Question

我想从this获得this JSON输出。遗憾的是，json_encode（）函数不会将数组编码为该格式。根本没有任何回报。这是我的代码。

$output = array(
    'responseData' => array(),
    'responseDetails' => null,
    'responseStatus' => 200
);

$x = 0;
while ($row = mysqli_fetch_assoc($result)) {
    foreach ($row as $k => $v) {
        $output['responseData']['result'][$x][$k] = $v;
    }
    $x++;
}

print_r($output);
header('Content-Type: application/json');
echo json_encode($output , JSON_FORCE_OBJECT);

我无法找到原因。有人请帮我找到解决方案。

编辑：抱歉。这是输出 -

预期的JSON输出 -

{
"responseData": {
    "results": [{
        "qid": 1,
        "qtitle": "When do we finish this project ?",
        "qimage_url": "http://www.wearesliit.com/example.png",
        "user": "samith",
        "date": "2016-01-01T02:15:12.356Z",
        "type": 1,
        "category": 5,
        "tags": ["common_senese", "truth", "bazsa_awsanna"],
        "note": "Sample quetion"
    }, {}, {}]
},
"responseDetails": null,
"responseStatus": 200 }

我根本没有得到任何JSON输出。但这是数组的print_r结果。

 Array(
[responseData] => Array
    (
        [result] => Array
            (
                [0] => Array
                    (
                        [question_ID] => 1
                        [question_Title] => Which shape does not belong with the other three shapes?
                        [question_Image_URL] => http://www.wearesliit.com/images/quiz/questions/1.jpg
                        [quetion_Note] => Easy IQ question.
                        [category_ID] => 7
                        [username] => samith
                        [added] => 2017-01-29 21:50:52
                    )

                [1] => Array
                    (
                        [question_ID] => 2
                        [question_Title] => Tim earns $10 per hour at his job.  When he gets paid on Friday, he is paid for 40 hours of work.  He then goes out and spends 10% of his earnings on entertainment that weekend.  How much money is he left with on Monday?
                        [question_Image_URL] => 
                        [quetion_Note] => Easy IQ question.
                        [category_ID] => 7
                        [username] => samith
                        [added] => 2017-01-29 21:50:52
                    )
            )

    )

[responseDetails] => 
[responseStatus] => 200 )

Answer 1

感谢@awiebe，我发现了确切的错误。它的

格式错误的UTF-8字符，可能编码错误

谢谢大家，我从另一个问题找到了解决方案。 'Malformed UTF-8 characters, possibly incorrectly encoded' in Laravel

Answer 2

当编码失败并且回显或打印中没有出现“假”结果时，

json_encode（）函数返回“false”。请参阅：http://php.net/manual/en/function.json-encode.php 处理此类问题的最佳方法是使用json_last_error_msg（）方法并根据找到的错误执行操作。见：http://php.net/manual/en/function.json-last-error-msg.php。一个例子是：

$show_json = json_encode($output , JSON_FORCE_OBJECT);
if ( json_last_error_msg()=="Malformed UTF-8 characters, possibly incorrectly encoded" ) {
    $show_json = json_encode($API_array, JSON_PARTIAL_OUTPUT_ON_ERROR );
}
if ( $show_json !== false ) {
    echo($show_json);
} else {
    die("json_encode fail: " . json_last_error_msg());
}

问题是，如果问题是编码字符，则不会显示。只是希望角色对你的工作不是必不可少的，或者如果你能在zigazillion字符串列表中找到不匹配输入，那就纠正它吧。其他类型的错误可以在这里找到：http://php.net/manual/en/json.constants.php。只需对您发现的错误应用if语句和更正。

我希望这有助于某人。

Answer 3

删除

header('Content-Type: application/json');

请记住，在发送任何实际输出之前，必须通过普通HTML标记，文件中的空行或PHP来调用header（）。使用include或require，函数或其他文件访问函数读取代码是一个非常常见的错误，并且在调用header（）之前输出空格或空行。使用单个PHP / HTML文件时存在同样的问题。

来自

的

和JSON_FORCE_OBJECT

echo json_encode($output , JSON_FORCE_OBJECT);

Answer 4

我想向您介绍这个问题，在link上我建议您使用像这样的json_encode包装器：

function safe_json_encode($value){
if (version_compare(PHP_VERSION, '5.4.0') >= 0) {
    $encoded = json_encode($value, JSON_PRETTY_PRINT);
} else {
    $encoded = json_encode($value);
}
switch (json_last_error()) {
    case JSON_ERROR_NONE:
        return $encoded;
    case JSON_ERROR_DEPTH:
        return 'Maximum stack depth exceeded'; // or trigger_error() or throw new Exception()
    case JSON_ERROR_STATE_MISMATCH:
        return 'Underflow or the modes mismatch'; // or trigger_error() or throw new Exception()
    case JSON_ERROR_CTRL_CHAR:
        return 'Unexpected control character found';
    case JSON_ERROR_SYNTAX:
        return 'Syntax error, malformed JSON'; // or trigger_error() or throw new Exception()
    case JSON_ERROR_UTF8:
        $clean = utf8ize($value);
        return safe_json_encode($clean);
    default:
        return 'Unknown error'; // or trigger_error() or throw new 
Exception();
}
}


function utf8ize($mixed) {
if (is_array($mixed)) {
    foreach ($mixed as $key => $value) {
        $mixed[$key] = utf8ize($value);
    }
} else if (is_string ($mixed)) {
    return utf8_encode($mixed);
}
return $mixed;
}

在定义了这些功能之后，您可以直接使用它，

echo safe_json_encode($response);

php JSON_encode不能正常工作

4 个答案: