使用inner join插入表时遇到的问题

INSERT INTO target_table (column1, column2, ...)
SELECT source_column1, source_column2, ...
FROM source_table
WHERE EXISTS (
    SELECT 1
    FROM another_table
    WHERE source_table.join_key = another_table.join_key
);

-- 创建一个临时表来存储需要插入的数据
CREATE TEMPORARY TABLE temp_table AS
SELECT source_column1, source_column2, ...
FROM source_table
INNER JOIN another_table ON source_table.join_key = another_table.join_key;

-- 将临时表中的数据插入目标表
INSERT INTO target_table (column1, column2, ...)
SELECT column1, column2, ...
FROM temp_table;

-- 删除临时表
DROP TEMPORARY TABLE temp_table;

WITH joined_data AS (
    SELECT source_column1, source_column2, ...
    FROM source_table
    INNER JOIN another_table ON source_table.join_key = another_table.join_key
)
INSERT INTO target_table (column1, column2, ...)
SELECT column1, column2, ...
FROM joined_data;

-- 使用子查询的方法
INSERT INTO target_table (id, name, age)
SELECT s.id, s.name, s.age
FROM source_table s
WHERE EXISTS (
    SELECT 1
    FROM another_table a
    WHERE s.join_key = a.join_key
);

-- 使用临时表的方法
CREATE TEMPORARY TABLE temp_table AS
SELECT s.id, s.name, s.age
FROM source_table s
INNER JOIN another_table a ON s.join_key = a.join_key;

INSERT INTO target_table (id, name, age)
SELECT id, name, age
FROM temp_table;

DROP TEMPORARY TABLE temp_table;

-- 使用CTE的方法
WITH joined_data AS (
    SELECT s.id, s.name, s.age
    FROM source_table s
    INNER JOIN another_table a ON s.join_key = a.join_key
)
INSERT INTO target_table (id, name, age)
SELECT id, name, age
FROM joined_data;