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Post a form in Ajax about mini3 HOT 6 CLOSED

panique avatar panique commented on August 15, 2024
Post a form in Ajax

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Comments (6)

crystlbrd avatar crystlbrd commented on August 15, 2024

What have you tried? How is your script looking right now? What isn't working? Any errors?

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hollow6 avatar hollow6 commented on August 15, 2024

I have two select boxes in a form.
I would like the second to display different items when the value of the first changes.
The example with Ajax shows how to send the url but not the variables.
If possible, please show a model where we send the url, variables in post method.

Sorry for my English, I'm french man.

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hollow6 avatar hollow6 commented on August 15, 2024

What have you tried? How is your script looking right now? What isn't working? Any errors?

I want something that looks like this :

$('.category_list').change(function(){
$.ajax({
url + "/banner/ajaxGetNews",
data:"category = " +category,
type:"POST"

})
.done(function(result) {
$('.news_list').html(result);
})
.fail(function() {
// this will be executed if the ajax-call had failed
})
.always(function() {
// this will ALWAYS be executed, regardless if the ajax-call was success or not
});
});

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ebitkov avatar ebitkov commented on August 15, 2024

data:"category = " +category,
type:"POST"

I assume your variable category is defined globally? Other than that your JS looks fine.

Next you should have a BannerController in your application\Controller directory. Something like this should work fine:

<?php

namespace Mini\Controller;

class BannerController {
    public function ajaxGetNews() {
        echo json_encode($_POST);
    }
}

this example should return your AJAX POST-Request as a JSON object.

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hollow6 avatar hollow6 commented on August 15, 2024

data:"category = " +category,
type:"POST"

I assume your variable category is defined globally? Other than that your JS looks fine.

Next you should have a BannerController in your application\Controller directory. Something like this should work fine:

<?php

namespace Mini\Controller;

class BannerController {
    public function ajaxGetNews() {
        echo json_encode($_POST);
    }
}

this example should return your AJAX POST-Request as a JSON object.

no, my category variable is not global.
if i complete the script like this, do you think it will work?

$('.category_list').change(function(){
var category = $(this).val();
$.ajax({
url + "/banner/ajaxGetNews",
data:"category = " +category,
type:"POST"
})
.done(function(result) {
$('.news_list').html(result);
})
.fail(function() {
// this will be executed if the ajax-call had failed
})
.always(function() {
// this will ALWAYS be executed, regardless if the ajax-call was success or not
});
});

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ebitkov avatar ebitkov commented on August 15, 2024

That should work, as long your element .category_list is a form input. Also, why aren't you just testing it :)

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