更快的mysql查询

Question

Is there a faster way to do this?

$data1 = mysql_query(
 "SELECT * FROM table1 WHERE id='$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;

Cheers!

Answer 1

查看您的查询，我认为您正在寻找这样的东西：

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

Answer 2

通常，如果您只关心匹配的行数，则可以将SELECT *缩短为例如SELECT COUNT(*) 。

Answer 3

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

This single query will produce a one-row result set with both values that you want.

Note that you should use a parameterized query instead of direct variable substitution to avoid SQL injection attacks.

Also, I'm guessing that SUM(type) isn't what you really want to do, since you could calculate it as (num_results * $type) without the second query.

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

一条线：

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