MySQL的存儲過程從幾秒鐘查詢返回變量的值？

Question

SET @v1 = ''; 
SET @v2 = ''; 

SET @Query = CONCAT('SELECT sum(colName1), sum(colName2) INTO @v1, @v2 FROM tableName WHERE id=1 '); 
PREPARE stmt FROM @Query; 

EXECUTE stmt; 
DEALLOCATE PREPARE stmt; 

SET @Query = CONCAT('SELECT id, name,',@v1,' as value1, ',@v2,' as value2 FROM tableName WHERE id=1 '); 
PREPARE stmt FROM @Query; 

EXECUTE stmt; 
DEALLOCATE PREPARE stmt; 

在讀取值時拋出異常 - 列'值1'不屬於表。

我怎樣才能得到@ v1和v2的@的價值。

請幫幫我。

Answer 1

的@ V2具有空值爲什麼存儲過程提供了錯誤。