我正在使用json的按钮。当用户正确点击它时，会增加相似数量，并在数据库中存储相应的id，但它显示浏览器中所有ID的增量是错误的。我想只显示用户喜欢的ID，但它显示给所有人。
测试网址为http://way2enjoy.com/app/check/test/indexp.php
indexp.php文件是
<?php
include('/home/xxx/con.php');
$query="Select * from users_jokes order by id desc limit 10";
$result=mysql_query($query);
?>
<html>
<head>
<meta charset="utf-8">
<script src="jquery-3.1.0.min.js"></script>
<script type="text/javascript">
var ajaxSubmit = function(formEl) {
var url = $(formEl).attr('action');
var comment=document.getElementById("jokes_comment").value;
var joke_id=document.getElementById("joke_id_hidden").value;
$.ajax({
url: url,
data:{
'action':'addComment',
'comment':comment,
'joke_id':joke_id
},
dataType: 'json',
type:'POST',
success: function(result) {
console.log(result);
$.ajax({
url: url,
data:{
'action':'getLastComment',
'joke_id':joke_id
},
dataType: 'json',
type:'POST',
success: function(result) {
$('#jokes_comment').val("");
console.log(result[0].description);
$("#header ul").append('<li>'+result[0].description+'</li>');
},
error: function(){
alert('failure');
}
});
},
error: function(){
alert('failure');
}
});
// return false so the form does not actually
// submit to the page
return false;
}
var ajaxLike=function()
{
var joke_id=document.getElementById("joke_id_hidden").value;
// setup the ajax request
$.ajax(
{
url: 'likeskk.php',
data:{
'action':'increment_like',
'joke_id':joke_id
},
dataType: 'json',
type:'POST',
success: function(result)
{
$.ajax(
{
url: 'likeskk.php',
data:{
'action':'display_like',
'joke_id':joke_id
},
dataType: 'json',
type:'POST',
success: function(result)
{
console.log(result);
$("label[for='like_counter']").text(result.likes);
},
error: function(result)
{
alert("error 2");
},
});
},
error: function()
{
alert('failure');
}
});
return false;
}
</script>
<p>commnet list</p>
<div id="header">
<ul id="commentlist" class="justList">
<?php
$query="Select * from comment where joke_id='2'";
$result=mysql_query($query);
while($data = mysql_fetch_array($result)){
$cont = $data['description'];
?>
<li><?php echo $cont;
?></li>
<?php
}
?>
</ul>
</div>
<?php
?>
<form method="post" action="processkk.php" onSubmit="return ajaxSubmit(this);">
<input type=text id="jokes_comment" name="jokes_comment">
</input>
<input type="submit" value="comment">
</form>
</body>
</html>
<?php
while($data = mysql_fetch_array($result)){
$id = $data['id'];
$cont = $data['content'];
$likes = $data['likes'];
?>
<p><?php echo $cont;?></p>
<input type="hidden" value="<?php echo $id ;?>" id="joke_id_hidden">
<p><button onClick="ajaxLike();">Like</button> <label for="like_counter"><?php echo $likes;?></label></p>
<?php }
?>
</body>
</html>
likeskk.php
<?php
include('/home/xxxxxxx/con.php');
$action=$_POST['action'];
if($action=="increment_like")
{
$joke_id=$_POST['joke_id'];
$query="update users_jokes set likes =likes+1 where id='".$joke_id."'";
$result=mysql_query($query);
// setup our response "object"
$retVal=array("Success"=>"true");
print json_encode($retVal);
}
if($action=="display_like")
{
$joke_id=$_POST['joke_id'];
$query = "select likes from users_jokes where id = '$joke_id'";
$qry = mysql_query($query);
while($rows = mysql_fetch_array($qry)){
$likes = $rows['likes'];
}
header('Content-Type: application/json');
// print json_encode(array('foo' => 'bar'));
print json_encode(array('success'=>'true','likes'=>$likes));
}
?>
当我点击一个像所有像增加。当我发布对其附加的一个ID的评论并显示在所有id
中答案 0 :(得分：1)
复制并粘贴。
<div class = "row">
<div class = "icon"> <img src = "icon.PNG" alt="Some of the interesting stuff I've worked on."> </div>
<div class = "icon"> <img src = "icon.PNG" alt="My past experiences."> </div>
</div>
答案 1 :(得分：0)
因为您使用相同的ID。 id必须是唯一的
<input type="hidden" value="4638" id="joke_id_hidden">
答案 2 :(得分：0)
您可以使用按钮本身上的数据集以及相应的ID，而不是为每个条目使用隐藏输入。单击该按钮时，javscript函数可以读取该数据集值并在ajax POST请求中使用该值。所以，作为一个例子：
<!doctype html>
<html>
<head>
<title>Like the jokes...</title>
</head>
<body>
<form id='bttns'>
<?php
/* pseudo generate some jokes with buttons and labels to emulate your page */
for( $i=1; $i < 20; $i++ ){
$random=rand(1,20);
echo "
<p>Funny joke...[{$i}]</p>
<input type='button' value='Like' data-id='$i' />
<label class='counter'>{$random}</label>";
}
?>
</form>
<script type='text/javascript'>
var col=document.querySelectorAll('#bttns input[type=\"button\"]');
for( var n in col )if(col[n].nodeType==1)col[n].onclick=function(e){
var el=typeof(e.target)!='undefined' ? e.target : e.srcElement;
var label=el.nextSibling.nextSibling;
var id=el.dataset.id;
label.innerHTML=parseInt( label.innerHTML )+1;
alert( 'use ajax to POST this id: '+id );
}
</script>
</body>
</html>
我或许应该添加一个ajax回调函数来为每个笑话而不是像这里更新likes的值的地方 - 它只是示例代码，展示了如何消除隐藏字段和问题重复的ids。