Select tag is inserting duplicate rows in sql
If click function is clicked for first time there are no duplicate rows
When it is clicked for second time 2 rows get inserted
for third time 3 rows and so on..
html
<select class="form-control" name="select" id="select">
<option value="pick">Select</option>
<option name="id" id="id" value="1">Name1</option>
<option name="id2" id="id2" value="2">Name2</option>
</select>
jquery..
$('#select').on('change', function (j){
$("#btn").click(function(e){
var selectedValue = $('#select').val();
$.ajax({
type: "POST",
url: "new.php",
data: '{"to":"'+selectedValue+'"}',
success: function(r){
}
});
});
});
Why is this happening..
jquery ajax
asked Nov 19 '18 at 8:42
SachinSachin
87
add a comment |
If click function is clicked for first time there are no duplicate rows
When it is clicked for second time 2 rows get inserted
for third time 3 rows and so on..
html
<select class="form-control" name="select" id="select">
<option value="pick">Select</option>
<option name="id" id="id" value="1">Name1</option>
<option name="id2" id="id2" value="2">Name2</option>
</select>
jquery..
$('#select').on('change', function (j){
$("#btn").click(function(e){
var selectedValue = $('#select').val();
$.ajax({
type: "POST",
url: "new.php",
data: '{"to":"'+selectedValue+'"}',
success: function(r){
}
});
});
});
Why is this happening..
jquery ajax
asked Nov 19 '18 at 8:42
SachinSachin
87
add a comment |
If click function is clicked for first time there are no duplicate rows
When it is clicked for second time 2 rows get inserted
for third time 3 rows and so on..
html
<select class="form-control" name="select" id="select">
<option value="pick">Select</option>
<option name="id" id="id" value="1">Name1</option>
<option name="id2" id="id2" value="2">Name2</option>
</select>
jquery..
$('#select').on('change', function (j){
$("#btn").click(function(e){
var selectedValue = $('#select').val();
$.ajax({
type: "POST",
url: "new.php",
data: '{"to":"'+selectedValue+'"}',
success: function(r){
}
});
});
});
Why is this happening..
jquery ajax
asked Nov 19 '18 at 8:42
SachinSachin
87
If click function is clicked for first time there are no duplicate rows
When it is clicked for second time 2 rows get inserted
for third time 3 rows and so on..
html
<select class="form-control" name="select" id="select">
<option value="pick">Select</option>
<option name="id" id="id" value="1">Name1</option>
<option name="id2" id="id2" value="2">Name2</option>
</select>
jquery..
$('#select').on('change', function (j){
$("#btn").click(function(e){
var selectedValue = $('#select').val();
$.ajax({
type: "POST",
url: "new.php",
data: '{"to":"'+selectedValue+'"}',
success: function(r){
}
});
});
});
Why is this happening..
jquery ajax
jquery ajax
asked Nov 19 '18 at 8:42
SachinSachin
87
asked Nov 19 '18 at 8:42
SachinSachin
87
asked Nov 19 '18 at 8:42
SachinSachin
87
asked Nov 19 '18 at 8:42
SachinSachin
87
asked Nov 19 '18 at 8:42
SachinSachin
87
87
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
you should unbind any existing event first
$("#btn").off().click(function(){
// ...
})
answered Nov 19 '18 at 8:46
Darryl CeguerraDarryl Ceguerra
885
add a comment |
You are assigning the button click handler inside the change handler. So you end up adding an extra handler every time.
You can get rid of the select change handler.
$("#btn").click(function(e){
var selectedValue = $('#select').val();
$.ajax({
type: "POST",
url: "new.php",
data: '{"to":"'+selectedValue+'"}',
success: function(r){
}
});
});
This much is sufficient.
answered Nov 19 '18 at 8:47
RoshanRoshan
1,7491921
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
you should unbind any existing event first
$("#btn").off().click(function(){
// ...
})
answered Nov 19 '18 at 8:46
Darryl CeguerraDarryl Ceguerra
885
add a comment |
you should unbind any existing event first
$("#btn").off().click(function(){
// ...
})
answered Nov 19 '18 at 8:46
Darryl CeguerraDarryl Ceguerra
885
add a comment |
you should unbind any existing event first
$("#btn").off().click(function(){
// ...
})
answered Nov 19 '18 at 8:46
Darryl CeguerraDarryl Ceguerra
885
you should unbind any existing event first
$("#btn").off().click(function(){
// ...
})
answered Nov 19 '18 at 8:46
Darryl CeguerraDarryl Ceguerra
885
answered Nov 19 '18 at 8:46
Darryl CeguerraDarryl Ceguerra
885
answered Nov 19 '18 at 8:46
Darryl CeguerraDarryl Ceguerra
885
answered Nov 19 '18 at 8:46
Darryl CeguerraDarryl Ceguerra
885
885
add a comment |
add a comment |
You are assigning the button click handler inside the change handler. So you end up adding an extra handler every time.
You can get rid of the select change handler.
$("#btn").click(function(e){
var selectedValue = $('#select').val();
$.ajax({
type: "POST",
url: "new.php",
data: '{"to":"'+selectedValue+'"}',
success: function(r){
}
});
});
This much is sufficient.
answered Nov 19 '18 at 8:47
RoshanRoshan
1,7491921
add a comment |
You are assigning the button click handler inside the change handler. So you end up adding an extra handler every time.
You can get rid of the select change handler.
$("#btn").click(function(e){
var selectedValue = $('#select').val();
$.ajax({
type: "POST",
url: "new.php",
data: '{"to":"'+selectedValue+'"}',
success: function(r){
}
});
});
This much is sufficient.
answered Nov 19 '18 at 8:47
RoshanRoshan
1,7491921
add a comment |
You are assigning the button click handler inside the change handler. So you end up adding an extra handler every time.
You can get rid of the select change handler.
$("#btn").click(function(e){
var selectedValue = $('#select').val();
$.ajax({
type: "POST",
url: "new.php",
data: '{"to":"'+selectedValue+'"}',
success: function(r){
}
});
});
This much is sufficient.
answered Nov 19 '18 at 8:47
RoshanRoshan
1,7491921
You are assigning the button click handler inside the change handler. So you end up adding an extra handler every time.
You can get rid of the select change handler.
$("#btn").click(function(e){
var selectedValue = $('#select').val();
$.ajax({
type: "POST",
url: "new.php",
data: '{"to":"'+selectedValue+'"}',
success: function(r){
}
});
});
This much is sufficient.
answered Nov 19 '18 at 8:47
RoshanRoshan
1,7491921
answered Nov 19 '18 at 8:47
RoshanRoshan
1,7491921
answered Nov 19 '18 at 8:47
RoshanRoshan
1,7491921
answered Nov 19 '18 at 8:47
RoshanRoshan
1,7491921
1,7491921
add a comment |
add a comment |
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