Solved

Radio button onChange not working when trying to submit form

Posted on 2010-11-19
8
857 Views
Last Modified: 2012-05-10
Hi,

The scenerio is as follows:

The form is loaded with radio buttons prepopulated.
I need to the form to submit if the user selects a new option.

Example code:

<form action="submit.php" method="post">
<input type="radio" name="mybutton" value="0" onChange="this.form.submit();" />
<input type="radio" name="mybutton" value="1" onChange="this.form.submit();" />
<input type="radio" name="mybutton" value="2" onChange="this.form.submit();" />
</form>

What have I done wrong?

Thanks
0
Comment
Question by:jagku
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8 Comments
 
LVL 13

Expert Comment

by:dsmile
ID: 34174811
Use onclick instead

<form action="submit.php" method="post">
<input type="radio" name="mybutton" value="0" onclick="this.form.submit();" />
<input type="radio" name="mybutton" value="1" onclick="this.form.submit();" />
<input type="radio" name="mybutton" value="2" onclick="this.form.submit();" />
</form>
0
 
LVL 52

Expert Comment

by:Ryan Chong
ID: 34174813
try use:

onClick="this.form.submit();"

instead?
0
 

Author Comment

by:jagku
ID: 34187524
Hi Guys,

I did try using onClick (actually, I should have mentioned that - sorry!).
However, the problem was that this was firing everytime the user clicked the radio button (even if it was already checked).

Thanks
Dipen
0
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LVL 52

Expert Comment

by:Ryan Chong
ID: 34187668
Try:

<form action="submit.php" method="post">
<input type="radio" name="mybutton" value="0" onClick="if ((!this.form.mybutton[0].checked)&&(!this.form.mybutton[1].checked)&&(!this.form.mybutton[1].checked)) this.form.submit();" />
<input type="radio" name="mybutton" value="1" onClick="if ((!this.form.mybutton[0].checked)&&(!this.form.mybutton[1].checked)&&(!this.form.mybutton[1].checked)) this.form.submit();" />
<input type="radio" name="mybutton" value="2" onClick="if ((!this.form.mybutton[0].checked)&&(!this.form.mybutton[1].checked)&&(!this.form.mybutton[1].checked)) this.form.submit();" />
</form>
0
 
LVL 52

Expert Comment

by:Ryan Chong
ID: 34187672
Try:

<form action="submit.php" method="post">
<input type="radio" name="mybutton" value="0" onClick="if ((!this.form.mybutton[0].checked)&&(!this.form.mybutton[1].checked)&&(!this.form.mybutton[2].checked)) this.form.submit();" />
<input type="radio" name="mybutton" value="1" onClick="if ((!this.form.mybutton[0].checked)&&(!this.form.mybutton[1].checked)&&(!this.form.mybutton[2].checked)) this.form.submit();" />
<input type="radio" name="mybutton" value="2" onClick="if ((!this.form.mybutton[0].checked)&&(!this.form.mybutton[1].checked)&&(!this.form.mybutton[2].checked)) this.form.submit();" />
</form>
0
 

Author Comment

by:jagku
ID: 34187835
Hi Ryancys,

Will the form submit if value = 0 is checked when we reach this screen and click on any of the buttons?

What I need (and I am starting to think that it isn't possible as the onClick event appears to fire after the radio button is changed (not before)) is to make sure that the form isn't submitted if (and only if) the radio button hasn't been changed.  I only want it to submit if there has been a radio button change.

0
 
LVL 52

Accepted Solution

by:
Ryan Chong earned 500 total points
ID: 34188229
you can try something like this instead:

<form action="submit.php" method="post">
<input type="radio" name="mybutton" value="0" onClick="if (x!=0) {this.form.submit();} else {x=0;}" />
<input type="radio" name="mybutton" value="1" onClick="if (x!=1) {this.form.submit();} else {x=1;}" />
<input type="radio" name="mybutton" value="2" onClick="if (x!=2) {this.form.submit();} else {x=2;}" />
</form>
<script language="javascript">
      var frm = document.forms[0];
      var x = -1;
      if (frm.mybutton[0].checked) x = 0;
      if (frm.mybutton[1].checked) x = 1;
      if (frm.mybutton[2].checked) x = 2;
</script>
0
 

Author Closing Comment

by:jagku
ID: 34189618
thanks
0

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