Undefined Index for Valid key from array in PHP

Question

While parsing a CSV file(I have used a Class found on Google), I ran into a problem. Here is an example of an array made from a .csv file(print_r):

  Array
    (
        [0] => Array
            (
                [Site] => ViralNova
                [Impressions] => 104719
                [CTR] => 0.30%
                [Clicks] => 311
                [Average CPC] => $ 0.400
                [CPM] => $ 1.19
                [Conversion Rate] => 1.29%
                [Actions] => 4
                [CPA] => $ 31.100
                [Spent] => $ 124.40
            )

        [1] => Array
            (
                [Site] => TMZ - Desktop
                [Impressions] => 103276
                [CTR] => 0.29%
                [Clicks] => 295
                [Average CPC] => $ 0.400
                [CPM] => $ 1.14
                [Conversion Rate] => 0.68%
                [Actions] => 2
                [CPA] => $ 59.000
                [Spent] => $ 118.00
            )
)

The Problem is that I cannot use the "Site" index. Whenever I try to, I get this notice:

<b>Notice</b>:  Undefined index: Site in <b>C:\xampp\htdocs\panel\update\assets\php\core\core-ajax.php</b> on line <b>104</b><br />

Line 104 being:

print $data[0]['Site'];

Also,

print $data[0]['Impressions'];

or

print_r($data[0]);

works without any problem and I get the correct value/array.

Any ideas?

Thanks!!

Do: var_dump($data) and look at the output in the source code! <- Show that — Rizier123
– Rizier123, Commented Apr 29, 2015 at 14:22
You have an extra space and therefore your index is ' Site' not 'Site'. Do echo $data[0][' Site'] and see if it works. — kidA
– kidA, Commented Apr 29, 2015 at 14:25
It should work if you use [' Site'] instead of ['Site']. Tested it on my local and worked fine. — kidA
– kidA, Commented Apr 29, 2015 at 14:36

user4844986 · Accepted Answer · 2015-04-29 17:25:26Z

1

Acc. to me your code is running fine. this problem is occur when in loop in a point your index is missing.

to resolve this issue..

use this..

<?php 
if($data[0]['Site']){
    print $data[0]['Site'];
}

// Also

if($data[0]['Impressions']){
   print print $data[0]['Impressions'];
}

?>

answered Apr 29, 2015 at 17:25

user4844986

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1 Comment

smottt Over a year ago

Surely you meant if (isset($data[0]['Site']))?

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