store a php array in session from input form

Question

I am having allot of trouble trying to get an array to work with sessions if anyone can help that would be great I'm not bothered about validation etc if I can just get it working i can then expand upon it.

HTML

<form method="post" action="array2.php">
    Select amount of tickets you require.
    <select name="options[]">
        <option value="1">1</option>
        <option value="2">2</option>
        <option value="3">3</option>
        <option value="4">4</option>
        <option value="5">5</option>
        <option value="6">6</option>
        <option value="7">7</option>
        <option value="8">8</option>
        <option value="9">9</option>
        <option value="10">10</option>
    </select>
    Select the acomidation you require.
    <select name="options2[]">
        <option value="camping">Camping</option>
        <option value="caravan">Caravan</option>
    </select>
    <input type="submit" value="Go!" />
</form>

array2.php

<?php
session_start();
$checked = $_POST['options'];
$checked2 = $_POST['options2'];
$_SESSION['user'] = true;
$_SESSION['checked'] = $checked;
$_SESSION['checked2'] = $checked2;
header('Location: array3.php');
?>

array3.php

<?php
session_start();
if(!isset($_SESSION['user'])){
    die("To access this page, you need to <a href='register.html'>LOGIN</a>");
}
$checked = $_SESSION['checked'];
$checked2 = $_SESSION['checked2'];
?>

<?php
    for($i=0; $i < count($checked && $checked2); $i++){
        echo "You have selected to receive " . $checked[$i] . " tickets<br/>";
        echo "And you have selected to receive " . $checked2[$i] . " for accommodation are you sure? <br/>";
    }
?>

The main problem is that the values are not being passed from array2 into array3, any help is welcomed.

EDIT - this worked fine until I tried to add in the sessions to get it working over multiple pages so I'm sure thats where the problem is

EDIT2 - thanks for all the help guys i got it working when I took out

$checked = $_POST['options'];
$checked2 = $_POST['options2'];

From array 3 it worked :) much appreciated!

Are you sure you getting value in array2 & not in array3 ? , if you can cross check for me. — Rikesh
– Rikesh, Commented Feb 26, 2013 at 10:43
use the power of print_r($_SESSION); after session_start(); in array3.php — DevMetal91
– DevMetal91, Commented Feb 26, 2013 at 10:44
I took out the redirect header on array2, and added in the for statement into array 2 and it works fine if that is what you mean — jphillip724
– jphillip724, Commented Feb 26, 2013 at 10:44
using print session i get Array ( [user] => 1 [checked] => Array ( [0] => 1 ) [checked2] => Array ( [0] => camping ) ) You have selected to recive tickets And you have selected to recive for accommodation are you sure? — jphillip724
– jphillip724, Commented Feb 26, 2013 at 10:45

DaveRandom · Accepted Answer · 2013-02-26 10:47:16Z

4

count($checked && $checked2)

...is your problem.

$checked && $checked2 is a logical expression will either be true or false, which means thats the count() call will always return false, and false will equate to 0 in a numeric comparison (the less than < comparison with $i), so the for loop won't perform any iterations.

However, the use of arrays in this situation is not appropriate anyway, because you have used a <select> element without a multiple property so it will only represent a single value, which means the array would only ever hold a single value. You should simplify this to just scalar values and you will find that a) it works and b) it's much easier in general.

answered Feb 26, 2013 at 10:47

DaveRandom

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2 Comments

jphillip724 Over a year ago

okay i will work on that, i got it working by removing the $checked = $_POST['options']; $checked2 = $_POST['options2']; and it works but i will look into what you are saying.

jphillip724 Over a year ago

yeh i will look into that, I've only been doing php for the last 6 weeks, really just trying to get the basics done the idea behind using both in the same statement was that if you have tickets selected then you must have accommodation hopefully i turn it into a better statement but really fow what i currently know i cant do much better than that for now.

Husman · Accepted Answer · 2013-02-26 10:47:50Z

2

Your for loop is written wrong.

for($i=0; $i < count($checked && $checked2); $i++)

That count statement will not work as you think. Break it into 2 for loops or an inner loop.

like this:

for($i=0; $i < count($checked); $i++){
  echo stuff here
}

for($i=0; $i < count($checked2); $i++){
  echo stuff here
}

answered Feb 26, 2013 at 10:47

Husman

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