Php 为什么json_encode返回空值？

我是PHP新手。我的问题是json_encode返回的空指针。我正在尝试从mamp服务器上的mysql数据库获取数据。我做错了什么？？我应该使用类似utf8\u编码的东西？？我已经试过了，没有什么特别的事情发生这是我的密码：

    <?php
    $servername = "localhost";
    $username = "root";
    $password = "root";
    $database = "authors";

    $db = new mysqli($servername, $username, $password, $database);

    if (!$db){
        die("Connection failed: " . mysqli_connect_error());
    }

    $sql = "SELECT * FROM names";

    if ($result = mysqli_query($db, $sql)){
        $resultArray = array();
        $tempArray = array();

        while($row = $result->fetch_object()){
            $tempArray = $row;
            array_push($resultArray, $tempArray);
        }
        header('Content-type: application/json');
        print_r( $resultArray );
        $s = json_encode($resultArray);
        echo $s;
        if( $s = NULL )
        {
            echo "Nullpointer";
        }
        elseif( $s == "" )
        {
            echo "Empty string";
        }
    }

    $db->close();
?>

---

更新代码：

<?php
    $servername = "localhost";
    $username = "root";
    $password = "root";
    $database = "authors";

    $db = new mysqli($servername, $username, $password, $database);

    if (!$db){
        die("Connection failed: " . mysqli_connect_error());
    }

    $sql = "SELECT * FROM names";

    if ($result = mysqli_query($db, $sql)){
        $resultArray = array();
        $tempArray = array();

        while($row = $result->fetch_object()){
            $tempArray = $row;
            array_push($resultArray, $tempArray);
        }
        header('Content-type: application/json');
        print_r( $resultArray );
        $s = json_encode($resultArray);
        echo $s;
        if( $s == false )
        {
            echo "Nullpointer";
        }
        elseif( $s == "" )
        {
            echo "Empty string";
        }
        else
        {
            echo "It's gooood";
        }
    }

    $db->close();
?>

---

你们能解决的问题是什么

探索错误

echo json_last_error();
echo json_last_error_msg();

使用JSON_UNESCAPED_UNICODE

$s = json_encode($resultArray, JSON_UNESCAPED_UNICODE);

---

如果（$s=NULL）

在这里应该是

==

instead@Adrian . json_encode从不返回

NULL

。它返回一个字符串，或FALSE。首先修复此问题，它可能会改变很多事情：）这样做会删除$s值并破坏测试逻辑。输出是：数组（[0]=>stdClass对象（[FirstName]=>Adrian[LastName]=>Kaczmarek）[1]=>stdClass对象（[FirstName]=>Krzysztof[LastName]=>Rózga））空指针$s==null，$s==“”与$s==false相同。请参阅严格等于（==vs.==）

echo json_last_error();
echo json_last_error_msg();

$s = json_encode($resultArray, JSON_UNESCAPED_UNICODE);