Question

我现在有点卡住了，希望你们给我打电话帮助我。

我的数据库中有一个表格如下：

Game - kickoff             - result
g1   - 2014-12-23 15:00:00 - 1
g2   - 2014-12-23 15:00:00 - 1
g3   - 2014-12-24 15:00:00 - 1
g4   - 2014-12-24 15:00:00 - 1
g5   - 2014-12-25 15:00:00 - 1

我想在我的页面中显示它，所以它看起来像这样：

23-12-2014:
G1 - 15:00
G2 - 15:00

24-12-2014
G3 - 15:00
G4 - 15:00

25-12-2014
G5 - 15:00

所以按日期安排，我可以弄清楚我应该怎么做。

这就是我的地方，但我不知道如何按日期分开。

$tips = $db -> select("SELECT * FROM games");

foreach($tips as $result) {

$game= $result['game'];
$kickoff= $result['kickoff'];
$result_end= $result['result'];

echo $game . $kickoff . $result_end;
}

Answer 1

您可以单独设置日期和时间的格式，以便也可以单独显示它们。然后，您可以保留一个变量来检查日期是否与上一个日期不同。

对查询进行排序也很重要（添加ORDER BY），否则同一日期可能会多次出现。

此代码假定kickoff是时间戳。如果它是一个字符串，您可能需要以另一种方式解析它，但这是一个特定的实现细节。

<?php

$tips = $db -> select("SELECT * FROM games ORDER BY kickoff");

$lastkickoffdate = '';
foreach($tips as $result) {

    $game= $result['game'];
    $kickoff= $result['kickoff'];
    $result_end= $result['result'];

    $kickoffdate = date("Y-m-d", $kickoff);
    $kickofftime = date('H:i', $kickoff);

    if ($kickoffdate !== $lastkickoffdate)
    {
        echo $kickoffdate . '<br>';
    }

    echo "$game - $kickofftime - $result_end";
}

Answer 2

我认为你需要改变

$tips = $db -> select("SELECT * FROM games");

到

$tips = $db -> select("SELECT * FROM games ORDER BY kickoff");

Answer 3

试试此代码

$tips = $db->select("SELECT * FROM games ORDER BY kickoff");

$lastkickoffdate = '';
foreach($tips as $result) {

    $game        = $result['game'];
    $kickoff     = $result['kickoff'];
    $kickoffdate = date("d-m-Y", strtotime($kickoff));
    $kickofftime = date('H:i', strtotime($kickoff));
    $result_end  = $result['result'];

    if($kickoffdate != $lastkickoffdate){
        echo '<p>'.$kickoffdate.'</p>';
    }
    echo ucfirst($game) .' - '. $kickofftime.'<br>';
    $lastkickoffdate = $kickoffdate;
}

Php - 显示按日期排列的sql数据

3 个答案: