MySQL 错误：查询未提供正确数据

Question

I do a mySQL query to get some data and then (for the purpose of debugging) print it out. In this particular sample there are 5 rows of data and each room_id in the database table has a value. However the print-out only shows the room_id of the first row.

$query_rooms = "SELECT room_id FROM lh_rooms WHERE hid = '$hid'";
$rooms = mysql_query($query_rooms, $MySQL) or die(mysql_error());
$row_rooms = mysql_fetch_assoc($rooms);
$numrows = mysql_num_rows($rooms);
    $i = 0;
while ($i < $numrows) {
    $room_id=$row_rooms['room_id'][$i];
echo $i." - ".$room_id."<br><br>";
   ++$i;
}

0 - 2

1 -

2 -

3 -

4 -

Can someone explain what is happening

Answer 1

You are fetching multiple rows.

So, you need to loop over the result set instead of fetching just one time.

Corrected code:

$query_rooms = "SELECT room_id FROM lh_rooms WHERE hid = '$hid'";
$rooms = mysql_query($query_rooms, $MySQL) or die(mysql_error());
$i=0;
while($row_rooms =  mysql_fetch_assoc($rooms)) {
    $room_id=$row_rooms['room_id'];
    echo $i." - ".$room_id."<br><br>";
    ++$i;
}

Note: Never use mysql_ , they are deprecated and will be removed in the upcoming versions. Use mysqli_ or PDO instead.

Answer 2

Try like this

$query_rooms = "SELECT room_id FROM lh_rooms WHERE hid = '$hid'";
$rooms = mysql_query($query_rooms, $MySQL) or die(mysql_error());
 $i = 0;
while ($row_rooms = mysql_fetch_assoc($rooms)) {
$room_id=$row_rooms['room_id'];
 echo $i." - ".$room_id."<br><br>";
$i++;
}

You are looping $i instead of looping the $row_rooms.