how to differenciate between onchange submit and button submit in spring controller

Question

I have the jsp code as :

    <s:url value="add" var="addUrl"/>
    <form:form name="add" action="${addUrl}" modelAttribute="content" method="post">
        <input type="hidden" name="add" value="" />
        <form:select path="id" class="dClass" onchange="submit()">
        <form:options items="${list}" /></form:select>
        <input type="submit" class="btn" value="Refresh Content" />
   </form:form>

And I have one controller method for both this call, is there any way to differentiate that from where the form is submitted? Either from the drop down or from the submit button?

@RequestMapping(value = "/add", method = RequestMethod.POST)
    public String getContentById(Model model, String id,Content content,HttpServletRequest request) {

if(call by onChange)

else if(call by submit button)

}

you have to make some unique sign to differentiate them, like add extra post param to `onchange()` — farmer1992, Nov 06 '13 at 08:22
how to add post param here? as i cant call the submit from javascript. because my controller returns a tiles name to render, and i dont have the ajax based view resolver for tiles in my project — Arpan Das, Nov 06 '13 at 11:47
I mean change js of `submit()`. add something in the function. — farmer1992, Nov 06 '13 at 11:50
but i dont have any submit() function in my .js file, its calling the default sumbit() method. — Arpan Das, Nov 06 '13 at 11:58

score 1 · Accepted Answer · edited May 23 '17 at 12:03

1

Kindly note that, onchange="submit()" will not submit your page. It will look for the function with name submit. Inside that function you can use code like this $('form#form_id').submit(); to submit. Check submit-a-form-using-jquery for more details.

edited May 23 '17 at 12:03

Community

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1

answered Nov 06 '13 at 12:12

Vinoth Krishnan

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score 0 · Answer 2 · answered Nov 06 '13 at 12:45

0

@Vinoth Krishnan Thanks for your answer , its working now :) added a filed in my form as "dropDownSubmit"

function submitForm(formId){
        var input = $("<input>").attr("type", "hidden").attr("name", "dropDownSubmit").val("true");
        $("#" + formId).append($(input));
        $("#" + formId).submit();
    }

Then in controller:

if("true".equalsIgnoreCase(form.getDropDownSubmit()){
 ..
}
else{
 .. button submit
}

answered Nov 06 '13 at 12:45

Arpan Das

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You are welcome. Mark it as answer. So others will get the benefits from that. – Vinoth Krishnan Nov 06 '13 at 14:22

how to differenciate between onchange submit and button submit in spring controller

2 Answers2