PHP 注意:未定义偏移量:1 读取数据时带数组 [英] PHP Notice: Undefined offset: 1 with array when reading data

问题描述

我收到此 PHP 错误:

I am getting this PHP error:

PHP Notice:  Undefined offset: 1

这是抛出它的 PHP 代码:

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_:
}

这个错误是什么意思?导致此错误的原因是什么?

What does this error mean? What causes this error?

推荐答案

更改

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

到

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

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

或者简单地说:

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

并非文件的每一行都有一个冒号，因此在它上面爆炸会返回一个大小为 1 的数组.

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

根据php.net 可能从explode返回值:

返回通过在分隔符形成的边界上拆分字符串参数而创建的字符串数组.

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

如果分隔符是一个空字符串 ("")，explode() 将返回 FALSE.如果分隔符包含字符串中不包含的值并且使用负限制，则返回空数组，否则返回包含字符串的数组.

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.

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