Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in D:\myapacheweb\test\self_info.php on line 7mysql_connect("localhost","root","1");
$query="select id from computer where user ='test2'";
$result=mysql_query($query);
echo "test2 使用了<br>";
while($row = mysql_fetch_array($result))
{
echo $row['id']."开始时间".$row[starttime]."<br>";
}求助啊
$query="select id from computer where user ='test2'";
$result=mysql_query($query);
echo "test2 使用了<br>";
while($row = mysql_fetch_array($result))
{
echo $row['id']."开始时间".$row[starttime]."<br>";
}求助啊
echo $row['id']."开始时间".$row[starttime]."<br>";
改为
echo $row['id']."开始时间".$row['starttime']."<br>";
SQL语句有错误,
或者数据库里没有这个数据造成的、
$result=mysql_query($query);
写成:
$result=mysql_query($query,$conn);
$result=mysql_query($query);
这个改成
$result=mysql_db_query('数据库名',$query);
mysql_select_db("dbname");