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Old June 6th, 2002, 12:30 PM
sharapov sharapov is offline
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Question [HELP PLEASE] Passing variable to new page

Hello,
I am having hard time passing the variable to the next page. May be you can help me. Let me explain:

I am using mySQL database to store information about images (ID, name, author, description, etc). I am pulling some of the information to create this (look at the example here: URL) Now, I want a new window appear when user clicks on the image. I have achieved it with the following script:

echo "<script language=\"JavaScript\">";
echo" function pop1() {";
echo" window.open(\"info.php?prod_id=$result[0]\"); }"; \\ $result[0] is variable that stores id of the image in database
echo "</script>";

I am calling this function in the following manner: echo"<img onclick=\"pop1();\"";

You can see the result if you click on the image. The new page opens up, BUT the id (product_id in this case) value is not passed to the next page correctly. If you click on the first or second image on the first page it shows that the ID is the same for both of them. However if you look at the source code you can see that the ids are assigned correctly. The same thing happens if you click on any image that says “no image available” (I am using different script to generate those). It seems that the script picks up and stores in the memory the value of an ID of the last image generated with the script (I don’t know if that make sense).

Oh, by the way the contents of info.php that I am calling in the script above are as following:

<?
$myid = $_GET['prod_id'];
echo" Product ID: $myid<br>";
?>

What am I doing wrong? Can you help?
Thank you.

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  #2  
Old June 6th, 2002, 06:02 PM
mytch mytch is offline
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Hey sharapov,
Looks like you've missed some curly braces, that's all. Just change the following line:

echo" window.open(\"info.php?prod_id=$result[0]\"); }"; \\ $result[0] is variable

to this:

echo" window.open(\"info.php?prod_id={$result[0]}\"); }"; \\ $result[0] is variable

and it should work perfectly.

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