偏移量 0 对 MySQL 结果索引 10 无效，无法从 mysql php 检索数据答案

【问题标题】：Offset 0 is invalid for MySQL result index 10 , cannot retrieve data from mysql php偏移量 0 对 MySQL 结果索引 10 无效，无法从 mysql php 检索数据
【发布时间】：2013-11-19 09:30:11
【问题描述】：

在脚本中，无论我输入正确还是错误的用户名，结果都是一样的。

是这样的：

{"success":1,"message":"登录成功！","user":{"customerID":null,"firstname":null,"lastname":null,"address":null,"邮政编码":null,"电话":null}}

我尝试使用 mysql_data_seek()，它返回 Offset 0 is invalid for MySQL result index 10.

我猜mysql_query() 没有获取任何内容。

谁能帮我解决这个问题？

 <?PHP

$username = "";
$password = "";
$re = array();


if ($_SERVER['REQUEST_METHOD'] == 'POST'){
$username = $_POST['username'];
$password = $_POST['password'];

$username = htmlspecialchars($username);
$password = htmlspecialchars($password);

//==========================================
//  CONNECT TO THE LOCAL DATABASE
//==========================================
$dbusername = ""; 
    $dbpassword = ""; 
    $host = ""; 
    $dbname = "";

$db_handle = mysql_connect($host, $dbusername, $dbpassword);
$db_found = mysql_select_db($dbname, $db_handle);

if ($db_found) {

    $username = quote_smart($username, $db_handle);
    $password = quote_smart($password, $db_handle);

    $query = "SELECT * FROM Customer WHERE username = $username AND password = md5($password)";
    $info = mysql_query($query);

//====================================================
//  CHECK TO SEE IF THE $result VARIABLE IS TRUE
//====================================================

    if ($info) {
$user = mysql_fetch_array( $info );
echo $user['firstname'];
    $re["success"]=1;
    $re["message"] = "Login successfully!";
//add parameters that will return to android
    $re["user"]["customerID"]=$user['customerID'];
    $re["user"]["firstname"]=$user['firstname'];
    $re["user"]["lastname"] = $user['lastname'];
    $re["user"]["address"]=$user['address'];
    $re["user"]["postcode"]=$user['postcode'];
    $re["user"]["phone"]=$user['phone'];

    $re= json_encode($re);
    echo $re;

        session_start();
        $_SESSION['login'] = "1";
        $_SESSION["re"]=$re;
        //header ("Location: userdata.php");

    }
    else {
    $re["success"]=0;
    $re["message"] = "User Not Found";
    $re = json_encode($re);
    echo $re;
    }
}

else {
    $re["success"]=0;
    $re["message"] = "Can not connect to database";
    $re= json_encode($re);
    echo $re;
}

}


?>


<html>
<head>
<title>Basic Login Script</title>
</head>
<body>

<FORM NAME ="Customer Login" METHOD ="POST" ACTION ="login.php">

Username: <INPUT TYPE = 'TEXT' Name ='username'  value="<?PHP print $username;?>"     maxlength="45">
Password: <INPUT TYPE = 'TEXT' Name ='password'  value="<?PHP print $password;?>"  maxlength="45">

<P align = center>
<INPUT TYPE = "Submit" Name = "Submit"  VALUE = "Login">
</P>

</FORM>

</body>
</html>

【问题讨论】：

标签： php html mysql sql json

【解决方案1】：

你可以试试这个，添加行数检查mysql_num_rows($info);

        $query = "SELECT * FROM Customer WHERE username = '".$username."' AND password = '".md5($password)."' ";

        $info = mysql_query($query) or die("MySQL ERROR: ".mysql_error()); 

        $count = mysql_num_rows($info); // added rows count check

        if($count>0){       

             $user = mysql_fetch_array( $info );
            echo $user['firstname'];
            $re["success"]=1;
            $re["message"] = "Login successfully!";
            //add parameters that will return to android
            $re["user"]["customerID"]=$user['customerID'];
            $re["user"]["firstname"]=$user['firstname'];
            $re["user"]["lastname"] = $user['lastname'];
            $re["user"]["address"]=$user['address'];
            $re["user"]["postcode"]=$user['postcode'];
            $re["user"]["phone"]=$user['phone'];    
            session_start();
            $_SESSION['login'] = "1";
            $_SESSION["re"]=$re;
            //header ("Location: userdata.php");
            $re= json_encode($re);
            echo $re;

        }
        else {
            $re["success"]=0;
            $re["message"] = "User Not Found";
            $re = json_encode($re);
            echo $re;
        }

注意：使用 mysqli_* 函数代替 mysql_* 函数（已弃用）

【讨论】：

我添加了 mysql_num_rows($info);无论输入是什么，结果都会变为 {"success":0,"message":"User Not Found"}。 mysql_query() 没有返回任何东西给我。我该如何解决这个问题？
echo $query string 并在mysql查询窗口中运行以检查表customer是否有数据
$info = mysql_query($query) or die("MySQL ERROR: ".mysql_error());更新我的错误
当然，它有数据。我已经仔细检查过了。
我添加 $info = mysql_query($query) 或 die("MySQL ERROR: ".mysql_error());但没有任何改变。还是 {"success":0,"message":"User Not Found"}

【解决方案2】：

您需要检查if是否有任何行返回。

session_start() 也必须在文件顶部

session_start();
$query = "SELECT * FROM Customer WHERE username = $username AND password = md5($password) LIMIT 1";
$info = mysql_query($query);
$found = mysql_num_rows($info);

if($found!=0)
{
    $user = mysql_fetch_array( $info );
    echo $user['firstname'];
    $re["success"]=1;
    $re["message"] = "Login successfully!";
    //add parameters that will return to android
    $re["user"]["customerID"]=$user['customerID'];
    $re["user"]["firstname"]=$user['firstname'];
    $re["user"]["lastname"] = $user['lastname'];
    $re["user"]["address"]=$user['address'];
    $re["user"]["postcode"]=$user['postcode'];
    $re["user"]["phone"]=$user['phone'];

    $re= json_encode($re);
    echo $re;

    $_SESSION['login'] = "1";
    $_SESSION["re"]=$re;
    //header ("Location: userdata.php");
}

注意： mysql_* 函数已弃用。继续mysqli_*功能尽快

【讨论】：

我添加了 mysql_num_rows($info);无论输入是什么，结果都会变为 {"success":0,"message":"User Not Found"}。 mysql_query() 没有返回任何东西给我。我该如何解决这个问题？

【解决方案3】：

如果结果集为空mysql_data_seek() 将失败并返回E_WARNING。我认为这发生在您的情况下，因为您在调用 mysql_data_seek() 之前没有检查结果集是否为空。

如果行数>=1，请始终检查结果，然后您可以安全地调用mysql_data_seek()

【讨论】：