Php Mysql登錄

我已經創建了兩個表名爲登錄和gotest.in gotest表我已經存儲了用戶的詳細信息和獨特的表中ID.in登錄表我存儲refid，用戶名和password.refid是主鍵在gotest table中包含相同的ID值。當它通過URl時，我從一個表單獲得ID。但當iam嘗試登錄時，它給我這個errpor「用戶名或密碼不正確！」。Php Mysql登錄

這裏是我的PHP代碼我的登錄頁面

<?php 

include_once 'dbconnect.php'; 

    $renewid = $_GET['ID']; 



    $query = "SELECT refid, username, password FROM ipay_login WHERE refid = '$renewid'"; 

    $result = mysql_query($query) or die(mysql_error()); 

    while ($row = mysql_fetch_array($result)) { 

     $renewid = $row['refid']; 
     $uname = $row['username']; 
     $upass = $row['password']; 

     echo $renewid . '<br />'; 
     echo $uname . '<br />'; 
     echo $upass . '<br />'; 

    } 

if(isset($_POST['btn-signup'])) { 


    $uname = $_POST['username']; 
    $upass = $_POST['password']; 


    /*echo $uname,$upass,$renewid;*/ 

    $result1 = mysql_query("SELECT * FROM ipay_login WHERE username = '$uname' AND password = '$upass'"); 

    if(mysql_num_rows($result1) > 0) 
    { 
     echo "sucesss"; 
    } 
    else 
    { 
     echo 'The Username or password are incorrect!'; 
    } 


} 
?> 

<html> 
<head></head> 
<body> 
<form id="convertion" method="post"> 


<!--<input type="hidden" id="refid" name="refid" value="<?php /*$_GET['refid']; */?>" /><br/>--> 
<input type="text" id="username" name="username" /><br/> 
<input type="text" id="password" name="password" /><br/> 

<button type="submit" id="btn-signup" name="btn-signup">SUBMIT</button> 
</form> 
</body> 
</html>

URL

http://xxx.yyy.example?ID=1000

來源

2016-04-29 colombo

如果用戶名和密碼都存儲在'ipay_login'表中，那麼爲什麼在SELECT查詢中使用'gotest'表？ –

你爲什麼認爲它應該輸出不同的東西？ –

請注意，'mysql_ *'函數在PHP 5.5中被棄用，並且在PHP 7.0中被完全刪除。使用['mysqli']（http://php.net/manual/en/book.mysqli.php）或['pdo']（http://php.net/manual/en/book.pdo.php）代替。 [這就是爲什麼你不應該使用'mysql_ *'函數]（http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php）。 –

嘗試所有這一切都在..first登錄更改REFID列ID.then運行下面的代碼

<?php 

include_once 'dbconnect.php'; 


$renewid = $_GET['ID']; 

$query = "SELECT * FROM login WHERE ID = '$renewid'"; 

$result = mysql_query($query) or die(mysql_error()); 

while ($row = mysql_fetch_array($result)) { 


    $uname = $row['username']; 
    $upass = $row['password']; 

    echo $uname . '<br />'; 
    echo $upass . '<br />'; 

} 

if(isset($_POST['btn-signup'])) { 


    $uname = $_POST['username']; 
    $upass= $_POST['password']; 


    $result1 = mysql_query("SELECT * FROM login WHERE username = '$uname' AND password = '$upass'"); 

    if(mysql_num_rows($result1) > 0) 
    { 

     echo "sucess"; 

    } 
    else 
    { 
     echo 'The username or password are incorrect!'; 
    } 
} 
?> 

<html> 
<head></head> 
<body> 
<form id="convertion" method="post"> 

    <input type="text" id="username" name="username" /><br/> 
    <input type="text" id="password" name="password" /><br/> 

    <button type="submit" id="btn-signup" name="btn-signup">SUBMIT</button> 
</form> 
</body> 
</html>

來源

2016-04-29 11:09:08 joseph

Php Mysql登錄

回答

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