PHP：JSON 返回 NULL 值答案

【问题标题】：PHP: JSON Returning NULL ValuesPHP：JSON 返回 NULL 值
【发布时间】：2014-10-06 17:30:53
【问题描述】：

我目前正在为我的网站开发搜索功能，并以 JSON 格式返回结果。但是，一些值在返回第一个结果后返回，如下所示：

{"fullname":"test name","occupation":"test","industry":"testing","bio":"i am testing stuff.","gender":"f","website":"http:\/\/yhisisasite.com","skills":["writing","reading","math","coding","baseball"],"interests":["coding","sampling","googling","typing","playing"]},{"fullname":null,"occupation":null,"industry":null,"bio":null,"gender":null,"website":null,"skills":["coding","docotrs","soeku","spelling"],"interests":["testing","wintro","skating","hockey","code"]}

我目前有一个类可以用作结果的模板，如下所示：

class SearchResultUserProfile {
    public $fullname = "";
    public $occupation = "";
    public $industry = "";
    public $bio = "";
    public $gender = "";
    public $website = "";
    public $skills = array();
    public $interests = array();

}

然后为了填充这些字段，我在 mysqli fetch 期间有几个循环：

while ($row = mysqli_fetch_array($result))
{   
    $max = sizeof($user_id_array);
    for($i = 0; $i < $max; $i++)
    {
    //create a new instance or object
    $searchResultUserProfile = new SearchResultUserProfile();
    $searchResultUserProfile->fullname = $row['fullname'];
    $searchResultUserProfile->occupation = $row['occupation'];
    $searchResultUserProfile->industry = $row['industry'];
    $searchResultUserProfile->bio = $row['bio'];
    $searchResultUserProfile->gender = $row['gender'];
    $searchResultUserProfile->website = $row['website'];

    //grab the interests and skills
    $skillDetails = fetchAllUserSkills($user_id_array[$i]);
    foreach($skillDetails as $row) {
        $thistest = $row['skills'];
        array_push($searchResultUserProfile->skills, $thistest);
    }

    $interestDetails = fetchAllUserInterests($user_id_array[$i]);
    foreach($interestDetails as $row) {
        $thistests = $row['interests'];
        array_push($searchResultUserProfile->interests, $thistests);
    }
    array_push($results, $searchResultUserProfile);

    }
    echo json_encode($results);
}

知道为什么会这样吗？是我如何迭代循环还是设置？我确信我忽略了一些简单的东西，但我无法弄清楚它是什么。

【问题讨论】：

您确定$results 在json_encode 之前不为空吗？
不相关，但要生成有效的 json，您需要将最后一条语句放在 while 循环之外（以防有超过 1 个结果）。

标签： php json for-loop mysqli while-loop

【解决方案1】：

问题是您在内部循环中覆盖了 $row 变量：

while ($row = mysqli_fetch_array($result))
       ^^^^ this is a result row from your query
{   
    $max = sizeof($user_id_array);
    for($i = 0; $i < $max; $i++)
    {
        $searchResultUserProfile = new SearchResultUserProfile();
        $searchResultUserProfile->fullname = $row['fullname'];

        ...

        foreach($skillDetails as $row) {
                                 ^^^^ here you are overwriting your query result
           ...
        }

        ...

    }
    echo json_encode($results);
}

因此，如果$max 大于 1，则从第二次迭代开始，您将使用上一个内部循环的最后一个结果。这不会是您期望的查询结果。

【讨论】：