PHP使用appserv-win32-2.5.9 ，测试如下程序出现如下错误，
感觉程序没什么错误，为什么会提示下面的错误信息啊？
appserv-win32-2.5.9 是我刚刚装的，请高手解决
一、程序部分：
<?
$mysql_server_name = "localhost";
$mysql_username = "root";
$mysql_password = "123454";
$mysql_database = "database";
$sql = "slecte * from table";
$conn = mysql_connect($mysql_server_name,$mysql_username,$mysql_password);
$result = mysql_db_query($mysql_database,$sql,$conn);
$row = mysql_fetch_row($result);
echo "<table border = 1 cellspacing = 0 cellpadding = 0>\n";
echo "<tr>\n";
for ($i = 0;$i < mysql_num_fields($result);$i++)
{
echo "<td nowrap>" . mysql_field_name($result,$i) . "</td>\n";
}
echo "</tr>\n";
mysql_data_seek($result,0);
while($row = mysql_fetch_row($result))
{
echo "<tr>\n";
for($i = 0;$i < mysql_num_fields($result);$i++)
{
echo "<td nowrap>$row[$i]</td>\n";
}
echo "</tr>\n";
}
echo "</table>";
mysql_free_result($result);
?>
二、相应错误提示
Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in E:\AppServ\www\sql.php on line 19
Warning: mysql_num_fields(): supplied argument is not a valid MySQL result resource in E:\AppServ\www\sql.php on line 22
Warning: mysql_data_seek(): supplied argument is not a valid MySQL result resource in E:\AppServ\www\sql.php on line 28
Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in E:\AppServ\www\sql.php on line 29
Warning: mysql_free_result(): supplied argument is not a valid MySQL result resource in E:\AppServ\www\sql.php on line 42
补充：二、程序部分
<?
$mysql_server_name = "localhost";
$mysql_username = "root";
$mysql_password = "123454";
$mysql_database = "mydatabase";
$sql = "slecte * from mytable where id = '1'";
$conn = mysql_connect($mysql_server_name,$mysql_username,$mysql_password);
//$result = mysql_db_query($mysql_database,$sql,$conn);
//选取您要处理的数据库
mysql_select_db($mysql_database,$conn);
//进行查询
$result = mysql_query($sql);
if($conn)
{
echo "连接成功！";
}
else
{
echo "连接失败！";
}
if ($result)
{
echo "连接成功！";
}
else
{
echo "程序出错！";
}
mysql_close($conn); //关闭连接
?>
测试结果仍然是：
连接成功！程序出错！
看来主要原因还是在这里，请高手帮忙解决啊？我是新手，确实搞不懂了啊追问：
太感谢，测试成功了！
?
测试了，变成了下面的错误：Warning: Wrong parameter count for mysql_select_db() in E:\AppServ\www\sql.php on line 19
Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in E:\AppServ\www\sql.php on line 21
测试了，错误提示还是没任何改变。
$sql = "select * from `table`";
答案：一个明显的错误：$sql = "slecte * from table"; 改成$sql = "select * from table";
其他程序，我仔细验证，没任何错误，您试试吧。
其他：hgj $result = mysql_db_query($mysql_database,$sql,$conn);
改成：
$result = mysql_select_db($mysql_database,$sql,$conn);
试试？ $sql = "slecte * from table";
table 使用了自带关键字，修改一下表名。 $DB_Server = "localhost";
$DB_Username = "root";
$DB_Password = "1234556";
$DB_DBName = "jishigou";
$Connect = @mysql_connect($DB_Server, $DB_Username, $DB_Password) or die("Couldn't connect.");
mysql_query('set names "gbk"');
mysql_select_db('jishigou');
链接数据库换成这种 $sql = "slecte * from table"; 应该是select。table是自己的表名，用户名密码什么的就不说了。
题目的错误在于一条资源都没有获取到，所以错误。 上一个：福州php程序员 转正后 工资大概在多少？
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