Ajax，JS函数返回未定义的值

Ajax, JS function returning undefined value

本文关键字：未定义 返回 函数 JS Ajax      更新时间：2023-09-26

我对这一切都很陌生，我正在学习Ajax和Javascript，我从php文件中获取值，但是当我尝试返回方法中的值时，我得到了一个未定义的值，我登录控制台，我尝试了很多事情但没有成功。有人可以教育我吗？请批评我的代码，我确定有很多不好的做法。

谢谢 

 function checkEmail(){
    var xhttp; 
    var status;
    xhttp = new XMLHttpRequest();
    var email = document.getElementById('email2').value;
    
    xhttp.onreadystatechange = function() {
      if (xhttp.readyState == 4 && xhttp.status == 200) {
            var xmlResponse = xhttp.responseText;      
            status = xmlResponse;
            
        }
    };
    xhttp.open("GET", "php/ajaxCom.php?email="+email, true);
    xhttp.send();
    return status;
 }

<?php
	include 'base.php';
	$status;
	$email_in_use = $_GET['email'];
	$query = mysqli_query($link, "SELECT * FROM users WHERE email='".$email_in_use."'");
	if(mysqli_num_rows($query) > 0){
	    $status = "false";
	}else{
		$status = "true";
	}
	echo $status;
?>

这就是我称之为检查电子邮件的地方 

function getStatus(field, name, value) {
	    var status = null;
	    if (!field.attr('required')) return null;
	    if (!value) status = 'Please fill out the required field: ' + name;
	    else if (emailField.test(name) || emailField.test(field.attr('type'))) {
	       var b = checkEmail();
	       console.log(b);
	      if (!emailValue.test(value)) {
	      	status = 'Please enter a valid email address for: ' + name;
	      	}else if ( b == "false"){
	      		alert("im here");
	    	 status = 'Please enter a valid email this email already has an acount';
	    	}
	    }
	    return status;
	  }

更新

统计信息警报显示正确，但控制台日志仍显示未定义 

function getStatus(field, name, value) {
	    var status = null;
	    if (!field.attr('required')) return null;
	    if (!value) status = 'Please fill out the required field: ' + name;
	    else if (emailField.test(name) || emailField.test(field.attr('type'))) {
	    	var b;
	    	
	       checkEmail(function(status) {
    			alert('Status: ' + status);
    			b = status;
			});
	       console.log(b);
	      if (!emailValue.test(value)) {
	      	status = 'Please enter a valid email address for: ' + name;
	      	}else if ( b == "false" || b == "true"){
	      		alert("im here");
	    	 status = 'Please enter a valid email this email already has an acount';
	    	}
	    }
	    return status;
	  }

您将立即返回状态，而实际的 AJAX 调用是异步返回的。

您需要更改流，以便 checkEmail 接受回调函数作为参数，而不是返回它。像这样：

function checkEmail(callback) {
    xhttp.onreadystatechange = function() {
        if (xhttp.readyState == 4 && xhttp.status == 200) {
        var xmlResponse = xhttp.responseText;      
        status = xmlResponse;
        callback(status);
    }
};
...
checkEmail(function(status) {
    alert('Status: ' + status);
});
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