Question

我是PHP新手，我有一个test project：

我想让代码更高效，因为生成需要一些时间。非常感谢，Nat

它使用的代码如下：

    $result1 = mysql_fetch_assoc(mysql_query("SELECT COUNT(*) AS `Rows`, `1` FROM `things` GROUP BY  `1` ORDER BY RAND() LIMIT 0 , 1"));
    $result2 = mysql_fetch_assoc(mysql_query("SELECT COUNT(*) AS `Rows`, `2` FROM `things` GROUP BY  `2` ORDER BY RAND() LIMIT 0 , 1"));
    $result3 = mysql_fetch_assoc(mysql_query("SELECT COUNT(*) AS `Rows`, `3` FROM `things` GROUP BY  `3` ORDER BY RAND() LIMIT 0 , 1"));
    $result4 = mysql_fetch_assoc(mysql_query("SELECT COUNT(*) AS `Rows`, `4` FROM `things` GROUP BY  `4` ORDER BY RAND() LIMIT 0 , 1"));

    echo $result1["1"];
    echo " ";
    echo $result2["2"];
    echo " ";
    echo $result3["3"];
    echo " ";
    echo $result4["4"];

    mysql_close($con);      
    ?>

Answer 1

我假设是善意的＆amp;使用您想要使用的方法以最佳方式清理代码。

<?

$count = 1;
$results = array();
for ($count; $count <= 4; $count++) {
  $query = sprintf("SELECT COUNT(*) AS `Rows`, `%d` FROM `things` GROUP BY  `1` ORDER BY RAND() LIMIT 1", $count)
  $results[] = mysql_fetch_assoc(mysql_query($query));
}

if (!empty($results)) {
  echo exlplode(" ", $results);
}

?>

如何使这个PHP / MYSQL代码更高效？

1 个答案: