Question

当某人点击心脏图标喜欢某张照片时，它会浏览其他照片，然后返回到该照片再次喜欢，因此它将在“新喜欢的照片”页面的第一张图像中再次显示

这是代码

function Favorite(userid,postid,status){
	   
			$.ajax({  
				url: '<?php echo site_url('Home/favorite'); ?>',
				type: "POST",
				data:{userid:userid,postid:postid,status:status}, 
			    success:function(responce){
				if(status==1){
					
				$('#showlikes'+postid).html('<span data-post-likes="'+postid+'"> '+responce+' </span><strong>Likes</strong>');
				
				$('#changelike'+postid).html('<span  onclick="return Favorite('+userid+','+postid+',0);" class="like-post active"><span class="like-icon heart"><div class="heart-animation-1 heart"></div><div class="heart-animation-2 heart"></div></span></span>');
				//alert('Post added in your Favorite list.');
				
				} else {
					
				$('#showlikes'+postid).html('<span data-post-likes="'+postid+'"> '+responce+' </span><strong>Likes</strong>');
				
				$('#changelike'+postid).html('<span  onclick="return Favorite('+userid+','+postid+',1);" class="like-post "><span class="like-icon heart"><div class="heart-animation-1 heart"></div><div class="heart-animation-2 heart"></div></span></span>');	
				//alert('Post remove in your Favorite list.');
				
				}
				
			}
			
		});

	}

<div class="sb_network_holder"><a class="sb_network_button fa fa-heart" target="_blank" href="//weheartit.com/heart-it/new_entry?encoding=UTF-8&amp;hearting_method=heart_it_button&amp;image_url=<?php echo $image_urls; ?>&amp;popup=1&amp;source_url=https://www.****.com/details/<?php echo $value['id']; ?>" data-network="heart" style="display: block; left: 3px; top: 3px;"></a><a class="sb_network_button fa fa-twitter" target="_blank" href="https://twitter.com/intent/tweet?url=https://www.****.com/details/<?php echo $value['id']; ?>&amp;text=<?php echo $value['description']; ?>" data-network="twitter" style="display: block; left: 3px; top: 3px;"></a><a class="sb_network_button fa fa-pinterest" target="_blank" href="//pinterest.com/pin/create/button/?url=https://www.****.com/details/<?php echo $value['id']; ?>&amp;media=<?php echo $image_urls; ?>&amp;description=<?php echo $value['description']; ?>" data-network="pinterest" style="display: block; left: 3px; top: 3px;"></a><a class="sb_network_button fa fa-facebook" target="_blank" href="//www.facebook.com/sharer.php?u=https://www.****.com/details/<?php echo $value['id']; ?>" data-network="facebook" style="display: block; left: 3px; top: 3px;"></a></div>
   </div>
		</div>
		</div>
	<div class="actions">
	    <span id="changelike<?php echo $value['id']; ?>">
		<span 
		<?php if($this->session->userdata('session_userId')==''){ ?> 
		onclick="alert('Login First');"
		<?php } else { 
		if($dataa == 0){ ?>
		onclick="return Favorite('<?php echo $this->session->userdata('session_userId'); ?>','<?php echo $value['id']; ?>',1);" 
		<?php } 
		else {
	    ?> 
		onclick="return Favorite('<?php echo $this->session->userdata('session_userId'); ?>','<?php echo $value['id']; ?>',0);"  <?php } } ?> class="like-post <?php if($dataa == 0){ ?> <?php } else { ?> active <?php } ?>" >
			<span class="like-icon heart" ><div class="heart-animation-1 heart"></div><div class="heart-animation-2 heart"></div></span>
		</span>
		</span>

我该如何解决？

当我回到照片时再次喜欢它

0 个答案: