> echo "";
> echo" function pop1() {";
> echo" window.open(\"info.php?prod_id=$result[0]\"); }";
> echo "";
> I am calling this function in the following manner: echo" onclick=\"pop1();\"";
OK, I tend not to put javascript inside php echo statements as it
confuses things - which variables
e [mailto:[EMAIL PROTECTED]]
> Sent: Thursday, June 06, 2002 2:50 PM
> To: [EMAIL PROTECTED]
> Subject: Re: [PHP] [PLEASE HELP] Passing variable to new page.
>
>
> > echo "";
> > echo" function pop1() {";
> > echo" window.open(\"info.php?pr
> echo "";
> echo" function pop1() {";
> echo" window.open(\"info.php?prod_id=$result[0]\"); }"; \\ $result[0] is
> variable that stores id of the image in database
> echo "";
>
> I am calling this function in the following manner: echo" onclick=\"pop1();\"";
^
You're not pass
ldn't have any problems.
http://www.gibsonusa.com/test/page/info.php?prod_id=35";
target="new">
-Kevin
- Original Message -
From: "Igor P." <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, June 06, 2002 12:37 PM
Subject: [PHP] [PLEASE
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:
http://www.g
5 matches
Mail list logo