[PHP] Creating pop-up window and passing variable to it?
How can I create a pop up window in my index.php (for example) when user clicks on the link and pass value of the variable from my index.php to this new window?
[PHP] Passing variable to the next page
Here is scenario that I am working on but have a problem. I have 10 thumbnail images on the page with information for them created from database. For example: + + + Some Image + + + Name of Image: Blah Author of Image: Blah (I am pulling more info about that image from database, but on this page I only display "Name of Image" and "Author of image," the rest of the info is stored in variables.) I want my users to click on the thumbnail. When they do I want a new window to open and I want information for that particular image that they clicked (along with other info that stored in variables for the same image) to be passed to that new opened page. How can I pass information about that particular image to the new page? Any ideas? Please Help.
[PHP] Passing variable to new page and pulling the rest of info from database
Hello, I am passing a variable to the new page, when user clicks on the link. Something like that: How can I extract all other information out of my database for that ID in the next page (showimage.php)? Thanks
[PHP] 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 "";
echo" function pop1() {";
echo" window.open(\"info.php?prod_id=$result[0]\"); }"; \\ $result[0]
is variable that stores id of the image in database
echo "";
I am calling this function in the following manner: echo"";
?>
What am I doing wrong? Can you help?
Thank you.
Sorry for lengthy e-mail.
[PHP] FW: NEED HELP (passing variable to new page)
So can anybody help me with my problem?
-Original Message-
From: Igor Portnoy
Sent: Thursday, June 06, 2002 9:36 AM
To: '[EMAIL PROTECTED]'
Subject: NEED HELP (passing variable to new page)
Hello,
I am having hard time passing the variable to the next page. May be you
can help me. Let me explain:
I am using mySQL database to store information about images (ID, name,
author, description, etc). I am pulling some of the information to
create this (look at the example here:
http://www.gibsonusa.com/test/page/index.php) Now, I want a new window
appear when user clicks on the image. I have achieved it with the
following script:
echo "";
echo" function pop1() {";
echo" window.open(\"info.php?prod_id=$result[0]\"); }"; \\ $result[0]
is variable that stores id of the image in database
echo "";
I am calling this function in the following manner: echo"";
?>
What am I doing wrong? Can you help?
Thank you.
Sorry for lengthy e-mail.
RE: [PHP] FW: NEED HELP (passing variable to new page)
It passes the wrong value to the next page.
Click on first or the second image on the first page. In the new window
the ID is set to 4, however the ID of first image is 2 not 4 (look at
the source code).
-Original Message-
From: Ray Hunter [mailto:[EMAIL PROTECTED]]
Sent: Thursday, June 06, 2002 11:32 AM
To: Igor Portnoy
Subject: RE: [PHP] FW: NEED HELP (passing variable to new page)
Your example works what is the problem...
Thank you,
RAY HUNTER
-Original Message-
From: Igor Portnoy [mailto:[EMAIL PROTECTED]]
Sent: Thursday, June 06, 2002 12:24 PM
To: [EMAIL PROTECTED]
Subject: [PHP] FW: NEED HELP (passing variable to new page)
So can anybody help me with my problem?
-Original Message-
From: Igor Portnoy
Sent: Thursday, June 06, 2002 9:36 AM
To: '[EMAIL PROTECTED]'
Subject: NEED HELP (passing variable to new page)
Hello,
I am having hard time passing the variable to the next page. May be you
can help me. Let me explain:
I am using mySQL database to store information about images (ID, name,
author, description, etc). I am pulling some of the information to
create this (look at the example here:
http://www.gibsonusa.com/test/page/index.php) Now, I want a new window
appear when user clicks on the image. I have achieved it with the
following script:
echo "";
echo" function pop1() {";
echo" window.open(\"info.php?prod_id=$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"";
?>
What am I doing wrong? Can you help?
Thank you.
Sorry for lengthy e-mail.
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[PHP] Problem running script after installing PHP 4.2.1
Hello, I am using MySQL to display some data to the users. I have upgraded to PHP Version 4.2.1 on my FreeBSD Apache server, and now data from the database is not displayed correctly. Well, the page with the data shows up, and it actually shows correct number of records. However, when I click on the "next" or "2" buttons the next page with records doesn't show up. See example here: http://www.gibsonusa.com/test2/page/index.php Before I upgraded to PHP 4.2.1 everything was working perfectly. Any ideas what could cause this, and how can it be fixed? Thanks.
