Question

我有一个名为events的mysql表，其中包含以下字段：id，date和name。日期字段的格式为yyyy-mm-dd hh :: mm：ss edit：表示它采用日期时间格式

我想在白天对活动进行分组，我不知道如何处理这个问题 - 有没有办法只从田间选择月份和日期？或者在我选择所有“事件”之后我应该使用PHP

我的最终目标是拥有这样的东西：

March 10th: 
  event1, 
  event2
March 11th: 
  event4, 
  event5

我找到MySQL select using datetime, group by date only，但我不确定如何实现它：

SELECT DATE_FORMAT(date, '%H%i'), DATE_FORMAT(date, '%M %D'), name FROM events ORDER BY date

谢谢！

编辑：

最终使用了这个：

$sql = "select team1, team2, DATE_FORMAT(date,'%Y-%m-%d') as created_day FROM games WHERE attack = '1' GROUP BY created_day";
    $result = mysql_query($sql);
    $curDate = "";

    while (list($team1, $team2, $date) = mysql_fetch_row($result))
    {
      if ($date != $curDate)
      {
        echo "$date --------\n";
        $curDate = $date;
      }

      echo "game data: $team1 $team2";
    }

Answer 1

如果您使用group by，则不会获得一行。{p>因此，通过Group By AFAIK无法实现您想要的方式。

$query = "SELECT distinct(DATE_FORMAT(date, '%M %D')) as d FROM yourtable";
$result = mysql_query($query);
while($row = mysql_fetch_assoc($result)) {
   echo $row['d']
   $sql = "SELECT * FROM yourtable WHERE DATE_FORMAT(date, '%M %D')='$row[d]'";
   $rs = mysql_query($query);
   while($r = mysql_fetch_assoc($rs)) {   
      echo "event";
   }
}

Answer 2

你确实应该使用php来完成这项工作。但由于大多数当前系统都是显示器中的sepate逻辑，我只使用一次传递而不是（NUMBER OF DAYS + 1）个SELECT，并准备一个我稍后可以重复用于显示的数组。

$query = "SELECT DATE_FORMAT(date, '%M %D') as d, name FROM yourtable ORDER BY date";
$foo=array();
$result = mysql_query($query);
while($row = mysql_fetch_assoc($result)) {
    //some logic to test if it's safe to add the name
    $foo[$row['d']][]=$row['name'];
}

然后当我需要它时（通过模板或“视图”）

foreach($foo as $date => $events) {
    echo $date . ":\n\t";          
    echo implode(",\n\t", $events);
    echo "\n";
}

因此它符合您为自己设定的格式。

希望有所帮助

Answer 3

我想从那个问题/答案中，你可以得到类似的东西

March 10th, event1
March 10th, event2
March 11th, event4
March 11th, event5

它并不真正将“日期”组合为您的愿望，但我认为您可以使用php继续此结果。

Answer 4

我同意Kharaone，与显示器分开的逻辑。话虽这么说，我认为类似于这个查询的东西可能就是你要找的东西：

SELECT A FROM 
(
    SELECT DATE_FORMAT(date,'%M %D:') AS A, DATE(date) AS B, 1 AS C FROM games GROUP BY DATE(date)
        UNION ALL
    SELECT name AS A, DATE(date) AS B, 2 AS C FROM games
) X
ORDER BY B, C ASC;

php mysql group by yyyy-mm-dd格式的日期

4 个答案: