对于MySQL结果索引10，偏移量0无效，无法从mysql php检索数据。
对于MySQL结果索引10，偏移量0无效，无法从mysql php检索数据。
提问于 2013-11-07 16:14:51
在脚本中，无论输入正确或错误的用户名，结果都是一样的。
它是这样的：
{“成功”：1，“消息”：“登录成功”，“用户”：{“customerID”：空，“首名”：空，“姓氏”：空，“地址”：空，“邮政编码”：空，“电话”：空}
我尝试使用mysql_data_seek()，它返回的偏移量0对MySQL结果索引10无效。
我想mysql_query()没有获取任何东西。
有人能帮我解决这个问题吗？
<?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 $password;?>" maxlength="45">
<P align = center>
<INPUT TYPE = "Submit" Name = "Submit" VALUE = "Login">
</P>
</FORM>
</body>
</html>回答 3
Stack Overflow用户
回答已采纳
发布于 2013-11-07 16:22:58
您能试一下吗，添加行计数检查mysql_num_rows($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_*函数而不是mysql_*函数(不推荐)
Stack Overflow用户
发布于 2013-11-07 16:22:04
您需要检查if --任何行返回与否。
另外，session_start()必须位于文件的顶部。
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:不推荐使用 mysql_*函数。尽快在mysqli_*函数上移动
Stack Overflow用户
发布于 2013-11-07 16:24:14
如果结果集为空，mysql_data_seek()将在E_WARNING中失败。这是我认为在您的情况下发生的，因为在调用mysql_data_seek()之前，您没有检查结果集是否为空。
如果行数为>=1，则始终检查结果，然后才能安全地调用mysql_data_seek()。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/19840981
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