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PHP variable in Select

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Nov 04.2008

Simple problem here I just can’t get it…

I have 2 session variables

$cat1 = $_SESSION[‘catCheck1’];
$cat2 = $_SESSION[‘catCheck2’];

In my SELECT, I have a for loop. All I want to do is get the variable names $cat1 and $cat2 passed to it like this:

[code]
for ( $i = 1; $i <= 2; $i++) {
$query = mysql_query(“SELECT * FROM category WHERE categoryID = $cat”.$i.” ORDER BY categoryName”);
[/code]

However, my select doesn’t like it and it select nothing. But if I put in the actual variable name like this, it works just great!

[code]
for ( $i = 1; $i <= 2; $i++) {

$query = mysql_query(“SELECT * FROM category WHERE categoryID =”. $cat1.” ORDER BY categoryName”);
while ($getName = mysql_fetch_array($query)) {
etc….
[/code]

What in the devil am I doing wrong with this select to get it to read the variable name right?

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PHP

2 Comments(s) ↴

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@bejitto101Nov 04.2008 — I don't believe you can call a variable by adding two variables together. Here's what your select is processing. It first looks at the $cat. Since this variable doesn't exist, you get nada. The $i returns either one or two. So your query is "SELECT * FROM category WHERE categoryID = 1 ORDER BY categoryName".

Here's a way to accomplish what I think you want:

[code=php]
$cat[] = $_SESSION['catCheck1'];
$cat[] = $_SESSION['catCheck2'];

for ($i = 0; $i <= 1; $i++) {
$query = mysql_query("SELECT * FROM category WHERE categoryID = ".$cat[$i]." ORDER BY categoryName");[/code]

Or keep the $cat the same and use:

[code=php]
foreach ($cat as $sub_cat) {
$query = mysql_query("SELECT * FROM category WHERE categoryID = $sub_cat ORDER BY categoryName");[/code]
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@Patty5authorNov 05.2008 — Putting $cat in an array and using the for loop worked like a charm! You are brilliant! Problem solved!

Thanks so much!
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