显示动态使用Ajax数据动态、数据、Ajax
在此code，点击等按钮后，该数据在数据库中添加了。我想现在要做的是，我想加入数据后，我就总喜欢询问所选择的项目，并且无需加载页面显示。
In this code, after clicking the like button, the data is added in the database already. What I wanted to do now is I would like after adding the data, I will query the total like of the chosen item and display it without loading the page.
这是我的code现在：
This is my code for now:
我的观点：
<p id='state'><i class='fa fa-thumbs-up'></i><span id="likeThis"><?php echo $countLike;?></span> likes • <i class='fa fa-thumbs-down'></i><?php echo $countDisLike;?> dislikes •<i class='fa fa-thumbs-down'></i><a href='<?php echo base_url();?>index.php/photoCheese/deleteUploadPic/<?php echo $row['uploadID'];?>'>Delete Picture</a></p>
<input type="button" onclick="getVal(this.value)" class='detailButton1' name='like_name' id='like_id' value='<?php echo $link;?>' title='Like this post'><i class='fa fa-thumbs-up'></i> Like</input>
JavaScript的：
Javascript:
函数getVal（值）
{
jQuery.ajax（{
键入：GET，
网址：？＆LT; PHP的回声BASE_URL（）;＆GT;的index.php / photoCheese / like_total /，
数据类型：JSON，
数据：{like_id：值}，
成功：函数（RES）{
警报（res.no_likes）;
如果（RES）{
jQuery的（＃likeThis）的HTML（res.no_likes）。
}
}
}）;
function getVal(value)
{
jQuery.ajax({
type:"GET",
url: "<?php echo base_url();?>index.php/photoCheese/like_total/",
dataType:'json',
data: {like_id : value},
success: function(res){
alert(res.no_likes);
if(res){
jQuery("#likeThis").html(res.no_likes);
}
}
});
控制器：
public function like_total(){
$id = $this->session->userdata('userID');
$upload = $this->input->get('like_id');
$data = array('like' => 1,
'userID'=>$id,
'uploadID'=>$_GET['like_id']);
$result = $this->photoCheese_model->get_like_total($data,$upload);
return json_encode($result);
}
型号：
public function get_like_total($data,$uplaod){
$success = $this->db->insert('tbl_like',$data);
if($success){
$this->db->select('uploadID,SUM(`like`) as no_likes',false);
$this->db->where('uploadID',$upload);
$this->db->where('like !=',2);
$query = $this->db->get();
}
return $query->result_array();
}
这code将不会显示total_likes。有什么不对的呢？
This code will not display the total_likes. What's wrong with this one?
推荐答案
在所有的帮助和研究。这是此问题的运行code
After all the helps and research. This is the running code of this problem.
在查看：
<p id='state'><i class='fa fa-thumbs-up'></i><span class="likeThis"><?php echo $countLike;?></span> likes • <i class='fa fa-thumbs-down'></i><?php echo $countDisLike;?> dislikes •<i class='fa fa-thumbs-down'></i><a href='<?php echo base_url();?>index.php/photoCheese/deleteUploadPic/<?php echo $row['uploadID'];?>'>Delete Picture</a></p>
<input type="button" onclick="getVal(this.value)" class='detailButton1' name='like_name' id='like_id' value='<?php echo $link;?>' title='Like this post'><i class='fa fa-thumbs-up'></i> Like</input>
JavaScript的：
Javascript:
＆LT;脚本类型=文/ JavaScript的＆GT;
功能getVal（值）
{
jQuery.ajax（{
键入：GET，
网址：？＆LT; PHP的回声BASE_URL（）;＆GT;的index.php / photoCheese / like_total /，
数据类型：JSON，
数据：{like_id：值}，
错误：函数（结果）{
$（'likeThis。'）追加。（'＆LT; P＆GT;再见了世界＆LT; / P＆GT;'）;
}，
成功：函数（结果）{
jQuery的（likeThis。）HTML（结果）。
}
}）;
}
＆LT; / SCRIPT＆GT;
<script type="text/javascript">
function getVal(value)
{
jQuery.ajax({
type:"GET",
url: "<?php echo base_url();?>index.php/photoCheese/like_total/",
dataType:'json',
data: {like_id : value},
error: function(result){
$('.likeThis').append('<p>goodbye world</p>');
},
success: function(result){
jQuery(".likeThis").html(result);
}
});
}
</script>
控制器：
public function like_total(){
$id = $this->session->userdata('userID');
$upload = $this->input->get('like_id');
$data = array('like' => 1,
'userID'=>$id,
'uploadID'=>$_GET['like_id']);
$result = $this->photoCheese_model->get_like_total($data,$upload);
$this->output->set_content_type('application/json');
$this->output->set_output(json_encode($result));
return $result;
}
型号：
public function get_like_total($data,$upload){
$success = $this->db->insert('tbl_like',$data);
//Query the total likes
if($success){
$this->db->select()->from('tbl_like');
$this->db->where('uploadID',$upload);
$this->db->where('like !=',2);
$query = $this->db->get();
return $query->num_rows();
}
return 0;
}
这code现在完美运行。感谢您的帮助反正。
This code runs perfectly now. Thanks for the help anyway.
