PHP Session Array Value keeps showing as "Array"

Question

When sending data from a form to a second page, the value of the session is always with the name "Array" insteed of the expected number.

The data should get displayed in a table, but insteed of example 1, 2, 3 , 4 i get : Array, Array, Array. (A 2-Dimensional Table is used)

Is the following code below a proper way to "call" upon the stored values on the 2nd page from the array ?

$test1 = $_SESSION["table"][0];
$test2 = $_SESSION["table"][1];
$test3 = $_SESSION["table"][2];
$test4 = $_SESSION["table"][3];
$test5 = $_SESSION["table"][4];

What exactly is this, and how can i fix this? Is it some sort of override that needs to happen?

Best Regards.

You should print_r those to see what's there.

nc3b
– nc3b

2010-05-18 17:52:07 +00:00
Commented May 18, 2010 at 17:52 — nc3b
– nc3b, Commented May 18, 2010 at 17:52

bimbom22 · Accepted Answer · 2010-05-18 18:50:57Z

7

You don't need any sort of override. The script is printing "Array" rather than a value, because you're trying to print to the screen a whole array, rather than a value within an array for example:

$some_array = array('0','1','2','3');

echo $some_array; //this will print out "Array"

echo $some_array[0]; //this will print "0"

print_r($some_array); //this will list all values within the array. Try it out!

print_r() is not useful for production code, because its ugly; however, for testing purposes it can keep you from pulling your hair out over nested arrays.

It's perfectly fine to access elements in your array by index: $some_array[2]

if you want it in a table you might do something like this:

<table>
    <tr>
    for($i = 0 ; $i < count($some_array) ; $i++) {
        echo '<td>'.$some_array[$i].'</td>';
    }
    </tr>
</table>

edited May 18, 2010 at 18:50

answered May 18, 2010 at 17:56

bimbom22

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1 Comment

Artefacto Over a year ago

I'll have to nitpick. PHP has nested arrays, not really multidimensional arrays.

user258082 · Accepted Answer · 2010-05-18 17:54:15Z

6

As noted, try

echo "<pre>";
print_r($_SESSION);
echo "</pre>";

That should show you what's in the session array.

answered May 18, 2010 at 17:54

user258082

2 Comments

user258082 Over a year ago

What does the output from the print_r show? That may help us with finding the issue.

Jonah Over a year ago

print_r is not to put into your page (as stated by jordanstephens in his answer). It's for debugging, so we can see the structure of the array.

Powerlord · Accepted Answer · 2010-05-18 18:00:00Z

0

A 2-dimensional table is just an array of arrays.

So, by pulling out $_SESSION["table"][0], you're pulling out an array that represents the first row of the table.

If you want a specific value from that table, you need to pass the second index, too. i.e. $_SESSION["table"][0][0]

Or you could just be lazy and do $table = $_SESSION["table"]; at which point $table would be your normal table again.

answered May 18, 2010 at 18:00

Powerlord

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Comments

ByteHamster · Accepted Answer · 2015-04-22 20:46:02Z

0

A nice way ...

<?php
    foreach ($_SESSION as $key => $value) {
        echo $key . " => " . $value . "<br>";
    }
?>

edited Apr 22, 2015 at 20:46

ByteHamster

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answered Apr 22, 2015 at 20:13

Donatello

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