我尝试更新表格,但出现此错误,请帮助我
INSERT INTO TBL_Attendance (columnName) VALUES (v1, v2...etc)
SELECT columnName
FROM TBL_Students
WHERE column = 'value'
答案 0 :(得分:0)
您需要将SQL查询视为字符串,并相应地连接变量:
"UPDATE posts SET title = '".$title."', content = '".$content."', date = '".$date."' , likes = '".$likes."', groups = '".$groups."' WHERE id = '".$id."'";
答案 1 :(得分:0)
您的where
子句有2个等号。将其更改为单个等号...而且,date是保留字,因此您需要将其包装在反引号中。
$sqlCommand = "UPDATE posts
SET title = '$title' ,
content = '$content',
`date` = '$date' ,
groups = '$groups'
WHERE id = 1";