[HELP] If $POST = Variable

[Solar]

What is the most effect approach to doing this?

 

<?php
$catid = $_POST['catid'];

if($catid = "Automotive"){ 
$catid = '3';
}
if($catid = "Business"){ 
$catid = '4';
}
if($catid = "Careers"){ 
$catid = '5';
}
if($catid = "Education"){ 
$catid = '6';
}
if($catid = "Financial"){ 
$catid = '7';
}
if($catid = "Government"){ 
$catid = '8';
}
if($catid = "Health"){ 
$catid = '9';
}
if($catid = "Mobile"){ 
$catid = '10';
}
if($catid = "Organization"){ 
$catid = '11';
}
if($catid = "Programming"){ 
$catid = '12';
}
if($catid = "Software"){ 
$catid = '13';
}
if($catid = "Travel"){ 
$catid = '14';
}
if($catid = "Web"){ 
$catid = '15';
}
if($catid = "Other"){ 
$catid = '16';
}
if($catid = "Gaming"){ 
$catid = '17';
} ?>

 

I tried this and it works;

<?php
$catid = "Automotive";
if($catid = "Automotive"){ 
$catid = '3';
}
echo $catid;
?>

 

But... When I add another if statement...

<?php
$catid = "Automotive";
if($catid = "Automotive"){ 
$catid = '3';
}
if($catid = "Gaming"){ 
$catid = '17';
}
echo $catid;
?>

 

It echos $catid as 17 and not 3 which is defined as a variable.

 

I hope there is a simple solution for this!

Thanks in advanced!

[smerny]

1) you should be using "elseif" for situations like this (when there can only be one answer)

 

2) the error is because you are using a single = sign, rather than a double ==, the double == is used for comparison, the single is used for setting...

 

your code explained:

<?php
$catid = "Automotive"; // set catid to "Automotive"
if($catid = "Automotive"){ // set catid to "Automotive", if set is successful...
$catid = '3'; // ...set catid to "3"
}
if($catid = "Gaming"){ //set catid to "Gaming", if set is successful...
$catid = '17'; // ...set catid to "17"
}
echo $catid;
?>

[.josh]

First off, in your conditions, = is the assignment operator, whereas == is the equality operator...you should be using == in your conditions.

 

But anyways, you can cut all those conditions out by using an array:

 

$catIDs = array(
  3 => "Automotive",
  4 => "Business",
  // etc...
);

$catid = array_search(trim($_POST['catid']), $catIDs);
echo $catid;

 

array_search will search the array and return the key (like 3,4, etc..) or if value not found, it will return boolean false.

 

[Solar]

Thanks for the fast replies! I figured it out using the manual way. No matter how many times I try doing arrays... They never work for me. My form selection is using Automotive, Business, Careers, etc... So I want them to transfer to numbers so when inserted with mysql, they display correctly in the blog id categories.

 

<?php if($_POST['catid']=="Automotive"){ 
$catid = '3';
}
if($_POST['catid']=="Business"){ 
$catid = '4';
}
if($_POST['catid']=="Careers"){ 
$catid = '5';
}
if($_POST['catid']=="Education"){ 
$catid = '6';
}
if($_POST['catid']=="Financial"){ 
$catid = '7';
}
if($_POST['catid']=="Government"){ 
$catid = '8';
}
if($_POST['catid']=="Health"){ 
$catid = '9';
}
if($_POST['catid']=="Mobile"){ 
$catid = '10';
}
if($_POST['catid']=="Organization"){ 
$catid = '11';
}
if($_POST['catid']=="Programming"){ 
$catid = '12';
}
if($_POST['catid']=="Software"){ 
$catid = '13';
}
if($_POST['catid']=="Travel"){ 
$catid = '14';
}
if($_POST['catid']=="Web"){ 
$catid = '15';
}
if($_POST['catid']=="Other"){ 
$catid = '16';
}
if($_POST['catid']=="Gaming"){ 
$catid = '17';
} ?>

 

If I were to do the array.... Would I want that backwards?

"Automotive" => 3,

"Business" => 4,

etc...

Or the way you have it is fine? I get an error.

Notice: Undefined index: catid

 

Warning: array_search() expects at least 2 parameters, 1 given

[.josh]

array_search() searches the array values and returns the key an array is array(key => value).  As far as the Notice...maybe you snagged the code before I edited it..I forgot to add the 2nd param when I first posted it.

[Solar]

I snagged before you had edit your post.

 

$catIDs = array(  
3 => "Automotive",
4 => "Business",
5 => "Careers",
6 => "Education",
7 => "Financial",
8 => "Government",
9 => "Health",
10 => "Mobile",
11 => "Organization",
12 => "Programming",
13 => "Software",
14 => "Travel",
15 => "Web",
16 => "Other",
17 => "Gaming"
);
$catid = array_search(trim($_POST['catid']), $catIDs);

 

This works WAY better. Cleaner, and nicer. This will open up more solutions around my website! Thankyou very much!