Question

我有以下代码

 public static $_myarray = array('This is S' => 'S',
                          'This is RC' => 'RC',
                          'This is RF' => 'RF',
                          'This is C' => 'C');

 $check_var = 'S';

我想要得到这个

echo $key_of_array =  array_search($check_var,$_myarray); // output should be 'This is S'

它返回空白值而不是“This is S”

我也试过以下。

- ＆GT;类型转换为$ check_var

- ＆GT; array_search（$ check_var，$ _ myArray的，真正的）;

但没有运气。有人可以帮我解决这个问题吗？

注意：它适用于其他值..问题只与数组的第一个元素..我在Linux上检查这个

Answer 1

直接来自docs，Returns the key for needle if it is found in the array, FALSE otherwise.所以这里是空字符串的来源。

$_myarray = [
    'This is S' => 'S',
    'This is RC' => 'RC',
    'This is RF' => 'RF',
    'This is C' => 'C'
];

$check_var = 'S';

echo array_search($check_var,$_myarray);
echo array_search('S',$_myarray);

适合我。

Answer 2

您的代码有效，但如果您不在课程范围内，则应该给出错误..

检查此fiddle

   <?php
    //echo phpinfo();
    /*public*/ $_myarray = array('This is S' => 'S', //note that public keyword causes an error out of class scope..
                              'This is RC' => 'RC',
                              'This is RF' => 'RF',
                              'This is C' => 'C');

    $check_var = 'S';
    echo $key_of_array =  array_search($check_var,$_myarray); // output should be 'This is S'

在php中打印array_search（）的第一个元素的键

2 个答案: