CODEWITHSUNDEEP
javascriptjquery
LetsSeo
LetsSeo

Reputation: 875

jQuery hotkeys works only once

I have 4 radio input.

And I have assigned them hotkeys. (Hotkeys are 1,2,3 and 4)

Here is my html code:

<label>
 <input type="radio" value="2" id="1" name="1">[c2]
</label>
<label>
 <input type="radio" value="4" id="2" name="1">[c4]
</label>
<label>
 <input type="radio" value="1" id="3" name="1">[c1]
</label>
<label>
 <input type="radio" value="3" id="4" name="1">[c3]
</label>

and javascript code:

$(document).keypress(function(e) {

  if(e.charCode == 49) {
    console.log("1");
    $('input:radio[id=1]').attr('checked',true);
  }

  if(e.charCode == 50) {
    console.log("2");
    $('input:radio[id=2]').attr('checked',true);
  }

  if(e.charCode == 51) {
    console.log("3");
    $('input:radio[id=3]').attr('checked',true);
  }

  if(e.charCode == 52) {
    console.log("4");
    $('input:radio[id=4]').attr('checked',true);
  }                  
});

You can also see, I put console.log events to script just to observe.

The strange thing for me, I can select redio buttons by their hotkeys only once per radio button.

*

If I press 1, it chooses the correct checkbox and gives me the correct console log.

Then I press 2, and works the same.

But then if I press 1 again, it gives me the console log but hotkey doesn't work.

Here is jsfiddle: https://jsfiddle.net/f07evno0/

How can I make hotkeys work more than once in one session?

Upvotes: 1

Views: 76

Answers (2)

Zay Lau
Zay Lau

Reputation: 1864

You may have to use 

$("input:radio[id=2]").prop("checked", true)


$(document).keypress(function (e) {

        if (e.charCode == 49) {
            console.log("1");
            $('input:radio[id=1]').prop('checked', true);
        }

        if (e.charCode == 50) {
            console.log("2");
            $('input:radio[id=2]').prop('checked', true);
        }

        if (e.charCode == 51) {
            console.log("3");
            $('input:radio[id=3]').prop('checked', true);
        }

        if (e.charCode == 52) {
            console.log("4");
            $('input:radio[id=4]').prop('checked', true);
        }
    });
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<label>
    <input type="radio" value="2" id="1" name="1">[c2]
</label>
<label>
    <input type="radio" value="4" id="2" name="1">[c4]
</label>
<label>
    <input type="radio" value="1" id="3" name="1">[c1]
</label>
<label>
    <input type="radio" value="3" id="4" name="1">[c3]
</label>

Try it on https://jsfiddle.net/f07evno0/1/

Upvotes: 4

Ying Yi
Ying Yi

Reputation: 782

I dont find out the reason,but this code worked:

$(document).keypress(function(e) {

  if(e.charCode == 49) {
    console.log("1");
    //$('input:radio[id=1]').attr('checked',true);
    $('input:radio[id=1]')[0].checked= "checked"
  }

  if(e.charCode == 50) {
    console.log("2");
    //$('input:radio[id=2]').attr('checked',true);
    $('input:radio[id=2]')[0].checked= "checked"
  }

  if(e.charCode == 51) {
    console.log("3");
    //$('input:radio[id=3]').attr('checked',true);
    $('input:radio[id=3]')[0].checked= "checked"
  }

  if(e.charCode == 52) {
    console.log("4");
    //$('input:radio[id=4]').attr('checked',true);
    $('input:radio[id=4]')[0].checked= "checked"
  }                  
});

Upvotes: 1

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