Question

我知道$ wins应该是3.因为我在表“rated_teams”的“win”列中有3行，带有整数“1”但是由于某种原因，这段代码不起作用。你能找到问题吗？另外，我知道其中一些已被删除。我会更新整个页面，一旦我得到它至少在工作状态。

<?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>

Answer 1

尝试

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

当你得到像

这样的时候

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

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

Answer 2

您没有设置$ row变量。编辑你的时间。

while ($row = mysql_fetch_array($query))

Answer 3

您需要为计算列提供别名。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>

Answer 4

请以正确的方式编写sql查询。写这样。

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

Answer 5

while ((mysql_fetch_array($query)))   {

应该是

while ($row = mysql_fetch_array($query) )  {

查询不返回任何结果

5 个答案: