AJAX不是分离JSON字符串字符串、不是、AJAX、JSON
我使用的星级评定系统来显示从SQL评级数据。这堪称每个项目都有独特的identifyer变量$ ID和评级TABL每个等级都有其独特identifyer $ storyidr。我想这个脚本显示：
的平均评分 的次数的项目已被评为该值是retirevable，但他们在网页上显示在一起，我看不出要分开它们。例如，对于具有4的平均等级，并已在额定的200倍的一个项目。当用户通过AJAX点击返回的数据看起来像：
对于response14200 对于响应24200我希望能够单独他们看起来像：
对于response14 对于响应2200HTML页
＆LT; DIV ID =产品的风格=＆GT;
＆LT; DIV CLASS =rateit数据storyidr =＆LT; PHP的echo $ ID;＆gt;中＆GT;
＆LT; / DIV＆GT;
＆LT; DIV CLASS =averagevote＆GT;
＆LT; DIV的风格=显示：块; ID =响应＆LT;？PHP的echo $ ID;＆gt;中＆GT;＆LT; PHP的echo $ avgratep; ？＆GT;＆LT; / DIV＆GT;＆LT; BR＆GT;
＆LT; DIV的风格=显示：块; ID =响应2＆LT;？PHP的echo $ ID;＆gt;中＆GT;额定＆LT;？PHP的echo $ rankcount; ？＆GT;时报＆LT; / DIV＆GT;
＆LT; / DIV＆GT;
＆LT; / DIV＆GT;
＆LT; PHP ENDWHILE; mysqli_close（$连接）; ？＆GT;
＆LT;脚本类型=文/ JavaScript的＆GT;
$（'＃currentslide .rateit）。绑定（额定复位'，功能（五）{
变种RI = $（本）;
VAR值= ri.rateit（'值'）;
变种storyidr = ri.data（'storyidr'）;
ri.rateit（只读，真正的）;
$阿贾克斯（{
数据类型：JSON，
网址：rate.php，
数据：{storyidr：storyidr，值：值}，
键入：POST，
成功：功能（数据）{
$（'＃响应+ storyidr）.replaceWith（平均评级+ data.avg +'/ 10'）;
$（'＃响应2+ storyidr）.replaceWith（'额定'+ data.cnt +'时代'）;
}，
错误：函数（jxhr，味精，ERR）{
$（'＃响应）追加。（'＆LT;李风格=颜色：红色＆GT;+味精+'＆LT; /李＆GT;'）;
}
}）;
}）;
＆LT; / SCRIPT＆GT;
PHP
＆LT; PHP
$ storyidr = $ _ POST ['storyidr'];
$ mysqli的= mysqli_connect（$ DBHOST，$数据库用户名，$ dbpasswd，$数据库名称）或死亡（无法连接到服务器。）;
如果（mysqli_connect_errno（$ mysqli的））
{
回声无法连接到MySQL：。 mysqli_connect_error（）;
}
$ SQL =INSERT INTO评级（storyidr，等级，ENTRY_DATE）VALUES（'$ _ POST [storyidr]'，'$ _ POST [值]'，现在（））;;
$ SQL =SELECT AVG（排名）AS avrank，COUNT（职级）AS countrank从收视率WHERE storyidr = $ storyidr。
如果（$ mysqli-＆GT; multi_query（$ SQL））
{$ mysqli-＆GT; next_result（）;
如果（$结果= $ mysqli-＆GT; store_result（））
{
$数据= mysqli_fetch_assoc（$结果）;
$ avrank = $数据['avrank'];
$ countrank = $数据['countrank'];
$ avrankr =圆（$ avrank，2）;
如果（is_null（$ avrank））{$ avrank =空;}
回声json_en code（阵列（'平均'=＆GT; $ avrankr，'CNT'=＆GT; $ countrank））;
}
}
？＆GT;
解决方案 
您应该只使用 json_en code（）一次，并且只能随声附和该函数的结果。不止一次做无效的JSON：
其他
{
$结果=阵列（）;
$结果['AV'] = $ avrankr;
$结果['CNT'] = $ countrank;
回声json_en code（$结果）;
}
然后，在你的JavaScript，您可以访问 data.av 和 data.cnt 直接：
$（'＃响应+ storyidr）.replaceWith（平均评级+ data.av +'/ 10'）;
$（'＃响应2+ storyidr）.replaceWith（data.cnt）;
您也可以设置在Ajax调用的的dataType 参数所提到的@barell，但通常jQuery将找出答案正确了。
编辑：：要避免你得到未定义错误，你应该做的是这样的：
$结果=阵列（状态=＆GT;「不及格」）;
...
