2014-04-01 10:55
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I am currently developing a website and was using strictly PHP, after a few hours of development I am noticing I really need to perform AJAX queries. However I am using CI and slightly confused on how I call Ajax queries - especially to a controllers method. I am looking to post data using the query,I have found the following using Jquery:

  type: "POST",
  url: url,
  data: data,
  success: success,
  dataType: dataType

but if I had a controller called main and method called postBack() - how would I pass postback() the data?

Could anyone point me in the right direction it would be much appreciated - apologies if this is extremely simple to say I am a newbie is an understatement.

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我目前正在开发一个网站并严格使用PHP,经过几个小时的开发后我注意到我真的需要 执行AJAX查询。 但是我使用CI并且稍微混淆了我如何调用Ajax查询 - 特别是对于控制器方法。 我希望使用查询发布数据,我使用Jquery发现了以下内容:

  $ .ajax({
 url:url  ,

但是如果我有一个名为main的控制器和方法 叫postBack() - 我如何传递postback()数据?

任何人都可以指出我正确的方向,我会非常感激 - 道歉,如果说这是一个非常简单的说我是一个新手是轻描淡写。

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3条回答 默认 最新

  • douzhi2760
    douzhi2760 2014-04-01 12:42

    Here is a Codeingiter/Jquery function pair that I wrote not long ago:


    var BASE_URL = <?php echo base_url(); ?>
    var stuff = "";
        if(stuff ){
            stuff = stuff + '-';
        stuff = stuff + $(this).html();
      type: "POST",
      url: BASE_URL+"gate",
      data: {'data':stuff},
      success: function(data){

    Codeigniter Controller:

    class Gate extends CI_Controller {
        public function index(){
                $data = explode('-', $_POST['data']);
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  • douzhi4991
    douzhi4991 2014-04-01 12:14

    In Ci u will use Ajax the for the URL part u will add the controller function name example var URL="HOME SITE URL"+"main/post Back"; //here main is controller and post Back is function that u want to call like below. $Ajax({ type: "POST",URL: URL,data: {star: 3},success: success,data Type: data Type }); and after that in post Back function in controller we used function post Back() { $variable1 = $_POST["star"]; //or $_GET["star"] or $this->input->post('star'); }

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  • dongzhen7108
    dongzhen7108 2014-04-01 12:24
    class MyClass extends MyController {
        function postBack(){
            //dosomething with post data
            $postdata = $this->input->post();
            $somearray = filter_something($postdata);
            //this is where you return the output to jQuery. you output a string in JSON format 
            echo json_encode($some_array);
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