mysql表存在存儲過程邏輯

Question

DECLARE FoundCount INT; 
    DECLARE db_Name VARCHAR(255); 

SET @Project_List_val = CONCAT(Project_Number_val, '_List'); 
SET @db_Name='test_rs'; 

    SET @sql1 = CONCAT('SELECT COUNT(1) INTO', @FoundCount ,' 
    FROM information_schema.tables 
    WHERE table_schema =',@db_Name,' 
    AND table_name =', @Project_List_val); 

PREPARE stmt FROM @sql1; 
EXECUTE stmt; 
DEALLOCATE PREPARE stmt; 

    IF FoundCount = 1 THEN 
SET @sql = CONCAT(' INSERT INTO test (Panel_Id) 
        SELECT Panel_Id 
        FROM ', @Project_List_val); 

PREPARE stmt FROM @sql; 
EXECUTE stmt; 
DEALLOCATE PREPARE stmt; 
END IF; 

這個存儲過程執行得很好。但是當試圖運行它時，它顯示的是錯誤。

1064 - 您的SQL語法錯誤;檢查與您的MySQL服務器版本相對應的手冊，以便在第1行的'NULL'附近使用正確的語法。

如果指定的表（@Project_List_val）存在，我需要運行第二個查詢。這樣@ FoundCount = 1這樣

Answer 1

檢查條件嘗試此

- DECLARE FoundCount INT;

DECLARE db_Name VARCHAR(255); 

SET @Project_List_val = CONCAT(Project_Number_val, '_List'); 
SET @db_Name='test_rs'; 

    SET @sql1 = CONCAT('SELECT COUNT(1) INTO ', @FoundCount ,' 
    FROM information_schema.tables 
    WHERE table_schema = ',@db_Name,' 
    AND table_name =', @Project_List_val); 

PREPARE stmt FROM @sql1; 
EXECUTE stmt; 
DEALLOCATE PREPARE stmt; 

    IF @FoundCount = 1 THEN 
SET @sql = CONCAT(' INSERT INTO test (Panel_Id) 
        SELECT Panel_Id 
        FROM ', @Project_List_val); 

PREPARE stmt FROM @sql; 
EXECUTE stmt; 
DEALLOCATE PREPARE stmt; 
END IF; 

使用這樣的：

SET @sql1 = CONCAT('SELECT COUNT(1) INTO ', @FoundCount ,' 
    FROM information_schema.tables 
    WHERE table_schema = ',@db_Name,' 
    AND table_name =', @Project_List_val); 

我有給空間後進入這個.copy和解決的是電話我

Answer 2

。對於@db_Name和@Project_List_Val語句中的那些變量缺少單引號