如果 （） {
...
如果（$结果）
{
$结果['状态'] ='成功';
$结果['AV'] = $ avrankr;
$结果['CNT'] = $ countrank;
}
}
回声json_en code（$结果）;
现在，你可以检查 data.status 先在你的AJAX调用的成功回调，并采取相应的操作：
成功：功能（数据）{
如果（data.status ==='失败'）{
//显示警告消息某处，这仅仅是一个例子
警报（找不到任何结果！）;
} 其他 {
$（'＃响应+ storyidr）.replaceWith（平均评级+ data.av +'/ 10'）;
$（'＃响应2+ storyidr）.replaceWith（'额定'+ data.cnt +'时代'）;
}
}，
I'm using a star ratings system to display rating data from SQL. Each item that can be rated has unique identifyer variable $id and each rating in ratings tabl has unique identifyer $storyidr. I would like this script to display:
the average rating the number of times the item has been rated.The values are retirevable but they display on the page together and I can't see how to seperate them. FOr example, for an item that has an average rating of 4 and has been rated 200 times. when user clicks the data returns via AJAX looking like:
For 'response1' 4"200" For 'response2' 4"200"I would like to be able to seperate them to look like:
For 'response1' 4 For 'response2' 200html page
<div id="products" style="">
<div class="rateit" data-storyidr="<?php echo $id; ?>">
</div>
<div class="averagevote">
<div style="display:block;" id="response<?php echo $id; ?>"><?php echo $avgratep; ?></div><br>
<div style="display:block;" id="response2<?php echo $id; ?>">RaTeD <?php echo $rankcount; ?> TiMeS</div>
</div>
</div>
<?php endwhile; mysqli_close($connection); ?>
<script type ="text/javascript">
$('#currentslide .rateit').bind('rated reset', function (e) {
var ri = $(this);
var value = ri.rateit('value');
var storyidr = ri.data('storyidr');
ri.rateit('readonly', true);
$.ajax({
dataType : 'json',
url: 'rate.php',
data: {storyidr: storyidr, value: value},
type: 'POST',
success: function (data) {
$('#response'+storyidr).replaceWith('Avg rating ' + data.avg + '/5');
$('#response2'+storyidr).replaceWith('Rated ' + data.cnt + ' times');
},
error: function (jxhr, msg, err) {
$('#response').append('<li style="color:red">' + msg + '</li>');
}
});
});
</script>
PHP
<?PHP
$storyidr=$_POST['storyidr'];
$mysqli = mysqli_connect($dbhost,$dbusername,$dbpasswd,$database_name) or die ("Couldn't connect to server.");
if (mysqli_connect_errno($mysqli))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql = "INSERT INTO ratings (storyidr, rank, entry_date) VALUES ('$_POST[storyidr]','$_POST[value]',now());";
$sql .= "SELECT AVG(rank) AS avrank, COUNT(rank) AS countrank FROM ratings WHERE storyidr = $storyidr";
if($mysqli->multi_query($sql))
{ $mysqli->next_result();
if ($result = $mysqli->store_result())
{
$data = mysqli_fetch_assoc($result);
$avrank = $data['avrank'];
$countrank = $data['countrank'];
$avrankr = round($avrank,2);
if(is_null($avrank)){$avrank ="null";}
echo json_encode(array('avg' => $avrankr, 'cnt' => $countrank));
}
}
?>
解决方案
You should only use json_encode() once and only echo the result of that function. Doing it more than once invalidates your json:
else
{
$results = array();
$results['av'] = $avrankr;
$results['cnt'] = $countrank;
echo json_encode($results);
}
Then, in your javascript, you can access data.av and data.cnt directly:
$('#response'+storyidr).replaceWith('Avg rating ' + data.av +'/5');
$('#response2'+storyidr).replaceWith(data.cnt);
You could also set the dataType parameter in your ajax call as mentioned by @barell, but normally jQuery will figure that out correctly already.
Edit: To avoid the undefined errors you are getting you should do something like:
$results = array('status' => 'fail');
...
if () {
...
if ($result)
{
$results['status'] = 'success';
$results['av'] = $avrankr;
$results['cnt'] = $countrank;
}
}
echo json_encode($results);
Now you can check for data.status first in the success callback of your ajax call and take the appropriate action:
success: function (data) {
if (data.status === 'fail') {
// show a warning message somewhere, this is just an example
alert('No results found!');
} else {
$('#response'+storyidr).replaceWith('Avg rating ' + data.av + '/5');
$('#response2'+storyidr).replaceWith('RaTeD ' + data.cnt + ' TiMeS');
}
},
