[英]AJAX POST not working to load div view in CODEIGNITER
幫我解決我的Ajax問題...好吧,我想做的是,我可能在主視圖中有一個側欄導航面板。.所有導航鏈接都是從database.now中獲取的。當我單擊任何鏈接時,我想要的是它應在相同的主視圖中但在中間面板中加載CI視圖div,如果單擊第二個鏈接,則應將新的div替換為舊視圖div,而不加載頁面。單擊后要加載的Divs放置在的views文件夾中項目(CI視圖文件夾)。 現在我嘗試的是單擊鏈接時,我將視圖名稱調用ajax,同時將數據傳遞到URL(/ controller / method),控制器加載該視圖,而ajax成功時,該視圖應以以下方式加載到中間面板中:新股
查看代碼:
腳本代碼:
<script type="text/javascript">
$(document).ready(function(){
$("a.links").click(function(e){
e.preventDefault();
alert("alert on click");
var platform_id = $(this).attr('id');
alert(platform_id);
$.ajax({
url: '<?php echo base_url() ?>admin/load_view',
type: 'POST',
datatype:'json',
data: {'view_param': platform_id},
success: function(response)
{
if (response.success == 1)
{
alert('success');
}
else{
alert('unseccessful'); //just a alert to see if response is successful
}
},
error: function(response){
alert(response.message);
}
});
})
})
</script>
<div id="left-column">
<h3>Header</h3>
<ul class="nav">
<?php foreach($ops as $operations)
{$op_name = $operations['admin_op'];
$op_link = $operations['link_to'];
?>
<li><a href="" id="<?php echo $op_name ; ?>" class="links"><?php echo $op_name ;?></a></li>
<?php } ?>
</ul>
<a href="http://all-free-download.com/free-website-templates/" class="link">Link here</a> <a href="http://all-free-download.com/free-website-templates/" class="link">Link here</a>
</div>
中間面板div:
<div id="mang_server"> </div>
控制器代碼:
public function load_view(){
$view_name = $this->input->post('view_param');
log_message('var dump', var_export($view_name));
$data['view_link'] = $this->admin_model->load_view_model($view_name);
$response['html_view'] = $this->load->view($data['view_link'],'',TRUE);
echo json_encode($response);
//$this->load->view($data);
}
我嘗試調試它,但同時我也觀察到它在ajax中成功了,但是它提醒了它的其他部分。
錯誤部分警報:
'Manage Servers'{"success":1,"message":"Some success message","html_view":"<table class=\"listing form\" cellpadding=\"0\" cellspacing=\"0\">\n <tr>\n <th class=\"full\" colspan=\"2\">Header Here<\/th>\n <\/tr>\n <tr>\n <td class=\"first\" width=\"172\"><strong>Lorem Ipsum<\/strong><\/td>\n <td class=\"last\"><input type=\"text\" class=\"text\" \/><\/td>\n <\/tr>\n <tr class=\"bg\">\n <td class=\"first\"><strong>Lorem Ipsum<\/strong><\/td>\n <td class=\"last\"><input type=\"text\" class=\"text\" \/><\/td>\n <\/tr>\n <tr>\n <td class=\"first\"><strong>Lorem Ipsum<\/strong><\/td>\n <td class=\"last\"><input type=\"text\" class=\"text\" \/><\/td>\n <\/tr>\n <tr class=\"bg\">\n <td class=\"first\"><strong>Lorem Ipsum<\/strong><\/td>\n <td class=\"last\"><input type=\"text\" class=\"text\" \/><\/td>\n <\/tr>\n <\/table>\n <p> <\/p>"}
在您的控制器中,您沒有分配前端所需的$response['success']
和$response['message']
。 將控制器代碼更改為
public function load_view()
{
$view_name = $this->input->post('view_param');
log_message('var dump', var_export($view_name, true));
$data['view_link'] = $this->admin_model->load_view_model($view_name);
$response['success'] = 1;
$response['message'] = "Some success message";
$response['html_view'] = $this->load->view($data['view_link'],'',TRUE);
echo json_encode($response);
exit;
}
希望這可以幫助 :)
注意:在log_message('var dump', var_export($view_name, true));
, var_export()
需要第二個參數來返回值。 否則,它將輸出文本並導致ajax響應出現問題。
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