> 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
3 matches
Mail list logo