PHP Notice: Undefined offset: 1 with array when reading data

Question

I am getting this PHP error:

PHP Notice:  Undefined offset: 1

Here is the PHP code that throws it:

$file_handle = fopen($path."/Summary/data.txt","r"); //open text file
$data = array(); // create new array map

while (!feof($file_handle) ) {
    $line_of_text = fgets($file_handle); // read in each line
    $parts = array_map('trim', explode(':', $line_of_text, 2)); 
    // separates line_of_text by ':' trim strings for extra space
    $data[$parts[0]] = $parts[1]; 
    // map the resulting parts into array 
    //$results('NAME_BEFORE_:') = VALUE_AFTER_:
}

What does this error mean? What causes this error?

How sure are you that every line in your file has a colon in it? — andrewsi, Jul 03 '13 at 19:16
do a `var_dump($parts)`. you'll probably find that the point you get that undefined offset, there is no `1` key in the parts array. — Marc B, Jul 03 '13 at 19:16
If it looks like all the lines have a colon, check for blank lines. — Barmar, Jul 03 '13 at 19:18
you just need to make sure that count($parts) == 2 prior to doing your assignment. — Orangepill, Jul 03 '13 at 19:18

score 146 · Accepted Answer · edited May 10 '17 at 03:03

Change

$data[$parts[0]] = $parts[1];

to

if ( ! isset($parts[1])) {
   $parts[1] = null;
}

$data[$parts[0]] = $parts[1];

or simply:

$data[$parts[0]] = isset($parts[1]) ? $parts[1] : null;

Not every line of your file has a colon in it and therefore explode on it returns an array of size 1.

According to php.net possible return values from explode:

Returns an array of strings created by splitting the string parameter on boundaries formed by the delimiter.

If delimiter is an empty string (""), explode() will return FALSE. If delimiter contains a value that is not contained in string and a negative limit is used, then an empty array will be returned, otherwise an array containing string will be returned.

score 21 · Answer 2 · edited May 23 '17 at 10:31

How to reproduce the above error in PHP:

php> $yarr = array(3 => 'c', 4 => 'd');

php> echo $yarr[4];
d

php> echo $yarr[1];
PHP Notice:  Undefined offset: 1 in 
/usr/local/lib/python2.7/dist-packages/phpsh/phpsh.php(578) : 
eval()'d code on line 1

What does that error message mean?

It means the php compiler looked for the key 1 and ran the hash against it and didn't find any value associated with it then said Undefined offset: 1

How do I make that error go away?

Ask the array if the key exists before returning its value like this:

php> echo array_key_exists(1, $yarr);

php> echo array_key_exists(4, $yarr);
1

If the array does not contain your key, don't ask for its value. Although this solution makes double-work for your program to "check if it's there" and then "go get it".

Alternative solution that's faster:

If getting a missing key is an exceptional circumstance caused by an error, it's faster to just get the value (as in echo $yarr[1];), and catch that offset error and handle it like this: https://stackoverflow.com/a/5373824/445131

score 4 · Answer 3 · answered Aug 23 '20 at 02:46

4

Update in 2020 in Php7:

there is a better way to do this using the Null coalescing operator by just doing the following:

$data[$parts[0]] = $parts[1] ?? null;

answered Aug 23 '20 at 02:46

Alladin

1,010
3
27
45

score 1 · Answer 4 · edited Feb 15 '14 at 03:54

1

This is a "PHP Notice", so you could in theory ignore it. Change php.ini:

error_reporting = E_ALL & ~E_NOTICE & ~E_STRICT & ~E_DEPRECATED

To

error_reporting = E_ALL & ~E_NOTICE

This show all errors, except for notices.

edited Feb 15 '14 at 03:54

Eric Leschinski

146,994
96
417
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answered Jul 16 '13 at 20:28

Orijmm

61
3

score 1 · Answer 5 · answered Dec 16 '18 at 10:03

1

my quickest solution was to minus 1 to the length of the array as

  $len = count($data);

    for($i=1; $i<=$len-1;$i++){
      echo $data[$i];
    }

my offset was always the last value if the count was 140 then it will say offset 140 but after using the minus 1 everything was fine

answered Dec 16 '18 at 10:03

Daniel Charles Mwangila

763
10
12

Actually you want to change `<=` to `<` and don't minus 1 your `$len` variable. – A1rPun Nov 28 '19 at 11:04
Thanks for heads i realised it recently when working on another project – Daniel Charles Mwangila Nov 28 '19 at 13:49

score 1 · Answer 6 · edited Jul 15 '19 at 02:00

1

The ideal solution would be as below. You won't miss the values from 0 to n.

$len=count($data);
for($i=0;$i<$len;$i++)
     echo $data[$i]. "<br>";

edited Jul 15 '19 at 02:00

Pang

9,564
146
81
122

answered Apr 05 '19 at 09:01

Sashi Giri

11
1

score 1 · Answer 7 · answered Aug 02 '20 at 16:53

In your code: $parts = array_map('trim', explode(':', $line_of_text, 2)); You have ":" as separator. If you use another separator in file, then you will get an "Undefined offset: 1" but not "Undefined offset: 0" All information will be in $parts[0] but no information in $parts[1] or [2] etc. Try to echo $part[0]; echo $part[1]; you will see the information.

score -1 · Answer 8 · answered Jul 30 '17 at 18:42

-1

I just recently had this issue and I didn't even believe it was my mistype:

Array("Semester has been set as active!", true)
Array("Failed to set semester as active!". false)

And actually it was! I just accidentally typed "." rather than ","...

answered Jul 30 '17 at 18:42

Erikas

1,006
11
22

score -1 · Answer 9 · edited Dec 22 '19 at 00:00

-1

The output of the error, is because you call an index of the Array that does not exist, for example

$arr = Array(1,2,3);
echo $arr[3]; 
// Error PHP Notice:  Undefined offset: 1 pointer 3 does not exist, the array only has 3 elements but starts at 0 to 2, not 3!

edited Dec 22 '19 at 00:00

Dharman

30,962
25
85
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answered Dec 21 '19 at 12:36

ZellaThor

1
1

PHP Notice: Undefined offset: 1 with array when reading data

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