very simple question

[inusrat]

Hi,

I am new to PHP. When a form is sumbited , how do we retrive variable in PHP on the other page..Like in ASP we do the following

birthcountry= Request.form("birthcountry")

***********

When you are coming through a link, how do we retrive the variables of the link ..like ASP we do ...


Tab = Request("tab")

Thanks

http://www.786Marriage.com

http://www.amasspharma.com

http://www.ambdex.com

 

[jimoblak]

$yourvariable=$_GET[userentereddata];

or

$yourvariable=$_POST[userentereddata];

You should search for ASP2PHP, a freeware that can convert ASP code to PHP. While it is not perfect, it is a good way to see how your ASP knowledge translates to PHP.


- - picklefish - -
Why is everyone in this forum responding to me as picklefish?

 

[inusrat]

Can you plz tell me what I am doing wrong here, I am trying to do construct a SQL statement with where clause.


________________
$yourvariable= "Women";

$result= mysql_query("SELECT * FROM products where gender= $yourvariable",$link)
or die("Couldn't execute query");
____________________

Thanks

http://www.786Marriage.com

http://www.amasspharma.com

http://www.ambdex.com

 

[danomac]

If gender is a text field you need the contents surrounded in quotes in the query statement.

$yourvariable= "Women";

$result= mysql_query("SELECT * FROM products where gender= [COLOR=red]'[/color]$yourvariable"[COLOR=red]'[/color],$link)
  or die("Couldn't execute query");

The above should work.

 

[ingresman]

Put the $yourvariable in to single quotes i.e.
SELECT * FROM products where gender= '$yourvariable'

 

[danomac]

Oops, ingresman, we posted at the same time, and mine with an error...

$result= mysql_query("SELECT * FROM products where gender= [COLOR=red]'[/color]$yourvariable[COLOR=red]'[/color]",$link)
  or die("Couldn't execute query");

Mine should be corrected now...

 

[inusrat]

Thanks a lot for everybody's reply, it works now..

http://www.786Marriage.com

http://www.amasspharma.com

http://www.ambdex.com