在驗證數據庫中值的心不是已經

我試圖驗證輸入的電子郵件地址是不是已經在我的數據基礎。我有以下代碼：在驗證數據庫中值的心不是已經

繼承人在我的HTML我的形式

<form action="signup.php" method="post" onsubmit="return validateForm(this);"> 
       <input type="text" name="First_name" placeholder="First Name" required pattern="[A-Za-z].{2,}" maxlength="20" oninvalid="this.setCustomValidity('Must be between 2 and 20 characters long and no numbers!')"/> 
       <input type="text" name="Last_name" placeholder="Last Name" required pattern="[A-Za-z']..{2,}" maxlength="20" oninvalid="this.setCustomValidity('Must be between 2 and 20 characters long! and no numbers')"/> 
       <input type="email" pattern="[a-z0-9._%+-][email protected][a-z0-9.-]+\.[a-z]{2,4}$" name="Email" required placeholder="Email Address" maxlength="30" oninvalid="this.setCustomValidity('Must be a validate Email Address!')"/> 
       <input type="email" pattern="[a-z0-9._%+-][email protected][a-z0-9.-]+\.[a-z]{2,4}$" name="Email2" required placeholder="Comfirm Email Address" maxlength="30" oninvalid="this.setCustomValidity('Must be a validate Email Address!')"/> 
       <input type="password" name="Password" placeholder="Password" required maxlength="20" required pattern=".{6,}" oninvalid="this.setCustomValidity('Must be between 6 and 20 characters long!')"/> 
       <input type="Submit" value="Submit" name="Submit" id="Submit"/> 
       </form>

我的繼承人在HTML頂部的jQuery：

<script type="text/javascript" src="js/jquery.js"></script> 
<script type="text/javascript"> 
$(document).ready(function() { 
    //result texts 
    var email_error = 'Emails do not match!'; 
    alert("Hi"); 
    //when button is clicked 
    $('#Submit').click(function(){ 
    //run the character number check 
    if($('#Email').val() != $('#Email2').val()){ 
     //if it's bellow the minimum show characters_error text ' 
     $('#emailValidate').html(email_error); 
    }else{ 
     //else show the cheking_text and run the function to check 
     check_availability(); 
    } 
    }); 

}); 

//function to check username availability 
function check_availability(){ 

    //get the username 
    var Email = $('#Email').val(); 

    //use ajax to run the check 
    $.post("checkmail.php", { Email: Email }, 
     function(result){ 
     //if the result is 1 
     if(result == 1){ 
      //show that the username is available 
      $('#emailValidate').html(Email + ' is Available'); 
     }else{ 
      //show that the username is NOT available 
      $('#emailValidate').html(Email + ' is not Available'); 
     } 
    }); 

} 
</script>

這是我的PHP文件

<?php 
$servername = "localhost"; 
$username = "root"; 
$password = "root"; 
$DB = "database"; 

// Create connection 
$conn = mysqli_connect($servername, $username, $password, $DB); 

// Check connection 
if ($conn->connect_error) { 
    die("Connection failed: " . $conn->connect_error); 
} 

$sql = "SELECT Email FROM Users WHERE Email = '".$_POST['Email']."'"; 
$select = mysqli_query($conn, $sql); 
$row = mysqli_num_rows($select); 

if($row > 0)) { 
    echo 1; 
}else 
echo 0; 
?>

現在，當我點擊提交沒有發生邏輯。由於某種原因，它沒有檢測到提交和不評估。我使用提交作爲ID，它不是拿起它。我錯誤地處理了這個問題嗎？

來源

2015-07-19 user081608

我沒有看到'validateForm（）'函數。你也沒有身份證。 –

由於'Email'是一個文本字段，你必須使用像引號：'WHERE電子郵件=「;'，注意到你的代碼也vurnable SQL注入 – Cyclonecode

$。員額「」 [電子郵件‘$ _ POST’。」「（「checkmail.php」？{電子郵件：電子郵件}，哪裏是電子郵件 – jrath

你可以把它使得SQL數據庫列必須是潮頭。因此，SQL會爲你做，並且你可以提交你的表單。如果存在與提交的電子郵件相匹配的電子郵件，則SQL將發出約束錯誤。

來源

2015-07-19 20:05:58 noahdotgansallo

我已經有了，但我想允許用戶在現場更改他們的電子郵件地址。這種方法可能嗎？ – user081608

@ user081608我會通過JavaScript/jQuery提交表單，這樣如果出現錯誤，您可以提醒用戶並允許他們更改電子郵件。 – noahdotgansallo

在驗證數據庫中值的心不是已經

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