json_encode无法在PHP中工作，意外结束数据

Question

我在这个php api中收到一个JSON.parse错误，即使数据库中的数据很好，它只是一个纯文本列，我不知道为什么PHP不解析简单文本的JSON，可以有人协助我下面的代码？

我从api获得状态200

  

错误：

SyntaxError：JSON.parse：JSON数据第2行第1列的意外数据结尾

  

api.php

    <?php
    $servername = "localhost";
    $username = "testuser";
    $password = "dbpass";
    $dbname = "testdb";

    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);
    // Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    }

    if (function_exists($_GET['method'])) {
        $_GET['method']();
    }

    function getAllBanks() {
        header('Content-Type: application/json');
        $banks_sql = mysql_query("SELECT DISTINCT BANK FROM BANK_MASTER;");
        $banks = [];
        while ($bank = mysql_fetch_object($banks_sql)) {
            $banks[] = $bank;
            $banks = json_encode($banks);
            echo $_GET['jsoncallback'] . '(' . $banks . ')';
        }
    }
    $conn->close();
    ?> 

  

以下是我从数据库中获取的数据

  

更新了新的mysqli_ api

    <?php
$servername = "localhost";
$username = "fayaz_api";
$password = "Fyzara$1994";
$dbname = "fayaz_BANK";

// Create connection

$conn = new mysqli($servername, $username, $password, $dbname);

// Check connection

if ($conn->connect_error)
    {
    die("Connection failed: " . $conn->connect_error);
    }

if (function_exists($_GET['method']))
    {
    $_GET['method']();
    }

function getAllBanks()
    {
    header('Content-Type: application/javascript');

    // $banks_sql = mysql_query("SELECT DISTINCT BANK FROM BANK_MASTER;");

    $banks = [];
    $sql = "SELECT DISTINCT BANK FROM BANK_MASTER";
    if ($result = mysqli_query($con, $sql))
        {



        while ($row = mysqli_fetch_array($result))
            {
              $banks[] = $row;
            }

        $banks = json_encode($banks);
        echo $_GET['jsoncallback'] . '(' . $banks . ')';

        // Free result set

        mysqli_free_result($result);
        }
    }

$conn->close();
?>