我正在寻找类似于MySQL像%term%的功能。根据本页的答案。我能够从文件中搜索JSON数组。

user_list。Json看起来像下面的示例:

{
  "user-23456": {
        "name": "John Doe",
        "age": "20",
        "email": "doe@sample.com",
        "user_id": "23456"
    },
    "user-09876": {
        "name": "Ronojoy Adams",
        "age": "35",
        "email": "joy@sample.com",
        "user_id": "09876"
    },
    "user-34890": {
        "name": "Will Artkin",
        "age": "16",
        "email": "will@sample.com",
        "user_id": "34890"
    },
}

/*
*search_key_like
*/

function search_key_like($value, $key, $array) {
     $results=array();
    $keyword = preg_quote($value, '~');
   foreach ($array as $k => $val) {
//if name a is spell John and keyword is sent as joh or JOH it will return null
//to fix the issue convert the string into lowercase and uppercase
       $data=array($val[$key],strtolower($val[$key]),strtoupper($val[$key]));
       if (preg_grep('~' . $keyword . '~', $data)) {
       array_push($results,$val[$key]);           
    }
   }
   return $results;
}

使用===提取JSON文件===

 $user_list_json='./user_list.json';
    if(file_exists($user_list_json) && file_get_contents($user_list_json)){
    $file_json_data=file_get_contents($user_list_json);
        
    $json_array_data=json_decode($file_json_data,true);
        
        $user_name_like = search_key_like('ron', 'name', $json_array_data);
   
         print "<pre>".print_r($user_name_like,true);
    }

如果你使用的是(PHP 5 >= 5.5.0)，你不需要自己写函数来实现这个功能，只需要写这一行就可以了。

如果你只想要一个结果:

$key = array_search(40489, array_column($userdb, 'uid'));

对于多个结果

$keys = array_keys(array_column($userdb, 'uid'), 40489);

如果你有一个关联数组，就像注释中指出的那样，你可以用:

$keys = array_keys(array_combine(array_keys($userdb), array_column($userdb, 'uid')),40489);

如果你使用的是PHP < 5.5.0，你可以使用这个backport，谢谢ramsey!

更新:我一直在做一些简单的基准测试，多重结果表单似乎是最快的，甚至比Jakub自定义函数还要快!

在Jakub的出色回答的基础上，这里有一个更通用的搜索，将允许指定键(不仅仅是uid):

function searcharray($value, $key, $array) {
   foreach ($array as $k => $val) {
       if ($val[$key] == $value) {
           return $k;
       }
   }
   return null;
}

用法:$results = searcharray('searchvalue'， searchkey， $array);

我想检查下面的数组$arr是否有'abc'存在于子数组中

$arr = array(
    array(
        'title' => 'abc'
    )
);

然后我可以用这个

$res = array_search('abc', array_column($arr, 'title'));
if($res == ''){
    echo 'exists';
} else {
    echo 'notExists';
}

我认为这是最简单的定义方式

这里有一句同样的句子，

$pic_square = $userdb[array_search($uid,array_column($userdb, 'uid'))]['pic_square'];

如果问题，即。

$a = [
     [
       "_id" => "5a96933414d48831a41901f2",
       "discount_amount" => 3.29,
       "discount_id" => "5a92656a14d488570c2c44a2",
     ],
     [
       "_id" => "5a9790fd14d48879cf16a9e8",
       "discount_amount" => 4.53,
       "discount_id" => "5a9265b914d488548513b122",
     ],
     [
       "_id" => "5a98083614d488191304b6c3",
       "discount_amount" => 15.24,
       "discount_id" => "5a92806a14d48858ff5c2ec3",
     ],
     [
       "_id" => "5a982a4914d48824721eafe3",
       "discount_amount" => 45.74,
       "discount_id" => "5a928ce414d488609e73b443",
     ],
    [
       "_id" => "5a982a4914d48824721eafe55",
       "discount_amount" => 10.26,
       "discount_id" => "5a928ce414d488609e73b443",
     ],
   ];

岁:

function searchForId($id, $array) {
    $did=0;
    $dia=0;
   foreach ($array as $key => $val) {
       if ($val['discount_id'] === $id) {
           $dia +=$val['discount_amount'];
           $did++;
       }
   }
    if($dia != '') {
        echo $dia;
        var_dump($did);
    }
   return null;
};
print_r(searchForId('5a928ce414d488609e73b443',$a));