Question

我发送ajax调用到控制器方法然后该方法检查数据是否存在然后echo true 如果在数据库中找不到数据然后echo false

问题是我无法在ajax成功函数中得到控制器方法的真假

控制器

interface Example {

    static void doJob1(String arg) {
        verifyArg(arg);
        ...
    }

    static void doJob2(String arg) {
        verifyArg(arg);
        ...
    }

    private static void verifyArg(String arg) {
        ...
    }
}

调用Ajax

public function ajax_load()

    {

            $project_id =$this->input->post('account');

             $this->db->select('Project_id');

            $this->db->from('proposal');

            $this->db->where('Project_id',$project_id);

            $query = $this->db->get()->row();

            if($query){
                echo "true";
            }else{
                echo "false";
            }

        }

Answer 1

您必须从控制器返回true或false值，如

if($query){
            $arr=["true"];
            echo json_encode($arr);
        }else{
            $arr=["false"];
            echo json_encode($arr);
        }

Ajax调用

$.ajax({
              url: "<?php echo base_url();?>proposal/ajax_load",
              type: 'POST',
              dataType: 'json',
              data : {"account" : project_id}, 
              success: function(result){
                  if(result[0]==="true"){
                    $("#err").show();
                    $("#err").html("Project id already exist");
                    $(".btn").prop('disabled', true);
                   }else if(result[0]==="false"){
                    $("#err").hide();
                    $(".btn").prop('disabled', false);
                   }
                }
    });

Answer 2

此代码适用于您，除了您正在撰写的查询类型之外。下次尝试显示完整代码以获得更快的帮助。 Stacoverflow充满了工程师，没有人窃取您的代码所以我想知道为什么要将它发送到中途。

好的确保你的后端工作正常

<html>
<head>
<body>

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
    <script>

         $(document).ready(function(){


         //$('#result').click(function(){

         var project_id = '1';





        $('#loader').fadeIn(400).html('Please Wait. Data is being Loaded');


         // assuming that you want query result by posting a variable
         var datasend = "account="+ project_id;

         $.ajax({

         type:'POST',
         url:'data.php',
         data:datasend,
         crossDomain: true,
         cache:false,
         success:function(msg){

//display image loader or text to alert the use that content is being loaded

         $('#loader').hide();

         if(msg=='true'){
alert('true');

// and display result

         $('#result').fadeIn('slow').prepend(msg);

}else{

alert('false');
}


         }

         });

         //})

         });


    </script>


<div id="loader"></div>
<div id="result"></div>
</body></html>

<强> data.php

public function ajax_load()

{

        $project_id =$this->input->post('account');

         $this->db->select('Project_id');

        $this->db->from('proposal');

        $this->db->where('Project_id',$project_id);

        $query = $this->db->get()->row();

        if($query){
            echo "true";
        }else{
           echo "false";
        }

    }

我无法在ajax调用中获得控制器方法的结果

2 个答案: