如何確定ajax的URL

Question

正如標題中所述......我無法確定我是否正確地執行了後期操作，或者如果操作正確執行，那麼爲什麼值是空的，因爲我檢查了通過它在術後前值...這裏是我的代碼：

腳本：

$.ajax({ 
    url: "<?php echo base_url();?>/Index/viewDayDocuments", 
    type: 'post', 
    data: {currentDay: 'currentDay', currentMonth: 'currentMonth', currentYear: 'currentYear'}, 
    success: function(result){ 
     $('.list').text(''); 
     $('.list').remove(); 
     $(".listIncoming").html("<p class = 'list'>This is the: "+ result +"</p>"); 
     $("#myform").show(500); 
    } 
}); 

會拋出回一個返回值

控制器代碼：

$data['day'] = $_POST['currentDay']; 
     $data['month'] = $_POST['currentMonth']; 
     $data['year'] = $_POST['currentYear']; 

     $date = $data['year']."-".$data['month']."-".$data['day']; 

     $this->load->model('search_form'); 
     $output['details'] = $this->search_form->searchDateRetrievedIncoming($date); 

     return $data; 

Answer 1

您的ajax請求需要一個字符串作爲字符串回顯。

$data['day'] = $_POST['currentDay']; 
    $data['month'] = $_POST['currentMonth']; 
    $data['year'] = $_POST['currentYear']; 

    $date = $data['year']."-".$data['month']."-".$data['day']; 

    $this->load->model('search_form'); 
    $output['details'] = $this->search_form->searchDateRetrievedIncoming($date); 

    echo json_encode($data); 

如果數組格式不是數據格式（例如JSON），則無法正確回顯該數組。

我們現在，把數據在JavaScript

$.ajax({ 
url: "<?php echo base_url();?>/Index/viewDayDocuments", 
type: 'post', 
data: {currentDay: 'currentDay', currentMonth: 'currentMonth', currentYear: 'currentYear'}, 
success: function(result){ 
    $('.list').text(''); 
    $('.list').remove(); 
    date = JSON.parse(result); 
    $(".listIncoming").html("<p class = 'list'>This is the: "+ date.month +"/"+date.day+ "/"+date.year +"</p>"); 
    $("#myform").show(500); 
    } 
}); 

在這個模塊中，我從改變你的PHP文件中的字符串轉換成JSON對象，以便它可以正確讀取。