如前所述，我得到了一个没有标识符的JSON字符串。我使用的代码是：
<html>
<head>
<script language="javascript" type="text/javascript" src="/content/scripts/jquery/v2.1.3/jquery-2.1.3.js"></script>
</head>
<body>
<h2> Client example </h2>
<h3>Output: </h3>
<?php
$host = "localhost";
$user = "{username}";
$pass = "{password}";
$databaseName = "{database}";
$tableName = "{table}";
$con = mysql_connect($host,$user,$pass);
$dbs = mysql_select_db($databaseName, $con);
$result = mysql_query("SELECT * FROM $tableName");
$array = mysql_fetch_row($result);
echo json_encode($array);
?>
</body>
我的结果是：
["2d791c8b-d8cf-11e4-a712-002590f3d888"，"********"，"****"]
我是PHP和MySQL的新手，所以要温和一点。
mysql_fetch_row返回枚举数组
[0] => "2d791c8b-d8cf-11e4-a712-002590f3d888"
mysql_fetch_array返回关联数组
["key"] => "2d791c8b-d8cf-11e4-a712-002590f3d888"
mysql_*函数现在已经失效，前面是MySQLi/PDO。
要具体回答您的问题，请尝试使用以下方法：
<?php
$host = "localhost";
$user = "{username}";
$pass = "{password}";
$databaseName = "{database}";
$tableName = "{table}";
$con = mysql_connect($host,$user,$pass);
$dbs = mysql_select_db($databaseName, $con);
$result = mysql_query("SELECT * FROM $tableName");
$array = mysql_fetch_array($result);
echo json_encode($array);
?>
但rockerBOO是对的，至少看看mysqli