Javascript ajax不工作，也不显示php数据

当一条记录显示数据时，当多条记录出现时，则不显示其他站点的数据。 ajaxx.php

<?php 
    include 'database.php';
    session_start();

    $post = $_POST;


        $search = $post['search'];
        $searchType = $post['searchType'];

        if ($searchType == 'all')
                {$sql = "SELECT DISTINCT title  FROM hadees WHERE title LIKE '$search%' AND (type='Bukhari' OR type='Muslim') ";}
            else
                {$sql = "SELECT DISTINCT title FROM hadees WHERE title LIKE '$search%' AND type='$searchType' ";}

        $result = mysqli_query($db,$sql);

        if ($result->num_rows > 0) {
            while($row = $result->fetch_assoc()) {
                $row['title'];
                echo json_encode($row);
            }
        } else 
            { echo "Not Found Result" ; }


?>

---

我认为你的问题在while循环中。您不想一行一行地编码每一行，但作为一个整体，就像这样

        $myResults = [];
        while($row = $result->fetch_assoc()) {
            $row['title'];
            $myResults[] = $row;
        }
        echo json_encode($myResults);

---

您正在使用

echo JSON\u encode（$row）生成无效的JSON在循环中

尝试创建一个行数组，然后显示它

if($result->num_rows > 0)
{
    $output = array();
    while($row = $result->fetch_assoc())
    {
        output[] = $row;
    }
    if($searchType == 'all')
    {
        echo json_encode($output);
    }
    else
    {
        echo json_encode(current($output)); // print just one
    }
}

if($result->num_rows > 0)
{
    $output = array();
    while($row = $result->fetch_assoc())
    {
        output[] = $row;
    }
    if($searchType == 'all')
    {
        echo json_encode($output);
    }
    else
    {
        echo json_encode(current($output)); // print just one
    }
}