Question

我在更改功能后获得警报值，但在成功功能之后，我没有得到任何值

我的ajax页面

$(document).ready(function(){
    $("#customer").change(function() {
        var customer_type = $(this).find(":selected").val();
        var dataString = 'customer_type='+ customer_type;
        $.ajax({
            url: 'http://localhost/capms_v3/ajax/getcust_type.php',
            dataType: "json",
            data: dataString,
            cache: false,
            success: function(customerData) {
                alert(data);
                alert("test");
                if(customerData) {
                    var customer = [customerData];
                    customerData.forEach(function(item) {
                        var data = '<tr>';
                        data+= '<td colspan="4"> </td>';
                        data+= '<td align="right">'+item.company_id+'</td>';
                        data+= '<td align="right">'+item.company_name+'</td>';
                        data+='</tr>';
                        $('.appendData').append(data);
                    });
                } else {
                    $("#heading").hide();
                    $("#records").hide();
                    $("#no_records").show();
                }
            }
        });
    });
});

我的数组值不是在成功函数之后但在getcusttype页面值中出现在数组中

getcusttype.php

<?php 
     //header("Content-type:application/json");
       include 'db.php';
       $db=DbConnect();

   if($_REQUEST['customer_type']) {
     $sql = "SELECT company_id,company_name FROM ca_customer WHERE customer_type ='".$_REQUEST['customer_type']."'";
     $result = mysql_query($sql) or die(mysql_error());
     $data = array();
     while( $rows = mysql_fetch_array($result) ) {
     $data[] = $rows;
              }
              echo json_encode($data);
              } else {
              echo 0;
              }?>

  //var customer =[{"0":"1","company_id":"1","1":"Win Win 
  web","company_name":"Win Win web"},{"0":"7","company_id":"7","1":"New 
  Company","company_name":"New Company"},
  {"0":"10","company_id":"10","1":"Murugan Super 
  Store","company_name":"Murugan Super Store"}];

成功之后：function（customerdata）如果我发出警报（数据）值警报我不知道我犯了什么错误。

视图

<select id="customer" name="customer_type" class="form-control">
 <option value="">Select Customer Type</option>
  <?php 
     foreach($all_ca_customer_type as $ca_customer_type)
      {
    $selected = ($ca_customer_type['customer_type_id'] == $this->input->post('customer_type')) ? ' selected="selected"' : "";
    echo '<option value="'.$ca_customer_type['customer_type_id'].'" '.$selected.'>'.$ca_customer_type['customer_type_name'].'</option>';
                            } 
                            ?>
                        </select>
 <tbody class="appendData">
    </tbody>

在成功函数之后没有得到值。如果有人面对这个问题，请帮助我。提前谢谢

Answer 1

在header("Content-type:application/json");

中取消注释getcusttype.php

Answer 2

您的alert()函数似乎正在引用data数据来自何处？尝试使用alert(customerData)代替alert(data)。

   success: function(customerData) {
        alert(data);
        alert("test");
        if(customerData) {
            var customer = [customerData];
            customerData.forEach(function(item) {
                var data = '<tr>';
                data+= '<td colspan="4"> </td>';
                data+= '<td align="right">'+item.company_id+'</td>';
                data+= '<td align="right">'+item.company_name+'</td>';
                data+='</tr>';
                $('.appendData').append(data);
            });
        } else {
            $("#heading").hide();
            $("#records").hide();
            $("#no_records").show();
        }
    }

如果您希望提醒您附加行，则应在alert(data)调用forEach()变量后将data调用移至customerData.forEach(function(item) { var data = '<tr>'; data+= '<td colspan="4"> </td>'; data+= '<td align="right">'+item.company_id+'</td>'; data+= '<td align="right">'+item.company_name+'</td>'; data+='</tr>'; alert(data); $('.appendData').append(data); });块。例如：

function num() {
  var number = prompt("Enter number");
  document.getElementById("number").innerHTML = number;

}

成功功能在ajax中不起作用

2 个答案: