这是我的表格:
<!DOCTYPE html>
<html>
<head>
</head>
<body>
<form action="register_ajax.php" method="get">
<input type="text" name="email">
<input type="submit" value="test">
</form>
</body>
</html>
这是我的php代码:
<?php
$dbc = mysqli_connect("localhost","root","*******","continental_tourism") OR die(mysqli_connect_error());
$email = $_GET['email'];
$query = "SELECT email FROM customer_info WHERE email = '$email' ";
$r = mysqli_query($dbc, $query) OR die(mysqli_error($dbc));
if($r)
echo "Email address exists!";
else
echo "sss";
?>
如果我输入正确的(db 上的现有电子邮件)$r 为真.但是如果我输入不存在的电子邮件，那么 $r 也是如此.这是为什么?基本上我想检测空集.我该怎么做?
If I enter a correct(Existing email on db) $r is true. But if I enter non existing email, then also $r is true. Why is that? Basically I want to detect the empty set. How can I do it?
谢谢！
$r 只有在出现 SQL 错误时才会为假.否则它将始终返回一个对象，即使您的 SELECT 语句没有返回任何行.
$r will only be false if there was SQL error. Otherwise it will always return an object, even if no rows are returned by your SELECT statement.
使用mysqli_num_rows() 计算返回的行数.零表示没有人使用该电子邮件地址.
Use mysqli_num_rows() to count how many rows are returned. Zero means no one is using that email address.
if(mysqli_num_rows($r))
echo "Email address exists!";
else
echo "sss";
这篇关于mysqli_query() 总是返回真的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！