选择进入错误

时间:2016-05-11 18:34:58

标签: sql

SELECT 
xfqti_virtuemart_products_pt_pt.virtuemart_product_id,   
xfqti_virtuemart_product_medias.virtuemart_media_id 
INTO #tempTable 
FROM xfqti_virtuemart_products_pt_pt

给出语法错误,我即将脱掉头发

2 个答案:

答案 0 :(得分:0)

插入并选择这些sintax

  Insert into your_Table (col1,col2)
  SELECT 
  xfqti_virtuemart_products_pt_pt.virtuemart_product_id,   
 xfqti_virtuemart_product_medias.virtuemart_media_id 

  FROM xfqti_virtuemart_products_pt_pt

for create table

  Create  your_Table as 
  SELECT 
  xfqti_virtuemart_products_pt_pt.virtuemart_product_id,   
 xfqti_virtuemart_product_medias.virtuemart_media_id 

  FROM xfqti_virtuemart_products_pt_pt

答案 1 :(得分:0)

作为Virtuemart,我猜这是一个MySQL数据库。如果是这样,创建临时表的正确语法是:

CREATE TEMPORARY TABLE IF NOT EXISTS tempTableName AS
(
   SELECT field1, field2
   FROM yourtable;
)

话虽如此,您的SELECT语句有两个不同表中的两个字段,但在语句的FROM子句中只提到了其中一个表。他们真的应该在那里和JOINed。类似的东西:

CREATE TEMPORARY TABLE IF NOT EXISTS tempTableName AS
(
   SELECT 
      t1.virtuemart_product_id,   
      t2.virtuemart_media_id 
   FROM 
      xfqti_virtuemart_products_pt_pt as t1 
      INNER JOIN xfqti_virtuemart_product_medias as t2 ON
          t1.product_id = t2.product_id
)

或者某种东西......我无法看到你的牌桌,而且自从我使用Virtuemart以来已经好几年了,所以它只是猜测牌桌关系。