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I'm writing an app with Phonegap and I have a register form that is sent through ajax.
It works fine when you hit the register button and execute the formcheck() function. However, when I hit the GO button from my android phone it submits the form instead of going through the formcheck() process. I tried:

<form id="RegForm" onsubmit="formcheck();return false;">

My form has no proper submit button but a button like this:

<input type="button" id="submitbtn" onclick="formcheck()"/>

I also tried to create a new OnSubmitForm() function that calls the formcheck() one but with no avail.
Thank you for helping.

jimpanzer
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Vincent
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2 Answers2

3

Found it!

1) Add this to the JS section:

$(document).ready(function() {
    $("#YourFormName").submit(function() {
    FormCheck();
    return false;
    });
});

function FormCheck() {
... validation process here ...
}

2) Make sure to include a Submit button in your form .. ( <input type="submit" )

Hope it'll help others so my 5 hour trying & testing time won't go wasted :)

Vincent
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1

Simply ensure your form tag is:

<form type='submit' onsubmit='return false;'></form>

This makes the submit action not do anything.

If you also need to hide the keyboard refer to How can I hide the Android keyboard using JavaScript?. A simple onsubmit='hideKeyboard(); return false;' will take care of this.

Community
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owencm
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