php explode returns wrong value

Question

I'm submitting a form and inserting the valuse into my mysql db. One of the form elements it a select with option values like 11#120#12 (id#cost#months).

I perform the following:

$plan = explode("#", $_POST['symb']);
$plan = $plan[0];
$cost = $plan[1];

and then submit to my db

echo $insertSQL = sprintf("INSERT INTO users (username, plan, cost) VALUES (%s, %s, %s)",
                       GetSQLValueString($_POST['username'], "text"),
                       GetSQLValueString($plan, "text"),
                       GetSQLValueString($cost, "int"));

The problem is that $cost is getting the wrong value, instead of 120 it's getting 1 as a value. Where's the error? (the posted sql statement is only part of the actual query, for demonstration purposes only)

You are overwriting $plan in line 2

Pekka
– Pekka

2012-05-29 09:31:31 +00:00
Commented May 29, 2012 at 9:31 — Pekka
– Pekka, Commented May 29, 2012 at 9:31

Quentin · Accepted Answer · 2012-05-29 09:31:58Z

4

Don't overwrite $plan with a string value before you try to extract the $cost from the array you originally stored in it.

Either change the first $plan in $plan = $plan[0]; (and change every reference to it later in the script) or move that line so it appears after $cost = $plan[1];.

(I'd recommend the former option for clarity).`

answered May 29, 2012 at 9:31

Quentin

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1 Comment

bikey77 Over a year ago

Wow! I guess I should take a break for not seeing the absolutely obvious... thanks for pointing it out for me.

Micku · Accepted Answer · 2012-05-29 09:36:36Z

0

$plan = explode('#',$_POST['symb']);
$p = $plan[0];
$cost = $plan[1];

answered May 29, 2012 at 9:36

Micku

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Comments

matchdav · Accepted Answer · 2012-05-29 09:45:07Z

0

On line 2, $plan is assigned "11" (which was $plan[0]), and on line 3, $plan[1] references the second character in the string which is '1'.

The comments above already have already told you how to fix it.

answered May 29, 2012 at 9:45

matchdav

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