2017-05-26 96 views
-2

我是編程新手。我其實想用數據庫中存儲的值檢查用戶輸入值。特別地,由用戶給出的電子郵件和密碼,打印相應電子郵件ID的詳細信息。這是我的代碼。請幫我糾正我的代碼。用PHP中的數據庫值檢查用戶輸入值

<?php 
    $conn = mysqli_connect('localhost', 'root', '', 'mydb'); 
    $sql = "SELECT email, first FROM register"; 
    $result = mysqli_query($conn, $sql); 

    if (mysqli_num_rows($result) > 0) { 
     while($row = mysqli_fetch_assoc($result)) { 
      echo "Welcome " . $row["first"]. "<br>". "Your Email Id is: " . $row["email"] . "<br>"; 
     }  
    } else { 
     echo "0 results"; 
    } 
    mysqli_close($conn); 
?> 
+0

由於沒有使用'WHERE'子句。它將顯示註冊表中的所有成員。 –

+1

*「我確實想檢查用戶輸入值」* - 我在這裏看不到用戶交互;嘗試一下。你的問題不支持任何內容,並且太寬泛。發佈相關的代碼,除非您希望我們爲您編寫代碼,但這不是這裏的工作方式。 –

回答

0
<?php 
    $conn  = mysqli_connect('localhost', 'root', '', 'mydb'); 
    $username = $_POST['username']; 
    $password = $_POST['password']; 
    $query  = "select * from register where email = '$username'"; 
    $resultSet = mysqli_query($query, $conn); 
    if(mysqli_num_rows($resultSet) > 0){ 
     $row = mysqli_fetch_assoc($resultSet); 
     if($row['password'] == $password){ // if you are using encryption like md5 or anything else then you have to add in this line accordingly 
      echo "Good, Logged In!"; 
     }else{ 
      echo "Oh No, password not correct!"; 
     } 
    }else{ 
     echo "Please enter correct email!"; 
    } 
+0

謝謝..這真的很有幫助.. –

+0

歡迎。你可以投票。 –

0
<?php 
     $conn = mysqli_connect('localhost','root','','mydb'); 
     $sql = "SELECT email,first FROM register where email=$_POST['form_ref_name']; 
     $result = mysqli_query($conn,$sql); 
     if($result){ 
     if (mysqli_num_rows($result) > 0) { 
     while($row = mysqli_fetch_assoc($result)) { 
     echo "Welcome " . $row["first"]. "<br>". "Your Email Id is: " . $row["email"] . "<br>"; 
      }  
     } 
     } 
     else { 
     echo "0 results"; 
     } 
     mysqli_close($conn); 
    ?> 

試試這可能幫助。 Insted你可以用$ row varible來包含來自數據庫的值。

0

你可以在下面提到的條件下使用。我認爲幫助下面的查詢。

<?php 
$email = $_POST['email']; 
$password= $_POST['password']; 

$sql = mysql_query("SELECT email,first FROM register WHERE email = $email and password = $password"); 
?>