我在允许提交ajax之前尝试验证表单,但是我有两个问题。第一个问题是我不知道在提交之前如何最好地进行验证(最专业的流程)。第二个问题是,什么阻止了我目前工作的验证码?希望提高效率,欢迎所有投入。非常感谢。
$('#form-reg').on('submit', function(){
var bool = false;
var name = document.getElementById('#name-reg');
var email = document.getElementById('#email-reg');
console.log(name);
console.log(email);
if(!/[^a-zA-Z]/.test(name)){
bool = true;
}
else{
bool = false;
}
if(bool == true){
console.log(document.getElementById('#name-reg'));
$('#form-reg').slideUp('slow');
// serialize the form
var formData = $(this).serialize();
console.log(formData);
$.ajax({
type : 'POST',
url : 'register.php',
data : formData,
success: function() {
alert("Success");
},
error: function(xhr) {
alert("fail");
}
})
.done(function (data) {
document.getElementById('form-reg').reset();
})
.fail(function (error) {
alert("POST failed");
});
//return false;
}
else {
alert('try again');
}
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js" type="text/javascript">
<form id = "form-reg">
<label id ="x" for="name">Name</label>
<input id="name-reg" name="name"></br>
<label id = "y" for="email">Email</label>
<input id="email-reg" name="email"></br>
<input type="submit" value="submit" id = "submit-reg">
</form>
&#13;
答案 0 :(得分:2)
您的代码存在问题,您必须停止提交表单,如:
for i in $(seq 0 0.05 1); do awk -v i=$i '{if ($7 <= i) print $0}' file.txt | wc -l ; done
并在检查验证后,如果验证成功,则提交表单。
答案 1 :(得分:0)
虽然这是一个小答案,但无论如何我都会发布。问题看起来像是
的使用var name = document.getElementById('#name-reg');
#导致第一个问题,应该是
var name = $('#name-reg')。val();
$('#form-reg').on('submit', function(){
var bool = false;
var name = $('#name-reg').val();
var email = $('#email-reg').val();
console.log(name);
console.log(email);
if(!/[^a-zA-Z]/.test(name)){
bool = true;
}
else{
bool = false;
}
if(bool == true){
console.log($('#name-reg').val());
$('#form-reg').slideUp('slow');
// serialize the form
var formData = $(this).serialize();
console.log(formData);
$.ajax({
type : 'POST',
url : 'register.php',
data : formData,
success: function() {
alert("Success");
},
error: function(xhr) {
alert("fail");
}
})
.done(function (data) {
document.getElementById('form-reg').reset();
})
.fail(function (error) {
alert("POST failed");
});
//return false;
}
else {
alert('try again');
}
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js" type="text/javascript">
<form id = "form-reg">
<label id ="x" for="name">Name</label>
<input id="name-reg" name="name"></br>
<label id = "y" for="email">Email</label>
<input id="email-reg" name="email"></br>
<input type="submit" value="submit" id = "submit-reg">
</form>