Question

我有以下sql查询，我想用PHP运行。为什么不起作用？

IF EXISTS(SELECT * FROM content WHERE site_id ='". $id ."') BEGIN
UPDATE content SET
titel = '". htmlentities($connect->real_escape_string($data[0])) ."',
content = '". htmlentities($connect->real_escape_string($data[1])) ."'
WHERE site_id ='". $id ."'
END
ELSE
BEGIN
INSERT INTO content (site_id, titel, content)
VALUES (
'". $id ."',
'". htmlentities($connect->real_escape_string($data[0])) ."',
'". htmlentities($connect->real_escape_string($data[1])) ."'
)
END

您的SQL语法有错误;查看与您的MySQL服务器版本对应的手册，以便在'IF EXISTS（SELECT * FROM content WHERE site_id ='1'）附近使用正确的语法。在第1行开始更新内容SET tit'

Answer 1

我认为site_id是表格中的唯一键。使用@Saty提出的解决方案

INSERT INTO content (site_id, titel, content)
VALUES (
  '". $id ."',
  '". htmlentities($connect->real_escape_string($data[0])) ."',
  '". htmlentities($connect->real_escape_string($data[1])) ."'
)
ON DUPLICATE UPDATE 
  titel = '". htmlentities($connect->real_escape_string($data[0])) ."',
  content = '". htmlentities($connect->real_escape_string($data[1])) ."';

SQL查询IF EXISTS无法正常工作

1 个答案: