Question

有更快的方法吗？

$data1 = mysql_query(
 "SELECT * FROM table1 WHERE id='$id' AND type='$type'"
) or die(mysql_error()); 

$num_results = mysql_num_rows($data1); 
$data2 = mysql_query(
 "SELECT sum(type) as total_type FROM table1 WHERE id='$id' AND type='$type'"
) or die(mysql_error()); 

while($info = mysql_fetch_array( $data2 )){
    $count = $info['total_type'];
} 
$total = number_format(($count/$num_results), 2, ',', ' ');
echo $total;

干杯！

Answer 1

查看您的查询，我认为您正在寻找类似的内容：

SELECT SUM(type) / COUNT(*) FROM table1 WHERE ...

Answer 2

一般情况下：SELECT *可以“缩短”为例如SELECT COUNT(*)，如果你关心的只是匹配的行数。

Answer 3

 SELECT COUNT(*) AS num_results, SUM(type) AS total_type FROM table1
    WHERE id = $id and type = $type

此单个查询将生成包含所需值的单行结果集。

请注意，您应该使用参数化查询而不是直接变量替换来避免SQL注入攻击。

另外，我猜测SUM（类型）不是你真正想做的，因为你可以在没有第二个查询的情况下将其计算为（num_results * $ type）。

Answer 4

$data1 = mysql_query("SELECT sum(type) as total_type,count(*) as num_rows FROM table1 WHERE id='$id' AND type='$type'"
) or die(mysql_error()); 
$info = mysql_fetch_array( $data1 );
$count = $info['total_type'];
$num_results = $info['num_rows'];
$total = ($count/$num_results); 
echo $total;

Answer 5

一行：

echo number_format(mysql_result(mysql_query("SELECT SUM(type) / COUNT(*) FROM table1 WHRE id = $id AND type = '$type'"), 0), 2, ',', ' ');

更快的mysql查询

5 个答案: