如何将查询结果输入到数组中,结果应该比较每个结果并合并为一个数组。
我有3个查询,我不知道我的编码有什么问题,我最好还是不够所以请各位帮助我,特别是sql和php的专家。
这是我的代码:
<?php
$a = $p['id'];
$query1 = mysql_query("SELECT id FROM prereservation where '$arrival' BETWEEN arrival and departure and room_id ='$a' and status = 'active'");
$rows1 = mysql_fetch_array($query1);
$query2 = mysql_query("SELECT id FROM prereservation where '$departure' BETWEEN arrival and departure and room_id ='$a' and status = 'active'");
$rows2 = mysql_fetch_array($query2);
$query3 = mysql_query("SELECT id FROM prereservation where arrival > '$arrival' and departure < '$departure' and room_id ='$a' and status = 'active'");
$rows3 = mysql_fetch_array($query3);
$sample = array_unique(array_merge($rows1, $rows2, $rows3));
?>
,显示如下:
<select id="select" name="qty[]" style=" width:50px;" onchange="checkall()">
<option value="0"></option>
<? $counter = 1; ?>
<? while ($counter <= ($p['qty']) - $????????) { ?>
<option value="<?php echo $counter ?>"><?php echo $counter ?></option>
<? $counter++;
}?>
</select>
所有我想要的例如输出是这样的:
$rows1 = array(id= 48, 55, 51, 53)
$rows2 = array(id= 48, 49, 51, 52)
$rows3 = array(id= 48, 49, 50, 51)
$sample = array_unique(array_merge($rows1, $rows2, $rows3))
所以样本的输出是这样的:
$sample = array(id= 48, 49, 50, 51, 52, 53)
然后在表格保留中,列为:' id ','到达','离开',' room_id < / strong>','数量'和'状态'。
在我的表格中保留每个 ID 不同 * qty *,所以我想总和 数量的数量的数量的数据:
id | qty
48 | 2
49 | 1
50 | 1
51 | 3
52 | 1
53 | 3
$total = sum(qty); which is 11
你们可以帮助我如何创建一个查询来获取每个 qty 的 总和 的 ID 的?另请检查第一个 3 查询,因为我知道每个查询都有错误。获取 array_merge 和 array_unique 。 非常感谢你们。
<select id="select" name="qty[]" style=" width:50px;" onchange="checkall()">
<option value="0"></option>
<? $counter = 1; ?>
<? while ($counter <= ($p['qty']) - $total) { ?>
<option value="<?php echo $counter ?>"><?php echo $counter ?></option>
<? $counter++;
}?>
</select>
这是我的数据库:
table rooms
id | type | qty
---|---------|-----
11 | single | 50
12 | double | 50
13 | deluxe | 20
table preservation
id | arrival | departure | room_id | qty | status
---|-------------|-------------|---------|-----|-------
48 | 05/11/2012 | 11/11/2012 | 11 | 2 | active
49 | 06/11/2012 | 11/11/2012 | 11 | 1 | active
50 | 06/11/2012 | 08/11/2012 | 11 | 1 | active
51 | 05/11/2012 | 07/11/2012 | 11 | 3 | active
52 | 06/11/2012 | 09/11/2012 | 11 | 1 | active
53 | 07/11/2012 | 07/11/2012 | 11 | 3 | active
所以在我的展示中,我有一个来自桌面房间(id = 11)的精选标签,选项是数量,扣除 <表保存(room_id = 11)
数量的强>总和答案 0 :(得分:0)
使用一个SQL语句进行选择,
"SELECT id, SUM(qty) FROM prereservation GROUP BY qty WHERE ( ( '$arrival' BETWEEN arrival and departure and room_id ='$a' ) OR ( '$departure' BETWEEN arrival and departure and room_id ='$a' ) OR ( arrival > '$arrival' and departure < '$departure' and room_id ='$a' ) ) AND status = 'active'";
编辑:
好吧这样的事情呢?
SELECT *, (SELECT SUM(preservation.qty) as sum FROM preservation WHERE rooms.id = preservation.room_id GROUP BY preservation.room_id ) as qty_reserved, rooms.qty - (SELECT SUM(preservation.qty) as sum FROM preservation WHERE rooms.id = preservation.room_id GROUP BY preservation.room_id ) as qty_available FROM rooms;
进入数据库一次,获得所需的所有信息并在整个应用程序中重复使用它会更加高效。如果仅显示这些变量,则不应该在php中使用任何逻辑。