Invoke form action method and then disable the button in JSP

Question

I have a JSP page where I have created a form and a button. Once user clicks on submit button, form action will call servlet doPost method and does the business logic. Now I need to disable the submit button to prevent user to click on that button twice till my business execution completes. My JSP pages is as shown below

<body>

        <form method="post" action="SubmitJob" id="myForm">
        <table>


                        <td><p>Please use the below button to run the job immediately</p> <br>
                            <div id="runNow">
                            <input type="submit" name="runNow" value="runNow" id="submitBtn">

                            <% out.println(request.getAttribute("status")); %>
                            </div>
                        </td>
                        </tr>
        </table>
    </form>

    </body>

To disable a submit button, I have used jquery and tried as follows

<script>
$(document).ready(function(){
    $('#submitBtn').click(function(){
        // $('form#myForm').submit();
        alert('after form submit');
                                        $('#submitBtn').prop("disabled",true); 
                                         alert('after button disabled');
    });
});
</script>

problem I am facing is, form action will not be invoked if button is disabled. Can some one suggest me, 1)how to invoke form action first then disable the button. 2) how to enable the submit button again once business logic in my servlet completes.3) How to call servlet doPost method from my java method

If you want to use `jquery` you do not need the button to be a submit button, a normal one is fine. — Scary Wombat, Jul 12 '17 at 06:49
Hi, I am using html form post method to submit the form. To disable the button, I am using Jquery — user8201218, Jul 12 '17 at 06:53

score 0 · Answer 1 · edited Jul 12 '17 at 06:46

0

1) You can call your form in another function like $("#myForm").submit();

2) Put this code in the success function in your server call:

$('#submitBtn').prop("disabled",true);

3) You can use the code below:

$.ajax({
            type: "POST",
            url: "",
            contentType: "application/json; charset=utf-8",
            success: function (response) {   
        }
    });

edited Jul 12 '17 at 06:46

Scary Wombat

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answered Jul 12 '17 at 06:38

solamente

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my bad was POST :) – solamente Jul 12 '17 at 06:45

score 0 · Answer 2 · answered Jul 12 '17 at 06:44

In your onclick function,you missed the key step :

$("#myForm").submit();

The premise is that the action attribute of your form label should define the url. Sample as below:

<form action="your url...." method="post" id="myForm">
</form>

In addition，your code

$('#submitBtn').prop("disabled",true);

already make sure that the button can't be clicked twice.

Invoke form action method and then disable the button in JSP

2 Answers2