bootstrap tooltip is not working with my jquery function

Question

If i removed my jquery link tooltip hover css background working fine but my jquery function not working ( obviously because of jquery link was removed) please look my code.... thanks `

  <div class="col-sm-6">
    <input type="text" id="id_part_pay"  value="<?php echo $listing['part_pay'];?>"   class="textbox" name="id_part_pay" <?php if($checked)echo 'style="display: block"'; else echo 'style="display: none"';?> />
    <a href="" class="fa fa-question-circle" data-toggle="tooltip" data-placement="right" title="Hooray!"></a>
    <span class="ft-s12"></span>
  </div>  
  <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js">    
    $(function() {
        $("#id_part_pay").next('span').hide(); 
        $("#id_part_pay").keyup(function() {
        var input = $(this).val();
        var v = input % 10;
        var span = $(this).next('span'); 
        if (v !== 0) {
        span.text("Enter Percentage in multiple of 10").show(); 
        return;
        }
        if (input < 20 || input > 100) {
         span.text("Percentage should be between 20 - 100").show();
        return;
        }
        span.text('').hide();//Clear Text and hide
       });
    });
  </script>

Arun P Johny · Accepted Answer · 2015-10-05 05:54:09Z

You can't have src and content for the same script tag

src
This attribute specifies the URI of an external script; this can be used as an alternative to embedding a script directly within a document. If a script element has a src attribute specified, it should not have a script embedded inside its tags.

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script>
    $(function () {
        $("#id_part_pay").next('span').hide();
        $("#id_part_pay").keyup(function () {
            var input = $(this).val();
            var v = input % 10;
            var span = $(this).next('span');
            if (v !== 0) {
                span.text("Enter Percentage in multiple of 10").show();
                return;
            }
            if (input < 20 || input > 100) {
                span.text("Percentage should be between 20 - 100").show();
                return;
            }
            span.text('').hide(); //Clear Text and hide
        });
    });
</script>

Shehary · Accepted Answer · 2015-10-05 05:54:40Z

You can't include <script> the way you are trying

first close </script> jQuery lib script

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

and then write custom script

<script>
$(function () {
    $("#id_part_pay").next('span').hide();
    $("#id_part_pay").keyup(function () {
        var input = $(this).val();
        var v = input % 10;
        var span = $(this).next('span');
        if (v !== 0) {
            span.text("Enter Percentage in multiple of 10").show();
            return;
        }
        if (input < 20 || input > 100) {
            span.text("Percentage should be between 20 - 100").show();
            return;
        }
        span.text('').hide(); //Clear Text and hide
    });
});
</script>

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