function isJson($string) {
   json_decode($string);
   return json_last_error() === JSON_ERROR_NONE;
}

应该是这样的:

 function isJson($string)
 {
    // 1. Speed up the checking & prevent exception throw when non string is passed
    if (is_numeric($string) ||
        !is_string($string) ||
        !$string) {
        return false;
    }

    $cleaned_str = trim($string);
    if (!$cleaned_str || !in_array($cleaned_str[0], ['{', '['])) {
        return false;
    }

    // 2. Actual checking
    $str = json_decode($string);
    return (json_last_error() == JSON_ERROR_NONE) && $str && $str != $string;
}

单元测试

public function testIsJson()
{
    $non_json_values = [
        "12",
        0,
        1,
        12,
        -1,
        '',
        null,
        0.1,
        '.',
        "''",
        true,
        false,
        [],
        '""',
        '[]',
        '   {',
        '   [',
    ];

   $json_values = [
        '{}',
        '{"foo": "bar"}',
        '[{}]',
        '  {}',
        ' {}  '
    ];

   foreach ($non_json_values as $non_json_value) {
        $is_json = isJson($non_json_value);
        $this->assertFalse($is_json);
    }

    foreach ($json_values as $json_value) {
        $is_json = isJson($json_value);
        $this->assertTrue($is_json);
    }
}

你真正需要做的就是…

if (is_object(json_decode($MyJSONArray))) 
{ 
    ... do something ...
}

这个请求甚至不需要一个单独的函数。只需将is_object包装在json_decode周围，然后继续。似乎这个解决方案让人们花了太多心思。

之前我只是检查一个空值，这实际上是错误的。

    $data = "ahad";
    $r_data = json_decode($data);
    if($r_data){//json_decode will return null, which is the behavior we expect
        //success
    }

上面的代码可以很好地处理字符串。然而，只要我提供号码，它就中断了。为例。

    $data = "1213145";
    $r_data = json_decode($data);

    if($r_data){//json_decode will return 1213145, which is the behavior we don't expect
        //success
    }

要修复它，我所做的很简单。

    $data = "ahad";
    $r_data = json_decode($data);

    if(($r_data != $data) && $r_data)
        print "Json success";
    else
        print "Json error";

只需添加这个条件:

check if the type is string and then json decode <?php $subject = ['description' => '200 extra contacts','value' => '15','product_code' => 'OS_CONT12']; $subject = '{"description":"200 extra contacts","value":15,"product_code":"OS_CONT12"}'; if(gettype($subject) == 'string'){ $data = json_decode($subject, true); print_r($data); } else{ print_r("saurabh kasmble"); } ?> OUTPUT : Array ( [description] => 200 extra contacts [value] => 15 [product_code] => OS_CONT12 )

问题的答案

函数json_last_error返回JSON编码和解码过程中发生的最后一个错误。因此，检查有效JSON的最快方法是

// decode the JSON data
// set second parameter boolean TRUE for associative array output.
$result = json_decode($json);

if (json_last_error() === JSON_ERROR_NONE) {
    // JSON is valid
}

// OR this is equivalent

if (json_last_error() === 0) {
    // JSON is valid
}

注意json_last_error仅在PHP >= 5.3.0中支持。

完整的程序来检查准确的错误

在开发期间了解准确的错误总是好的。下面是基于PHP文档检查确切错误的完整程序。

function json_validate($string)
{
    // decode the JSON data
    $result = json_decode($string);

    // switch and check possible JSON errors
    switch (json_last_error()) {
        case JSON_ERROR_NONE:
            $error = ''; // JSON is valid // No error has occurred
            break;
        case JSON_ERROR_DEPTH:
            $error = 'The maximum stack depth has been exceeded.';
            break;
        case JSON_ERROR_STATE_MISMATCH:
            $error = 'Invalid or malformed JSON.';
            break;
        case JSON_ERROR_CTRL_CHAR:
            $error = 'Control character error, possibly incorrectly encoded.';
            break;
        case JSON_ERROR_SYNTAX:
            $error = 'Syntax error, malformed JSON.';
            break;
        // PHP >= 5.3.3
        case JSON_ERROR_UTF8:
            $error = 'Malformed UTF-8 characters, possibly incorrectly encoded.';
            break;
        // PHP >= 5.5.0
        case JSON_ERROR_RECURSION:
            $error = 'One or more recursive references in the value to be encoded.';
            break;
        // PHP >= 5.5.0
        case JSON_ERROR_INF_OR_NAN:
            $error = 'One or more NAN or INF values in the value to be encoded.';
            break;
        case JSON_ERROR_UNSUPPORTED_TYPE:
            $error = 'A value of a type that cannot be encoded was given.';
            break;
        default:
            $error = 'Unknown JSON error occured.';
            break;
    }

    if ($error !== '') {
        // throw the Exception or exit // or whatever :)
        exit($error);
    }

    // everything is OK
    return $result;
}

使用有效的JSON INPUT进行测试

$json = '[{"user_id":13,"username":"stack"},{"user_id":14,"username":"over"}]';
$output = json_validate($json);
print_r($output);

有效的输出

Array
(
    [0] => stdClass Object
        (
            [user_id] => 13
            [username] => stack
        )

    [1] => stdClass Object
        (
            [user_id] => 14
            [username] => over
        )
)

使用无效JSON进行测试

$json = '{background-color:yellow;color:#000;padding:10px;width:650px;}';
$output = json_validate($json);
print_r($output);

无效的输出

Syntax error, malformed JSON.

额外注意(PHP >= 5.2 && PHP < 5.3.0)

由于PHP 5.2中不支持json_last_error，因此可以检查编码或解码是否返回布尔值FALSE。这里有一个例子

// decode the JSON data
$result = json_decode($json);
if ($result === FALSE) {
    // JSON is invalid
}