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

问题

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?

回答1:

Change

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

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.

回答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

回答3:

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

error_reporting = E_ALL & ~E_NOTICE

This show all errors, except for notices.

回答4:

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

回答5:

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>";

回答6:

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 ","...

回答7:

Hide php warnings in file

error_reporting(0);

来源：https://stackoverflow.com/questions/17456325/php-notice-undefined-offset-1-with-array-when-reading-data

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php

error-handling

syntax-error

undefined

fatal-error