$query = "SELECT * FROM student_info WHERE name = '".$login_

来源：学生作业帮编辑：拍题作业网作业帮分类：综合作业时间：2024/04/28 02:09:19

$query = "SELECT * FROM student_info WHERE name = '".$login_name."'"; $result = mysql_query($query)
$query = "SELECT * FROM student_info WHERE name = '".$login_name."'";
\x05$result = mysql_query($query) or die("Could not execute query.");
WHERE name = '".$login_name."'" 我不把这些打上的话.result会运行前半段.有了where后就是不执行这个语句了.

一、不执行这个语句是什么意思?MySQL错误?还是没有记录?
把$result = mysql_query($query) or die("Could not execute query.");改成
$result = mysql_query($query) or die(mysql_error());
看看MySQL报什么错.
二、把$query = "SELECT * FROM student_info WHERE name = '".$login_name."'";
打印出来看看有没有问题,
echo $query = "SELECT * FROM student_info WHERE name = '".$login_name."'";
如果不能目视出错误,把它放在mysql命令里面查询一下看看,报什么错.如果不报错,说明没有问题,应该是没有该条记录.

$query = "SELECT * FROM student_info WHERE name = '".$login_ $query = mysql_query( select * from 表名 where id='$_GET[id]' SELECT name FROM [BookStore].[dbo].[user] WHERE userid=?AND Query query = session.createQuery("from User"); $id="select id from account where name='".$_SESSION['login'] string strsql1 ="select * from product where name=session[pr PHP数据库的mysql_query("SELECT * FROM list WHERE name=这里怎么写能让nam 英语翻译select top 4 * from shop_books where anclassid = (select select * from book where 1=1 select * from student where name like '%[^ja,as,on]%' select id,name from student where id in ( select id from sco String sql = "select * from tb_consumer where account='" + a