生手上路(关于mysql_fetch_array)
时间:2021-07-01 10:21:17
帮助过:17人阅读
新手上路(关于mysql_fetch_array)
include 'con_db.php';
$sqlsl="SELECT * FROM `phptestdb`.`inputtest`";
$dbdate=mysql_query($sqlsl)or die(mysql_error());
while($list=mysql_fetch_array($dbdate)){
?>
=$list[content]?> |
用户:=$list[user]?> |
时间:=$list[date]?> |
___________________________________________________________________________________________________________________________________________________
测试提示:
undefined constant content - assumed 'content'
Use of undefined constant user - assumed 'user'
Use of undefined constant date - assumed 'date'
才刚上路,求知道的人指点,万分感谢!
------解决思路----------------------=$list['content']?>
加上引号,其他类同