Question

SQL查询：

Select count(distinct signature)
from (Select table_name
      from information_schema.tables
      where table_schema="Log" and table_name like "%ERROR_%")
AS mytable;

预期输出：不同的＆＃34;签名＆＃34;来自所有表格。

实际输出：未知列＆＃39;签名＆＃39;在＆＃39;字段列表＆＃39;;

注意：我的所有表都有签名和字段名称

Answer 1

好吧，你的子查询就像这样开始：

Select table_name from information_schema.tables

因此您可以使用的单个列是“table_name”。

因此，如果您尝试从该子查询中选择签名，它会告诉您没有这样的列。

您可能想要做这样的事情（伪代码）：

query(
    Select table_name 
    from information_schema.tables 
    where table_schema="Log" and table_name like "%ERROR_%"
)
$sql = '';
$sep = '';
while ($row = fetch_assoc()) {
    $sql .= $sep."SELECT signature FROM $row[table_name] ";
    $sep = 'UNION ALL ';
}
$sql = "SELECT COUNT(DISTINCT signature) as c FROM ($sql)";
query($sql);
$row = fetch_assoc()
echo "There are $row[c] distinct signatures.<br />\n";

Answer 2

看起来你在mysql端尝试动态查询。请检查一下......

-- Step1. prepare your select query here
set @sqlQuery := 
(select
 GROUP_CONCAT(
  concat(' SELECT \'',table_name,'\' as col1,  COUNT(*) as countItem FROM ', table_name)
  SEPARATOR ' UNION ALL '
)  from information_schema.tables as t
where t.table_schema="Log" and t.table_name like "%ERROR_%");

-- Step2. execute the dynamic query
prepare stmt from @sqlQuery;
execute stmt;
drop prepare stmt;

查询以计算多个表上的字段上的不同值

2 个答案: