大家好，我是想传个id参数；然后输出有关参数的内容；但显示错误？？
不知道哪里错了，刚刚接触麻烦大家告一下。谢了！！
mysql_result函数不是读取相应字段的内容么??
<?php
$connect = mysql_pconnect("localhost","root","");
mysql_select_db($connect);
$sql = "select * from test where id=".$_GET["id"];
echo $_GET["id"];echo $sql;
$result = mysql_query($sql);
$username1= mysql_result($result,0,"username");
$id1 = mysql_result($result,0,"id");
$password1 = mysql_result($result,0,"password");
$name1 = mysql_result($result,0,"name");
echo "ID:$id1,用户名：$username1,";
echo "密码：$password1,真名：$name1<br>";
mysql_close($connect);

?>
===错误信息
1select * from test where id=1
Warning: mysql_result(): supplied argument is not a valid MySQL result resource in d:\usr\www\html\phpexe\display.php on line 7

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in d:\usr\www\html\phpexe\display.php on line 8

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in d:\usr\www\html\phpexe\display.php on line 9

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in d:\usr\www\html\phpexe\display.php on line 10
ID:,用户名：,密码：,真名：

好象用这种输出,好象也存在上面的问题??
for($i=0;$i<$rows;$i++)
{
mysql_data_seek($result,$i);
$data = mysql_fetch_array($result);
echo "ID:<a href='display.php?id=$data[id]'>$data[id]</a>,用户名：$data[username],";
echo "密码：$data[password],真名：$data[name]<br>";
}
这个究竟是什么原因??

好象用这种输出,好象也存在上面的问题??
for($i=0;$i<$rows;$i++)
{
mysql_data_seek($result,$i);
$data = mysql_fetch_array($result);
echo "ID:<a href='display.php?id=$data[id]'>$data[id]</a>,用户名：$data[username],";
echo "密码：$data[password],真名：$data[name]<br>";
}
这个究竟是什么原因??

如果改成下面这些：
$connect = mysql_pconnect("localhost","root","");
mysql_select_db($connect);
$sql = "select * from test where id=".$_GET["id"];
echo $_GET["id"];echo $sql;
$result = mysql_query($sql);
$rows = @mysql_num_rows($result);
$username1= @mysql_result($result,0,"username");
$id1 = @mysql_result($result,0,"id");
$password1 = @mysql_result($result,0,"password");
$name1 = @mysql_result($result,0,"name");
echo "ID:$id1,用户名：$username1,";
echo "密码：$password1,真名：$name1<br>";
mysql_close($connect);
只输出 id:用户名：密码：真名:
$username1这些变量好象没有东西一样啊！！？？？

昨天晚上3点多,又来看了看这个程序.
终于是发现问题的所在了,mysql_select_db($connect);这句我没有选择数据库.
所以sql语句是执行不出来.看来是太马虎了!!!

....你好努力呀！！
^_^
加油

等下学期,我再学php.你们先学