... especially when the error is not even with PHP :-) Contrary to what some newbies seem to think, this list is not a free consulting service... none of us are obligated to help everyone with every problem.
We're all volunteers -- sometimes we just don't have the time or knowledge to answer every question, especially when the questions aren't related to PHP, or are really vauge or common or just plain silly :-) --- Scott Hurring Systems Programmer EAC Corporation [EMAIL PROTECTED] Voice: 201-462-2149 Fax: 201-288-1515 > -----Original Message----- > From: 1LT John W. Holmes [mailto:[EMAIL PROTECTED]] > Sent: Thursday, June 06, 2002 2:40 PM > To: Igor Portnoy > Cc: [EMAIL PROTECTED] > Subject: Re: [PHP] FW: NEED HELP (passing variable to new page) > > > Asking twice doesn't help, bud. It's just going to piss people off. > > If someone has an answer they'll let you know, otherwise > look for other > resources. > > ---John Holmes... > > ----- Original Message ----- > From: "Igor Portnoy" <[EMAIL PROTECTED]> > To: "[EMAIL PROTECTED]" <'[EMAIL PROTECTED]'> > Sent: Thursday, June 06, 2002 2:24 PM > Subject: [PHP] FW: NEED HELP (passing variable to new page) > > > So can anybody help me with my problem? > > > > -----Original Message----- > From: Igor Portnoy > Sent: Thursday, June 06, 2002 9:36 AM > To: '[EMAIL PROTECTED]' > Subject: NEED HELP (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: > http://www.gibsonusa.com/test/page/index.php) 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. > > Sorry for lengthy e-mail. > > > > > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
