如何在onsubmit函数中使用jquery post方法

Question

如何合并： alidateForm（） - 我用它来验证表单

与

valimail（电子邮件，通行证） - 我用它来检查电子邮件是否已经在数据库中，是否不插入数据

valimail 应该以某种方式返回 true 如果电子邮件地址已经在数据库中并返回 false ，如果没有并将数据插入数据库

HTML：

<form method="POST" action="sin" name="register_form" id="regform" onsubmit="return validateForm();">
<input name="email" id="email" placeholder="E-Mail" autocomplete="off" class="" value='' type="text"><br>
<input name="pass" type="password" id="pass" placeholder="Password" autocomplete="off" value='' class=""><br>
<input name="pass1" type="password" id="pass1" placeholder="Repeat Password" autocomplete="off" value='' class=""><br>

JS：

function validateForm() {
    var email = document.forms["register_form"]["email"].value;
    var pass = document.forms["register_form"]["pass"].value;
    var pass1 = document.forms["register_form"]["pass1"].value;
    var atpos = email.indexOf("@");
    var dotpos = email.lastIndexOf(".");
    if (atpos< 1 || dotpos<atpos+2 || dotpos+2>=email.length) {
        alert("This is not email");
        return false;
    }else{
    if(pass.length<5||pass1.length<5 ){
        alert("Password too short");
        return false;
    }else{
        if (pass!=pass1){
            alert("Passwords are not the same");
            return false;
        }else{
            if (**here function valimail**){return false;}
        }
    }
    }
}
function valimail(email,pass) {
            $.post('checkemail.php', {'email' : email,'pass' : pass}, function(data) {
             if(data=='true'){
                alert("This email is already in use");
             }
             });
}

PHP：

$con=mysqli_connect("xxx.xxx.xxx.xxx","login","pass","db");

$email = mysqli_real_escape_string($con, $_POST['email']);
$haslo = mysqli_real_escape_string($con, $_POST['pass']);

$query = mysqli_query($con, "SELECT * FROM users WHERE email='".$email."'");

if(mysqli_num_rows($query) > 0){
    echo 'true';
}else{
    $sql="INSERT INTO users (email, password) VALUES ('$email', '$pass')";
    if (!mysqli_query($con,$sql)) {
    die('Error: ' . mysqli_error($con));
    }
}

mysqli_close($con);