Question

如上所述，我收到一个没有标识符的JSON字符串。我正在使用的代码是：

<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：

[＆＃34; 2d791c8b-d8cf-11e4-a712-002590f3d888＆＃34;＆＃34; ********＆＃34;＆＃34; ******** ＆＃34;]

我是PHP和MySQL的新手，所以请保持温和。

Answer 1

mysql_fetch_row返回枚举数组

[0] => "2d791c8b-d8cf-11e4-a712-002590f3d888"

mysql_fetch_array返回一个关联数组

["key"] => "2d791c8b-d8cf-11e4-a712-002590f3d888"

mysql_ *函数现已不存在，先于MySQLi / PDO。

Answer 2

要具体回答您的问题，请尝试使用：

<?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);
?>

但是摇滚乐队是对的，至少看看mysqli

尝试使用php从mysql中选择记录，结果是在没有标识符的json中

2 个答案: