Question

我有一个表单，在您提交后会显示感谢信息，但是，如果您提交的电子邮件已经存在于数据库中，我希望它提醒“无效电子邮件”。

现在，即使电子邮件已经存在于数据库中，它也只会转到感谢信息（并且不显示警报）。我使用ajax将表单数据发送到数据库。我在代码中做错了什么？

我的ajax代码：

$(document).ready(function(){
$("#submit").click(function(){

var name = $("#fullname2").val();
var email = $("#fullemail2").val();
var state = $("#selectstate").val();
// Returns successful data submission message when the entered information is stored in database.
var dataString = 'FullName='+ name + '&email='+ email + '&SovereignState='+ state;
if(name==''||email==''||state=='')
{
alert("Please Fill All Fields");
}
else
{
// AJAX Code To Submit Form.
$.ajax({
type: "POST",
url: "demo.php",
data: dataString,
cache: false,
success: function( phpSays ){
    if ( phpSays == "OK"){
        alert("invalid email");
    }
    else {
        $('#sinatra2').hide();
        $('#thanks').show();
    }
}
});
}
return false;
});
});

这是php：

<?php

define('DB_NAME', '');
define('DB_USER', '');
define('DB_PASSWORD','');
define('DB_HOST', '');

$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD) ;

if (!$link) {
    die('Could not connect: ' . mysql_error ()) ;
}

$db_selected = mysql_select_db (DB_NAME, $link) ;

if (!$db_selected) {
    die('Cant use ' . DB_NAME . ': ' . mysql_error()) ;
    }

$new_email = $_POST ['email'];

$sql2 = mysql_query ("SELECT email from Data_2 WHERE email = '$new_email");
$row = mysql_fetch_row($sql2);

if(mysql_num_rows($row) > 0) {

echo 'OK' ;

}
else {

$value = $_POST ['FullName'];
$value2 = $_POST ['email'];
$value3 = $_POST ['SovereignState'];

$sql = "INSERT INTO Data_2 (FullName, email, SovereignState) VALUES     ('$value', '$value2', '$value3')";

if (!mysql_query ($sql)) {
    die('Error: ' . mysql_error());
}
}

mysql_close();
?>

Answer 1

试试这个：

$sql2 = mysql_query("SELECT COUNT(`email`) AS count from Data_2 WHERE email = $new_email");

if ($sql2)
    {
    $count = mysql_fetch_assoc($sql2) ['count'];
    if ($count > 0)
        {
        echo 'email exists';
        }
      else
        {
        echo 'email does not exist';
        }
    }

Answer 2

您错过了SQL中的结尾引用。您需要转义电子邮件以防止SQL注入。你使用mysql_num_rows（）错误，它接受资源。以下是修复所有问题的示例：

$result = mysql_query ("SELECT email from Data_2 WHERE email = '".mysql_real_escape_string($new_email)."'");
if (mysql_num_rows($result) > 0) {
    echo 'OK' ;
}

Answer 3

$sql2 = mysql_query ("SELECT email from Data_2 WHERE email = '$new_email");

替换为：

$sql2 = mysql_query ("SELECT email from Data_2 WHERE email = '$new_email' ");

另外

if(mysql_num_rows($row) > 0) {
  echo 'OK' ;

}

替换为：

if(mysql_num_rows($sql2) > 0) {
   echo 'OK' ;
}

Answer 4

将此行var dataString = 'FullName='+ name + '&email='+ email + '&SovereignState='+ state;替换为var dataString = {FullName: name, email: email, SovereignState: state};。可以解决问题的家。

使用Ajax查看数据库中是否已存在电子邮件

4 个答案: