PHP PDO - Using MySQL Variables(PHP PDO - 使用 MySQL 变量)
问题描述
我正在尝试使用 PDO 在 PHP 中运行查询.该查询在顶部有一些变量来确定排名,除了在 $sql 中使用 SET @var 时,它返回一个空行集.但是,如果我删除有问题的 SQL,它会返回正常.
I'm trying to run a query in PHP using PDO. The query has some variables at the top to determine a rank, except the when using the SET @var in the $sql, it returns an empty rowset. If I remove the offending SQL however, it returns fine.
我不想在脚本中返回 @prev_value、@rank_count 或 @rank_increasing,只返回它在 SELECT 中创建的排名.
I don't want to return @prev_value, @rank_count or @rank_increasing in my script, only the rank it creates in the SELECT.
你能告诉我我做错了什么吗?
Can you let me know what I am doing wrong please?
谢谢
$sql = "
SET @prev_value = NULL;
SET @rank_count = 0;
SET @rank_increasing = 0;
SELECT a.*
, @rank_increasing := @rank_increasing + 1 AS row_num
, CASE
WHEN @prev_value = score
THEN @rank_count
WHEN @prev_value := score
THEN @rank_count := @rank_increasing
END AS rank
FROM (
-- INLINE VIEW --
) a
";
try {
$sth = $dbh->prepare($sql);
$sth->execute(array($var1, $var2));
return $sth->fetchAll(PDO::FETCH_ASSOC);
} catch (Exception $e) {
return $e;
}
推荐答案
在这里找到解决方案:https://stackoverflow.com/a/4685040/1266457
谢谢:)
修复:
// Prepare and execute the variables first
$sql = "
SET @prev_value = NULL;
SET @rank_count = 0;
SET @rank_increasing = 0;
";
$sth = $dbh->prepare($sql);
$sth->execute();
// Run the main query
$sql = "
SELECT a.*
, @rank_increasing := @rank_increasing + 1 AS row_num
, CASE
WHEN @prev_value = score
THEN @rank_count
WHEN @prev_value := score
THEN @rank_count := @rank_increasing
END AS rank
FROM (
-- INLINE VIEW --
) a
"; ...
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本文标题为:PHP PDO - 使用 MySQL 变量
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