如果表单中输入的电子邮件地址已存在，则显示错误

Question

如何使用“emailaddress”作为唯一提供错误的重复条目？

我发现很多答案都使用mysql_query，但我想使用mysqli。

<?php
$con=mysqli_connect("localhost","","","");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

$sql="INSERT INTO entry (firstname, lastname, emailaddress, favoritesong) VALUES ('$_POST[firstname]','$_POST[lastname]','$_POST[emailaddress]','$_POST[favoritesong]')";

if (!mysqli_query($con,$sql))
  {
  die('Error: ' . mysqli_error($con));
  }
echo "1 record added";

mysqli_close($con);
?>

Answer 1

警告！已发布的代码和此答案（因为我现在只解决问题）包含大量SQL注入泄漏。请阅读SQL注入并使用转义或准备好的语句。

<?php

$con = mysqli_connect("localhost", "", "", "");
// Check connection
if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
    exit();
}
$existsQuery = "select count(*) as count from entry where emailaddress like '".$_POST[emailaddress]."'";
$existsResult = mysqli_query($con, $existsQuery);

if($existsResult->fetch_object()->count > 0)
{
    echo "email already exist";
}
else
{
    $sql = "INSERT INTO entry (firstname, lastname, emailaddress, favoritesong) VALUES ('$_POST[firstname]','$_POST[lastname]','$_POST[emailaddress]','$_POST[favoritesong]')";

    if (!mysqli_query($con, $sql)) {
        die('Error: ' . mysqli_error($con));
    }
    echo "1 record added";
}

mysqli_close($con);
?>

Answer 2

只需在表格中设置emailaddress unique即可。