我正在建立一个帐户注册页面,检查用户的电子邮件地址与数据库,以确保所说的用户没有创建重复的帐户。使用ajax报告数据库中是否存在电子邮件。当在表格中输入电子邮件地址时,我总是会得到“电子邮件是ok的”。这意味着它不存在于数据库中。然而,这是不正确的。如果我在php解析器中将电子邮件指定为变量,比如$email="email@exists.com“,那么它实际上会返回正确的结果。我猜想解析器没有从表单中获得值来添加到查询中。你知道为什么这不管用吗?
<input type="text" id="password" class="form-control" name="password" value="" pattern="(?=.*\d)(?=.*[a-z])(?=.*[A-Z]).{8,}" placeholder="password" autocomplete="off" required />
Ajax:
<script>
function ajax_post(){
// Create our XMLHttpRequest object
var hr = new XMLHttpRequest();
// Create some variables we need to send to our PHP file
var url = "email_check.php";
var fn = document.getElementById("firstname").value;
var ln = document.getElementById("lastname").value;
var e = document.getElementById("email").value;
var pwd = document.getElementById("password").value;
var vars = "firstname="+fn+"&lastname="+ln;
hr.open("POST", url, true);
// Set content type header information for sending url encoded variables in the request
hr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
// Access the onreadystatechange event for the XMLHttpRequest object
hr.onreadystatechange = function() {
if(hr.readyState == 4 && hr.status == 200) {
var return_data = hr.responseText;
document.getElementById("status").innerHTML = return_data;
}
}
// Send the data to PHP now... and wait for response to update the status div
hr.send(vars); // Actually execute the request
document.getElementById("status").innerHTML = "processing...";
}
</script>
和解析器:
<?php
include 'db.php';
$test= $_POST['email'];
$sql="SELECT email FROM users where email = '$test' LIMIT 1";
$result = mysqli_query($conn,$sql);
if(mysqli_num_rows($result) > 0)
{
echo 'email is in use.';
exit();
} else if(mysqli_num_rows($result) < 1){
echo 'email is ok';
exit();
}
?>
发布于 2018-06-09 06:31:26
在发送请求时,您没有将电子邮件字段添加到您的vars中,如下所示:
var vars = "firstname="+fn+"&lastname="+ln;
发布于 2018-06-09 09:53:00
您需要向解析器发送电子邮件,而不是名字/姓氏
var vars = "email="+e;
https://stackoverflow.com/questions/50769143
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