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?

Answer 1

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.

Answer 2

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