用符号（MySQL/PHP）替换DB查询中的冗余日期

我有一个数据库查询，它按时间顺序显示历史事件列表，如下所示：

(URL = MySite/Calendar/January_1)<br>
On this day in history...<br>
1968 - A volcano erupted.<br>
1968 - A country was invaded.<br>
1968 - Someone had a hit song.<br>
1970 - A famous person was born.

1968 - A volcano erupted.<br>
&#8226; A country was invaded.<br>
&#8226; Someone had a hit song.<br>
1970 - A famous person was born.

然后，我显示了一个历史事件列表，像这样

echo join( $Brief, '<br>' );

---

不管怎样，我的目标是一样的；用某种“丁巴”或符号替换多余的日期（年）。

看起来您需要在输出前对结果集进行分组：

$events = array();

while($row = $stm->fetch()){
    $year = $row['Year']; // current year

    if(!isset($events[$year]){   // if no such group
        $events[$year] = array();   // create group
    }

    $events[$year][] = $row['Brief']; // add data to year group
}

// Output:

foreach($events as $year => $event){
    echo $year, ':<br>'; // show year;

    foreach($event as $data){
        echo $data, '<br>';   // show row;
    }
}

$events=array（）；
而（$row=$stm->fetch（））{
$year=$row['year']；//当前年份
如果（！isset（$events[$year]）{//如果没有这样的组
$events[$year]=array（）；//创建组
}
$events[$year][=$row['short']；//将数据添加到年份组
}
//输出：
foreach（$events作为$year=>$event）{
echo$year'：
；//显示年份；
foreach（$event as$data）{
echo$data，“
”；//显示行；
}
}

此外，您还可以轻松地更改模型的输出：

foreach($events as $year => $event){
    echo $year, ' &#8226; ', implode('<br>&#8226; ', $event);
}

foreach（$events作为$year=>$event）{
echo$year，•；'，内爆（“
•；”，$event）；
}

---

另外，您不需要在字符串的每一端连接

”

。

使用一个变量来保存前一行中的年份。每次通过循环，检查当前年份是否与变量相同；如果是，则显示丁巴特而不是年份。这不是很简单吗？对您来说很简单；我仍在试图找出如何照你的建议去做。也许我玩一会儿后会得到它。（好吧，那是另一种方法。）++@DavidBlomstrom Barmar的一个更快。但我的产品能更好地控制产量。如果包装在

标签中。选择解决问题的任何答案作为解决方案，如果其中有帮助的话。干杯

$events = array();

while($row = $stm->fetch()){
    $year = $row['Year']; // current year

    if(!isset($events[$year]){   // if no such group
        $events[$year] = array();   // create group
    }

    $events[$year][] = $row['Brief']; // add data to year group
}

// Output:

foreach($events as $year => $event){
    echo $year, ':<br>'; // show year;

    foreach($event as $data){
        echo $data, '<br>';   // show row;
    }
}

foreach($events as $year => $event){
    echo $year, ' &#8226; ', implode('<br>&#8226; ', $event);
}

$prevYear = null;
while ($row = $stm->fetch())
{
    $Year = $row['Year'];
    if ($Year == $prevYear) {
        $YearStr = '• ';
    } else {
        $YearStr = $Year . ' $#8211; ';
        $prevYear = $Year;
    }
    $Brief[] = $YearStr.$row['Brief'];
}