我有下一个代码:
SET @rownum=0;
UPDATE product_images AS t, (SELECT @rownum:=@rownum+1 rownum, id, rel
FROM product_images WHERE product_id='227') AS r
SET t.rel = r.rownum
WHERE t.id = r.id
这在phpmyadmin中工作得很好
但是......下一个代码(女巫实际上是相同的)但放在PHP代码中
mysql_query ("
SET @rownum=0;
UPDATE product_images AS t,
(SELECT @rownum:=@rownum+1 rownum, product_images.*
FROM product_images WHERE product_id='$pid') AS r
SET t.rel = r.rownum WHERE t.id = r.id ") or die(mysql_error());
GIVES ME ERROR:""你的SQL语法有错误; 检查与MySQL服务器版本对应的手册,以便在'UPDATE product_images AS t附近使用正确的语法(SELECT @rownum:= @ rownum + 1 rownum,product_images.*'at line 1""
请帮忙.谢谢.