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Ajax Returning to text, not html
How can I get jQuery to perform a synchronous, rather than asynchronous, Ajax request?How to manage a redirect request after a jQuery Ajax callAbort Ajax requests using jQueryevent.preventDefault() vs. return falseGet selected text from a drop-down list (select box) using jQueryjQuery AJAX submit formjQuery Ajax File UploadAjax request returns 200 OK, but an error event is fired instead of successIs Safari on iOS 6 caching $.ajax results?How do I return the response from an asynchronous call?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;
I have a complete code for getting the values from PHP through Jquery AJAX with JSON datatype.
But after ajax done, it doesn't response to html. Its become text.
the result become like this.
$("#category li").one('click', function()
var catId = $(this).attr("id");
var that = this;
$.ajax(
type: 'GET',
url: '<?php echo base_url()."dashboard/getsubcategory/"; ?>'+catId,
dataType: 'html',
data: category: catId,
beforeSend: function()
$("#loader").show();
)
.done(function(data)
var categories = JSON.parse(data)
$.each(categories, function(k,category)
that.after('<li class="collection-item">'+category.id+'</li>');
);
)
.fail(function()
alert( "Fetch failed." );
);
);
JSON output :
[
"id": "7",
"name": "Blazer & Suits"
,
"id": "8",
"name": "Blouses & Shirts"
,
"id": "9",
"name": "Friendly URL in CodeIgniter"
]
Why it has returning to :
<li class="collection-item">7</li>
<li class="collection-item">8</li>
<li class="collection-item">9</li>
output that I expected is
<li class="collection-item">7</li>
<li class="collection-item">8</li>
<li class="collection-item">9</li>
preview :
http://prntscr.com/n1x0ey
Please help ..
Thank a lot.
jquery ajax
add a comment |
I have a complete code for getting the values from PHP through Jquery AJAX with JSON datatype.
But after ajax done, it doesn't response to html. Its become text.
the result become like this.
$("#category li").one('click', function()
var catId = $(this).attr("id");
var that = this;
$.ajax(
type: 'GET',
url: '<?php echo base_url()."dashboard/getsubcategory/"; ?>'+catId,
dataType: 'html',
data: category: catId,
beforeSend: function()
$("#loader").show();
)
.done(function(data)
var categories = JSON.parse(data)
$.each(categories, function(k,category)
that.after('<li class="collection-item">'+category.id+'</li>');
);
)
.fail(function()
alert( "Fetch failed." );
);
);
JSON output :
[
"id": "7",
"name": "Blazer & Suits"
,
"id": "8",
"name": "Blouses & Shirts"
,
"id": "9",
"name": "Friendly URL in CodeIgniter"
]
Why it has returning to :
<li class="collection-item">7</li>
<li class="collection-item">8</li>
<li class="collection-item">9</li>
output that I expected is
<li class="collection-item">7</li>
<li class="collection-item">8</li>
<li class="collection-item">9</li>
preview :
http://prntscr.com/n1x0ey
Please help ..
Thank a lot.
jquery ajax
1
please post the code. otherwise, nobody can help.
– Ashkan Mobayen Khiabani
Mar 23 at 16:36
add a comment |
I have a complete code for getting the values from PHP through Jquery AJAX with JSON datatype.
But after ajax done, it doesn't response to html. Its become text.
the result become like this.
$("#category li").one('click', function()
var catId = $(this).attr("id");
var that = this;
$.ajax(
type: 'GET',
url: '<?php echo base_url()."dashboard/getsubcategory/"; ?>'+catId,
dataType: 'html',
data: category: catId,
beforeSend: function()
$("#loader").show();
)
.done(function(data)
var categories = JSON.parse(data)
$.each(categories, function(k,category)
that.after('<li class="collection-item">'+category.id+'</li>');
);
)
.fail(function()
alert( "Fetch failed." );
);
);
JSON output :
[
"id": "7",
"name": "Blazer & Suits"
,
"id": "8",
"name": "Blouses & Shirts"
,
"id": "9",
"name": "Friendly URL in CodeIgniter"
]
Why it has returning to :
<li class="collection-item">7</li>
<li class="collection-item">8</li>
<li class="collection-item">9</li>
output that I expected is
<li class="collection-item">7</li>
<li class="collection-item">8</li>
<li class="collection-item">9</li>
preview :
http://prntscr.com/n1x0ey
Please help ..
Thank a lot.
jquery ajax
I have a complete code for getting the values from PHP through Jquery AJAX with JSON datatype.
But after ajax done, it doesn't response to html. Its become text.
the result become like this.
$("#category li").one('click', function()
var catId = $(this).attr("id");
var that = this;
$.ajax(
type: 'GET',
url: '<?php echo base_url()."dashboard/getsubcategory/"; ?>'+catId,
dataType: 'html',
data: category: catId,
beforeSend: function()
$("#loader").show();
)
.done(function(data)
var categories = JSON.parse(data)
$.each(categories, function(k,category)
that.after('<li class="collection-item">'+category.id+'</li>');
);
)
.fail(function()
alert( "Fetch failed." );
);
);
JSON output :
[
"id": "7",
"name": "Blazer & Suits"
,
"id": "8",
"name": "Blouses & Shirts"
,
"id": "9",
"name": "Friendly URL in CodeIgniter"
]
Why it has returning to :
<li class="collection-item">7</li>
<li class="collection-item">8</li>
<li class="collection-item">9</li>
output that I expected is
<li class="collection-item">7</li>
<li class="collection-item">8</li>
<li class="collection-item">9</li>
preview :
http://prntscr.com/n1x0ey
Please help ..
