连接数据库是没有问题的,是用include("link_db.inc");
插入数据正常,就是调出来有问题。出错信息是:
Warning: Unable to jump to row 0 on MySQL result index 2 in /www/office/gzgrain.com/pic_news/pic_news_info.php on line 6Warning: Unable to jump to row 0 on MySQL result index 2 in /www/office/gzgrain.com/pic_news/pic_news_info.php on line 7Warning: Unable to jump to row 0 on MySQL result index 2 in /www/office/gzgrain.com/pic_news/pic_news_info.php on line 8Warning: Unable to jump to row 0 on MySQL result index 2 in /www/office/gzgrain.com/pic_news/pic_news_info.php on line 9Warning: Unable to jump to row 0 on MySQL result index 2 in /www/office/gzgrain.com/pic_news/pic_news_info.php on line 10
6,7,8,9,10分别就是上面我给出的3-7行
插入数据正常,就是调出来有问题。出错信息是:
Warning: Unable to jump to row 0 on MySQL result index 2 in /www/office/gzgrain.com/pic_news/pic_news_info.php on line 6Warning: Unable to jump to row 0 on MySQL result index 2 in /www/office/gzgrain.com/pic_news/pic_news_info.php on line 7Warning: Unable to jump to row 0 on MySQL result index 2 in /www/office/gzgrain.com/pic_news/pic_news_info.php on line 8Warning: Unable to jump to row 0 on MySQL result index 2 in /www/office/gzgrain.com/pic_news/pic_news_info.php on line 9Warning: Unable to jump to row 0 on MySQL result index 2 in /www/office/gzgrain.com/pic_news/pic_news_info.php on line 10
6,7,8,9,10分别就是上面我给出的3-7行
$result=mysql_db_query("tennic",$query);
后$result所得记录集的行数为0,请检查$id的值,并确保表中有该行
其实我就用$query="select * from pic_news where id='$id'";不能取得提交来这个页面的id值吗?如果不行,用session可以吗?如果用session,请问格式(用session记录和调出session)应该怎么样?
$id=$HTTP_GET_VALS["id"];php4.1.0以上
$id=$_GET["id"];
parse error in /www/office/gzgrain.com/pic_news/pic_news_info.php on line 6
请问是什么原因呢?
应该这样写sql
$query="select * from pic_news where id=$id";