0

我知道 $wins 应该是 3。因为我在表“rated_teams”的“win”列中有 3 行整数“1”但由于某种原因，这段代码不起作用。请问你能找到问题吗？另外，我知道其中一些已被废弃。我会更新整个页面，一旦我得到它至少处于工作状态。

<?php
$sql = "SELECT SUM(win) FROM rated_teams WHERE server='$server' AND name='$myteam'";
$query = mysql_query($sql, $con) 
or die('A error occured: ' . mysql_error());    
while ((mysql_fetch_array($query)))     {                           
$wins = $row['SUM(win)'];                    
}
?>
<h3>Total Wins: <?php echo $wins?> </h3>
4

5 回答 5

2

尝试

$sql = "SELECT SUM(win) as sum FROM rated_teams WHERE server='$server' AND name='$myteam'";

当你得到像

while ($row = mysql_fetch_array($query)) {                           
    $wins = $row['sum'];                    
}

我的建议是尽量避免mysql_*使用函数，因为它们已被弃用。而是使用mysqli_*函数或PDO statements.

于 2013-08-03T05:11:17.890 回答
0

您不设置 $row 变量。编辑你的时间。

while ($row = mysql_fetch_array($query))
于 2013-08-03T05:12:49.463 回答
0

您需要为计算列指定别名。试试这个：

<?php
$sql = "SELECT SUM(win) as sumwin FROM rated_teams WHERE server='$server' AND name='$myteam'";
$query = mysql_query($sql, $con) or die('A error occured: ' . mysql_error());    
while ($row = mysql_fetch_array($query))     {                           
$wins = $row['sumwin'];                    
}  
?>
<h3>Total Wins: <?php echo $wins?> </h3>
于 2013-08-03T05:12:57.823 回答
0

请以正确的方式编写sql查询。这样写。

$sql = "SELECT SUM(win) as sumwin FROM rated_teams WHERE server='".$server."' AND name='".$myteam."'";
于 2013-08-03T05:19:18.120 回答
0 
while ((mysql_fetch_array($query)))   {

应该

while ($row = mysql_fetch_array($query) )  {
于 2013-08-03T05:36:51.493 回答