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我正在使用 php 和 MSSQL 开发 Web 服务。我可以连接 db 并检索值。但是当我使用 json 时，它不会返回任何值。

用于 Web 服务的 ajax 代码是

    $.ajax({
        url: 'http://209.249.81.138/dev/logintalk.php',
        type: 'POST',
        data: "username=" + $("#username").val()+"&password=" + $("#password").val(),
        dataType: 'json',
        success: function(data){
            alert(data);
        },
        error: function(){
            alert("failure...");
        }
       });
}

使用的php代码是

<?php

    $server = "server";
    $username = "username";

    $password = "secret";
    $database = "db";

    $con = mssql_connect($server, $username, $password) or die("Couldn't connect to SQL Server on $Server");

    $users = $_POST['username'];
    $passwd = $_POST['password'];

    $selected = mssql_select_db($database, $con) or die("Couldn't open database $database");

    mssql_query('SET CHARACTER SET utf8'); 
    $query = "SELECT * FROM zNewUsers WHERE memberid = '$users' AND userid = '$passwd'";
    $result = mssql_query($query) or die ("Unable to verify user because ");
    $row = mssql_fetch_array($result);
    $count = mssql_num_rows($result);

    if($count == 0){
        $data=array(0=>"Incorrect values...");
        $value[]=$data[0];
        echo json_encode($value);
    }
    else{
    $val[]=$row['SessionID'];
    echo json_encode($val);
    }  
    mssql_close($con);

?>
4

1 回答 1

0

毫无疑问的一些提示：

  1. success: function(data){
    console.log(data);
    //This is better for debugging if you use firebug, if you don't do it.
},

  2. if($count == 0){
    echo json_encode(array(0=>"Incorrect values..."));
    //Why create new variables? Pass what you want directly to json_encode()
}
else{
    echo json_encode($row['SessionID']);
}

回到问题：

var_dump() 在编码之前要编码的数据并确保它没问题，然后检查你在浏览器中获得的console.log(data)信息并与我们分享这些信息，以便我们提供更好的帮助。

于 2012-09-27T12:44:30.373 回答