Thank a lot.
jquery ajax
jquery ajax
edited Mar 23 at 16:53
Henry
asked Mar 23 at 16:34
HenryHenry
32
32
1
please post the code. otherwise, nobody can help.
– Ashkan Mobayen Khiabani
Mar 23 at 16:36
add a comment |
1
please post the code. otherwise, nobody can help.
– Ashkan Mobayen Khiabani
Mar 23 at 16:36
1
1
please post the code. otherwise, nobody can help.
– Ashkan Mobayen Khiabani
Mar 23 at 16:36
please post the code. otherwise, nobody can help.
– Ashkan Mobayen Khiabani
Mar 23 at 16:36
add a comment |
1 Answer
1
active
oldest
votes
You are storing in the variable that the reference to the Element and using it to append to the document. You should wrap the element before calling the method after.
Furthermore, in your ajax call you are setting the value dataType to html when it seems that you expect a json.
So, wrapping up, your code should look like:
$("#category li").one('click', function()
var $that = $(this);
var catId = $that.attr("id");
$.ajax(
type: 'GET',
url: '<?php echo base_url()."dashboard/getsubcategory/"; ?>'+catId,
dataType: 'json',
data: category: catId,
beforeSend: function()
$("#loader").show();
)
.done(function(data)
data.forEach(function(category)
$that.after('<li class="collection-item">'+category.id+'</li>');
);
)
.fail(function()
alert( "Fetch failed." );
);
);
Thank @mgarcia you save my day..
– Henry
Mar 23 at 18:10
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
oldest
votes
active
oldest
votes
active
oldest
votes
You are storing in the variable that the reference to the Element and using it to append to the document. You should wrap the element before calling the method after.
Furthermore, in your ajax call you are setting the value dataType to html when it seems that you expect a json.
So, wrapping up, your code should look like:
$("#category li").one('click', function()
var $that = $(this);
var catId = $that.attr("id");
$.ajax(
type: 'GET',
url: '<?php echo base_url()."dashboard/getsubcategory/"; ?>'+catId,
dataType: 'json',
data: category: catId,
beforeSend: function()
$("#loader").show();
)
.done(function(data)
data.forEach(function(category)
$that.after('<li class="collection-item">'+category.id+'</li>');
);
)
.fail(function()
alert( "Fetch failed." );
);
);
Thank @mgarcia you save my day..
– Henry
Mar 23 at 18:10
add a comment |
You are storing in the variable that the reference to the Element and using it to append to the document. You should wrap the element before calling the method after.
Furthermore, in your ajax call you are setting the value dataType to html when it seems that you expect a json.
So, wrapping up, your code should look like:
$("#category li").one('click', function()
var $that = $(this);
var catId = $that.attr("id");
$.ajax(
type: 'GET',
url: '<?php echo base_url()."dashboard/getsubcategory/"; ?>'+catId,
dataType: 'json',
data: category: catId,
beforeSend: function()
$("#loader").show();
)
.done(function(data)
data.forEach(function(category)
$that.after('<li class="collection-item">'+category.id+'</li>');
);
)
.fail(function()
alert( "Fetch failed." );
);
);
Thank @mgarcia you save my day..
– Henry
Mar 23 at 18:10
add a comment |
You are storing in the variable that the reference to the Element and using it to append to the document. You should wrap the element before calling the method after.
Furthermore, in your ajax call you are setting the value dataType to html when it seems that you expect a json.
So, wrapping up, your code should look like:
$("#category li").one('click', function()
var $that = $(this);
var catId = $that.attr("id");
$.ajax(
type: 'GET',
url: '<?php echo base_url()."dashboard/getsubcategory/"; ?>'+catId,
dataType: 'json',
data: category: catId,
beforeSend: function()
$("#loader").show();
)
.done(function(data)
data.forEach(function(category)
$that.after('<li class="collection-item">'+category.id+'</li>');
);
)
.fail(function()
alert( "Fetch failed." );
);
);
You are storing in the variable that the reference to the Element and using it to append to the document. You should wrap the element before calling the method after.
Furthermore, in your ajax call you are setting the value dataType to html when it seems that you expect a json.
So, wrapping up, your code should look like:
$("#category li").one('click', function()
var $that = $(this);
var catId = $that.attr("id");
$.ajax(
type: 'GET',
url: '<?php echo base_url()."dashboard/getsubcategory/"; ?>'+catId,
dataType: 'json',
data: category: catId,
beforeSend: function()
$("#loader").show();
)
.done(function(data)
data.forEach(function(category)
$that.after('<li class="collection-item">'+category.id+'</li>');
);
)
.fail(function()
alert( "Fetch failed." );
);
);
answered Mar 23 at 17:44
mgarciamgarcia
3967
3967
Thank @mgarcia you save my day..
– Henry
Mar 23 at 18:10
add a comment |
Thank @mgarcia you save my day..
– Henry
Mar 23 at 18:10
Thank @mgarcia you save my day..
– Henry
Mar 23 at 18:10
Thank @mgarcia you save my day..
– Henry
Mar 23 at 18:10
add a comment |
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1
please post the code. otherwise, nobody can help.
– Ashkan Mobayen Khiabani
Mar 23 at 16:36