【发布时间】:2014-11-13 05:37:10
【问题描述】:
我有以下代码,效果很好。如果他输入的电子邮件地址已经存在于数据库中,我希望用户收到一条错误消息! 谢谢!
$con=mysqli_connect("xxxx.com","xxxx_xxx","xxxx","yyyyy_");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$firstname = mysqli_real_escape_string($con, $_POST['firstname']);
$lastname = mysqli_real_escape_string($con, $_POST['lastname']);
$password = mysqli_real_escape_string($con, $_POST['password']);
$email = mysqli_real_escape_string($con, $_POST['email']);
$address = mysqli_real_escape_string($con, $_POST['address']);
$postcode = mysqli_real_escape_string($con, $_POST['postcode']);
$country = mysqli_real_escape_string($con, $_POST['country']);
$phonenumber = mysqli_real_escape_string($con, $_POST['phonenumber']);
$rating = mysqli_real_escape_string($con, $_POST['rating']);
$sql="INSERT INTO customers (firstname, lastname, password, email, address, postcode, country, phonenumber, rating)
VALUES ('$firstname', '$lastname', '$password', '$email', '$address', '$postcode', '$country', '$phonenumber', '$rating')";
if (!mysqli_query($con,$sql)) {
die('Error: ' . mysqli_error($con));
}
echo "1 record added";
mysqli_close($con);
【问题讨论】:
-
你在问之前没有谷歌这个吗?使用
mysqli_num_rows() -
snicker 你太可笑了@Fred-ii- Google 太难了。
-
@JayBlanchard 那不是巧克力棒吗?现在你已经完成了,我现在有零食了。 士力架
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我曾经在巧克力吧点了一杯饮料...@Fred-ii-
-
@MixalisPapoulakis:*咳嗽* 独特的*咳嗽*。
标签: php mysql mysqli duplicates