display array only if exists

Question

I made a page where users can see data entered by them in the database.
I used:

$select = "SELECT * FROM texts WHERE user='".$user."' ORDER BY date DESC, id DESC";
$result = mysql_query($select);
$array = array();
while($show = mysql_fetch_assoc($result))
{
$array[] = $show;
}
echo "<strong>".$array[0]['id']."</strong><br />";
echo "<strong>".$array[1]['id']."</strong><br />";
echo "<strong>".$array[2]['id']."</strong><br />";
echo "<strong>".$array[3]['id']."</strong><br />";
echo "<strong>".$array[4]['id']."</strong><br />";

Code works, but sometimes I have less than 10 values to return, sometimes even more.
If I use this and I have only 2 arrays to return, I get:

Notice: Undefined offset: 2 in ownposts.php on line 15
Notice: Undefined offset: 3 in ownposts.php on line 16
Notice: Undefined offset: 4 in ownposts.php on line 17

How it's possible to echo $arrray[4]['id] only if exist $array[4]?
I've tried with:

$zero = $array[0];
if(!empty($zero))
{
echo "<strong>".$zero['id']."</strong><br />";
}
$four = $array[4];
if(!empty($four))
{
echo "<strong>".$five['id']."</strong><br />";
}

But doesn't work as I excepted and still return Notice: Undefined offsed: 4 in ownposts.php on line 17.

Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial. — h2ooooooo
– h2ooooooo, Commented Sep 4, 2013 at 11:08
Try not to use 'mysql_' functions anymore, it's deprecated and should not be used in new applications. Instead use Mysqli or PDO — Edd
– Edd, Commented Sep 4, 2013 at 11:09
Deprecated bla bla bla... There's absolutly no point wasting cpu time in preparing a statement that will only be ran once, and deleted on request end, as long as you properly escape input using mysql_real_escape_string. — Virus721
– Virus721, Commented Sep 4, 2013 at 11:15

h2ooooooo · Accepted Answer · 2013-09-04 11:12:12Z

3

Instead of what you currently do, this:

while($show = mysql_fetch_assoc($result))
{
$array[] = $show;
}
echo "<strong>".$array[0]['id']."</strong><br />";
echo "<strong>".$array[1]['id']."</strong><br />";
echo "<strong>".$array[2]['id']."</strong><br />";
echo "<strong>".$array[3]['id']."</strong><br />";
echo "<strong>".$array[4]['id']."</strong><br />";

Why not just show whatever is found by mysql as you find it:

while($show = mysql_fetch_assoc($result))
  {
    echo "<strong>".$show['id']."</strong><br />";
  }

Or if you have other things to do with the array than the code you showed us, use a loop on the newly created array, ie a foreach.

edited Sep 4, 2013 at 11:12

h2ooooooo

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answered Sep 4, 2013 at 11:09

James

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Comments

Liam Allan · Accepted Answer · 2013-09-04 11:08:46Z

1

try

foreach($array as $data){
echo '<strong>'.$data['id'].'</strong><br />';
}

answered Sep 4, 2013 at 11:08

Liam Allan

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Comments

anon · Accepted Answer · 2013-09-04 11:12:21Z

0

I'd rewrite it as follows:

$i = 0;
while($show = mysql_fetch_assoc($result))
{
    $array[] = $show;
    if (isset($array[$i])) {
        echo "<strong>".$array[$i]['id']."</strong><br />";
    }
    $i++;
}

answered Sep 4, 2013 at 11:12

anon

Comments

Hardik · Accepted Answer · 2013-09-04 11:12:38Z

0

there is function in php for that to check, array_key_exists. You can check that at, http://php.net/manual/en/function.array-key-exists.php

answered Sep 4, 2013 at 11:12

Hardik

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