPT>");[/code]

this should then redirect me to the page with the following code

<?php include("header.inc");

//open db connection
$link=mysqli_connect('localhost','root','voBtGM','JobBase');

$id= $_GET['ReferenceCode'];

//build and execute query
$select="SELECT JobTitle, Salary, Location, Description FROM Jobs Where ReferenceCode=$id";
$result=mysqli_query($link,$select);

$row=mysqli_fetch_assoc($result) or die("query failed".mysql_error());

$JobTitle=$row['JobTitle'];
$Salary=$row['Salary'];
$Location=$row['Location'];
$Description=$row['Description'];

//close db connection
mysqli_close($link);?>

<br>
<TABLE border="1">
  <TR><TH>Job Title</TH>
      <TH>Salary</TH>
      <TH>Location</TH>
      <TH>Description</TH>
  </TR>

<TR><TD><? echo $JobTitle?></TD>
    <TD><? echo $Salary?></TD>
    <TD><? echo $Location?></TD>
    <TD><? echo $Description?></TD>
</TR>
</TABLE>



</body>
</HTML>

i know this code is correct as if i type in a reference code in the URL it loads the correct data. However, when i try and get it to use the reference code passed from the previous page i get the following error:

Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, boolean given

So i am guessing i am not passing the variable through the URL properly. Can anyone help me to work out why, if i have not provided the relevent code can someone advise me where to look.

Thanks for any help

bradgrafelman

First of all, your PHP code won't work. You can't just put "echo $id" inside of a print() statement and expect PHP to parse that as a command. If you're using double quotes ( " ), you can simply just put the variable inside the string and PHP will parse the string and put in the variable. Alternatively, you can escape out of the string and add the variable using periods. Examples:

print "<SCRIPT LANGUAGE=\"JavaScript1.1\">
location = \"jobedited.php?ReferenceCode=$id\"; 
</SCRIPT>";

// or...

print '<SCRIPT LANGUAGE="JavaScript1.1"> 
location = "jobedited.php?ReferenceCode=' . $id . '"; 
</SCRIPT>';

Also, when you get MySQL errors, use some error debugging:

$result=mysqli_query($link,$select) or die('Error, MySQL said: ' . mysqli_error($link));