数据库结构:
ID | USERID | ENTRYDATE | EXITDATE | PRICE | ROOMID
-----------------------------------------------------
1 | 2 | 2012-10-1 | 2012-10-4| 100 | 1
2 | 2 | 2012-10-1 | 2012-10-4| 100 | 2
3 | 2 | 2012-10-1 | 2012-10-4| 100 | 3
4 | 2 | 2012-10-6 | 2012-10-9| 100 | 4
5 | 2 | 2012-10-6 | 2012-10-9| 100 | 55我需要一些帮助。我想显示此数据库中的一组记录。我希望显示具有相同用户in的所有文件,但将具有相同ENTRYDATE和EXITDATE的文件分组到同一DIV中。例如,结果应该是这样的.
ID | USERID | ENTRYDATE | EXITDATE | PRICE | ROOMID
------------------------------------------------------
1 | 2 | 2012-10-1 | 2012-10-4| 100 | 1
2 | 2 | 2012-10-1 | 2012-10-4| 100 | 2
3 | 2 | 2012-10-1 | 2012-10-4| 100 | 3ID | USERID | ENTRYDATE | EXITDATE | PRICE | ROOMID
----------------------------------------------------
4 | 2 | 2012-10-6 | 2012-10-9| 100 | 4
5 | 2 | 2012-10-6 | 2012-10-9| 100 | 5发布于 2013-01-23 03:10:42
我不确定我是否正确地理解了你的问题。我假设您有一个字符串索引的数组,并且您希望根据ENTRYDATE和EXITDATE值对其进行拆分。可能是这样的:
$currentEntry = $data[0]['ENTRYDATE'];
$currentExit = $data[0]['EXITDATE'];
$splitted = array();
$i = 0;
foreach ($data as $row) {
if ($row['ENTRYDATE'] == $currentEntry && $row['ENTRYDATE'] == $currentExit) {
$splitted[$i] []= $row;
} else {
$currentEntry = $row['ENTRYDATE'];
$currentExit = $row['EXITDATE'];
$i++;
}
}https://stackoverflow.com/questions/14465644
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