按特定日期进行排序

Question

我有一个从mysql数据库收集数据的简单代码：

$get = mysql_query("SELECT * FROM table ORDER BY dato ASC") or die(mysql_error());
if(mysql_num_rows($get)) {
while($show = mysql_fetch_array($get)) {

}
} else {
echo "No results";
}

但名单上现在有200多个项目，我想按日期对它们进行排序，例如。

26/08-2015
1. line
2. line
3. line
4. line

27/08-2015
5. line
6. line
7. line

而不是

1. line
2. line
3. line
4. line
5. line
6. line
7. line

我现在明白了

我排序后的日期是mktime约会。

$dato = mktime(0, 0, 0, date('m'), date('d'), date('y'));

Answer 1

用这个，

$get = mysql_query("SELECT DISTINCT(DATE(`date`)) as dt, t.* FROM table t ORDER BY 1") or die(mysql_error());
if(mysql_num_rows($get)) {
    $prevDT = "";
    while($show = mysql_fetch_array($get)) {
        if (strlen($prevDT) < 0 || strcmp($prevDT, $show['dt']) != 0){
            echo("<br />");
            echo($show['dt']."<br />");
            $prevDT = $show['dt'];
        }
        echo($show['field']."<br />");
    }
} else {
    echo "No results";
}

Answer 2

试试看这个条件在哪里

$dato = mktime(0, 0, 0, date('m'), date('d'), date('y'));

$get = mysql_query("SELECT * FROM table WHERE date > $dato ORDER BY date ASC") or die(mysql_error());
if(mysql_num_rows($get)) {
    while($show = mysql_fetch_array($get)) {

    }
} else {
    echo "No results";
}

Answer 3

使用这个

$currentDate = null;
$get = mysql_query("SELECT * FROM table ORDER BY dato ASC") or die(mysql_error());
if(mysql_num_rows($get)) {
while($show = mysql_fetch_array($get)) {
if($currentDate != $show['dato'])
     echo "<BR><BR>";
$currentDate = $show['dato'];
}
} else {
echo "No results";
}