I have a web form to let user enter some information and submit. I want such behavior that if the submission is successful, the it will go to a page. But if there are some form submission errors (in my case is due to existing ID), I want the form stay there without going to the next page and alert the user "ID already exists". I used jQuery to handle the success and fail events:
function checkForm() {
$.post("/myapp/rest/customer/created", $("form").serialize(), function(
data, status) { })
.done(function() {alert("post success");})
.fail(function() {alert("post error");});
return false;
}
My form is:
<form action="/myapp/rest/customer/created"
onsubmit="return checkForm();" method="POST">
<table border="1">
<tr>
<td>Customer name:</td>
<td><input type="text" id="name" name="name"></td>
</tr>
<tr>
<td>Customer ID:</td>
<td><input type="text" id="id" name="id"></td>
</tr>
<tr>
<td>Customer DOB:</td>
<td><input type="text" id="dob" name="dob"></td>
</tr>
</table>
<br /> <input type="submit" value="Submit">
</form>
After submission, no matter whether the submission is a success or failure, I only see the popup message box saying "post success" or "post error". But it will never go to the next page if it's successful. I even tried a different approach like the code below but it didn't work either:
function checkForm() {
$.post("/myapp/rest/customer/created", function(data, status) {
if (status === "success") {
alert("post success");
} else {
alert("post error");
}
});
return false;
}
So how to correct it to make it redirect to the next page (/myapp/rest/customer/created) if the submission is successful?
