ajax php星级系统(仅显示mysql数据库中的平均日期)
我需要一个评级系统,显示在php的平均评级。我完成了评级过程(保存和更新过程)。我只需要用php显示平均评级(使用PHPAJAX评级系统) 当我从数据库中检索数据时,出现了错误。代码如下:ajax php星级系统(仅显示mysql数据库中的平均日期),php,javascript,ajax,Php,Javascript,Ajax,我需要一个评级系统,显示在php的平均评级。我完成了评级过程(保存和更新过程)。我只需要用php显示平均评级(使用PHPAJAX评级系统) 当我从数据库中检索数据时,出现了错误。代码如下: <?php $con = mysql_connect("localhost","root",""); if(!$con){ echo "Connection to the Database Consol
<?php
$con = mysql_connect("localhost","root","");
if(!$con){
echo "Connection to the Database Console was Unsuccessful";
}
$select = mysql_select_db("oilandgas13",$con);
if(!$select){
echo "Connection to the Database was Unsuccessful";
}
$add_coun= "SELECT sum(rating) sum, count(id) count from comments WHERE item_id = $itemID AND status=1";
$result = mysql_query($add_coun,$con);
if(!$result)
{
echo "query was not successfully";
}
$result = mysql_fetch_object($result);
$sum = $result->sum;
$count = $result->count;
$rating = $sum / $count;
echo $rating;
?>
我犯了这样的错误:
警告:mysql\u fetch\u object():在第19行的C:\wamp\www\final work\u apr51\final work\u apr51\calculation.php中,提供的参数不是有效的mysql结果资源
警告:第23行C:\wamp\www\final work\u apr51\final work\u apr51\calculation.php中被零除
也许这能帮到你
$link = mysqli_connect("localhost","mysqlusername","mysqlpassword","dbname");
$rating = mysql_real_escape_string($_GET['id']);
$q = mysqli_query($link,"SELECT * FROM ratings WHERE id='{$rating}'"); //Get our ratings by the page that has rated
//Die if id dont exist!
if(mysqli_num_rows($q) == 0) die("Wrong page id!");
//Select good & bad ratings
$good = mysqli_query($link,"SELECT * FROM ratings WHERE id='{$rating}' AND value ='yes'");
$bad = mysqli_query($link,"SELECT * FROM ratings WHERE id='{$rating}' AND value ='no'");
//Count good & bad ratings
$gcnt = mysqli_num_rows($good);
$bcnt = mysqli_num_rows($bad);
//Calculate
$totalVotes = $gcnt + $bcnt;
if($totalVotes == 0){
echo $totalVotes." votes";
}
if($totalVotes > 0){
echo "<font color='green'>".$totalVotes." votes</font>";
}
if($totalVotes < 0){
echo "<font color='red'>".$totalVotes." votes</font>";
}
$link=mysqli_connect(“localhost”、“mysqlusername”、“mysqlpassword”、“dbname”);
$rating=mysql\u real\u escape\u字符串($\u GET['id']);
$q=mysqli_查询($link,“SELECT*FROM ratings,其中id='{$rating}')//通过已评级的页面获取我们的评级
//如果我不存在就死吧!
如果(mysqli_num_rows($q)==0)死亡(“错误的页面id!”);
//选择好的和坏的评级
$good=mysqli_查询($link,“SELECT*FROM ratings,其中id='{$rating}'和value='yes');
$bad=mysqli_查询($link,“SELECT*FROM ratings,其中id='{$rating}'和value='no');
//计算好的和坏的评分
$gcnt=mysqli\u num\u行($good);
$bcnt=mysqli\u num\u行($bad);
//算计
$TotalVoces=$gcnt+$bcnt;
如果($TotalVoces==0){
echo$TotalVoces.“投票”;
}
如果($TotalVoces>0){
回声“$TotalVoces.”选票;
}
如果($totalvoces<0){
回声“$TotalVoces.”选票;
}
但你的问题是什么?你有什么问题?你尝试过哪些方法不起作用?你应该考虑使用MySQL或PDO而不是贬低的MySQL函数。您还应该添加错误处理来记录您收到的错误。谢谢您没有删除它。投票是真正的MVP。我会把我的留给你的,@OrelBiton