JSON对象添加到Javascript数组？

Question

我在解码JSON对象时遇到了问题。我使用Ajax调用来调用PHP脚本。PHP脚本返回一个JSON编码的对象，我可以读取它，但只能读取第一条记录。如何解码JSON对象？我使用的是JQUERY Mobile。

PHP脚本：

<?php

    $host = "localhost";
    $user = "root";
    $pass = "[password]";

    $databaseName = "zombieSurvival";
    $tableName = "TBLusers";

    //Connect to mysql database

    include 'DB.php';
    $con = mysql_connect($host,$user,$pass);
    $dbs = mysql_select_db($databaseName, $con);

    // 2) Query database

    $result = mysql_query("SELECT * FROM $tableName");
    $array = mysql_fetch_array($result);
    echo json_encode($array);

?>

下面是Ajax调用：

$.ajax({
       url: '/PHP/getUserMarkers.php',
       data: "",
       dataType: 'json',
       success: function(data){

       //How can I treat 'data' variable to make it a
       //javascript array?


       }
       });

Answer 1

问题出在PHP脚本返回的JSON对象上。如下所示调整了PHP脚本，现在我可以访问所有返回的行和单个字段名称。

<?php

    $host = "localhost";
    $user = "root";
    $pass = "muertealregueton";

    $databaseName = "zombieSurvival";
    $tableName = "TBLusers";


    include 'DB.php';
    $con = mysql_connect($host,$user,$pass);
    $dbs = mysql_select_db($databaseName, $con);


    $result = mysql_query("SELECT * FROM $tableName");
    $array = array();
    while(($row = mysql_fetch_array($result))) {
        $array[] = $row;
    }
    echo json_encode($array);

?>

然后返回ajax代码，该代码在后面的记录中调用警报：

$.ajax({
       url: '/PHP/getUserMarkers.php',
       data: "",
       dataType: 'json',
       success: function(data){


       //now I can access each row and field
       //like here I access the 3rd field of the 15th record
       alert(data[15][2]);



       }
       });