Question

我将访问次数和用户时间保存在我的页面中。 mySQL如下所示：

id  - user - page - timestamp
340 - 1 - page1 - 2009-05-18 22:11:11
339 - 1 - page3 - 2009-05-18 22:10:01
338 - 2 - page4 - 2009-05-18 22:08:05
337 - 1 - page2 - 2009-05-18 22:06:00
336 - 2 - page1 - 2009-05-18 22:15:10
...
...
280 - 1 - page5 - 2009-05-17 14:10:10
279 - 2 - page 3 - 2009-05-17 12:05:00

如何使用PHP获得以下输出？

预期的outpout (show.php?user=1)

2009-05-18

page 1 - 2009-05-18 22:11:11 last page
page3 - 2009-05-18 22:10:01 stayed for 1 minute 10 seconds
page2 - 2009-05-18 22:06:00 stayed for 4 minutes 1 second

2009-05-17

page 5 - 2009-05-17 14:10:10 last page
page 3 - 2009-05-17 12:05:00 stayed for 2 hours 5 minutes 10 seconds

到目前为止我做了什么

$sql= mysql_query(SELECT user, page, timestamp FROM paths WHERE DATE(`date`) = '2009-05-18'); // must not set a date, but get all the dates

while($row = mysqli_fetch_array($sql)) {
        echo $row["page"]. " - " . $row["timestamp"]. "<br>";
//echo this timestamp - previous timestamp
//if different day then start again
    }

Answer 1

实际上我对此有一个想法，但它对时间没用;

sql = "SELECT user, page, timestamp FROM paths WHERE DATE(`date`) = '2009-05-18')" ORDER BY page ASC,timestamp DESC;

$a = null;
while($row = mysqli_fetch_array($sql)) {
    $a[] = $row;
}

$i = 0;

while(i<sizeof($a)){
    if($a[i]->page == $a[i+1]->page){
        $sub = (int) $a[i+1]->timestamp - (int) $a[i]->timestamp;
        echo '' .$a[i]->page. '-' .$a[i]->timestamp.' stayed for '.$sub. 'miliseconds';
    }
    i = i+2;
}

Answer 2

试试这个。我没有经过测试。但它应该工作。如果有任何错误意味着让我知道..

// query to fetch the dates first.
$date_query = mysql_query ( "SELECT DISTINCT(DATE(timestamp)) FROM paths order by timestamp asc;" );

while ( $row = mysql_fetch_array ( $date_query ) ) {
    echo "<h1>".$row ['timestamp']."</h1><br/>";
    // query to fetch the related datas based on the date from the table.
    $related_day_result = mysql_query("SELECT * from path where DATE(timestamp) = '".$row['timestamp']."'");
    while ( $results = mysql_fetch_array ( $related_day_result ) ) {
        echo $results ["page"] . " - " . $results ["timestamp"] . "<br>";
    }
}

Answer 3

此查询几乎可以满足您的所有需求...... SqlFiddle Demo

SELECT page,max(`timestamp`) `timestamp` from paths
group by date(`timestamp`),page order by `timestamp` desc

但如果您需要显示停留时间，则需要计算。如果有任何关于计算的查询我在那里解释

$sql = "SELECT page,max(`timestamp`) `timestamp` from paths
group by date(`timestamp`),page order by `timestamp` desc";

$res = mysql_query($sql) or die(mysql_error());

$lastDate = null;
$diff = 'last page';
$lastVisitTime = '';
while($row = mysql_fetch_array($res))
{
    $time = strtotime($row['timestamp']);
    $thisDate = date('Y-m-d',$time);
    //echo "   ".$thisDate;

    if($lastDate == $thisDate)
    {
        $date1 = new DateTime($lastVisitTime);
        $date2 = new DateTime($row['timestamp']);
        $interval = $date2->diff($date1);
        $diff = "  stayed ". $interval->h . " hours ".$interval->i . " minutes ".$interval->s." seconds";

        /*other method
        $time1 = $lastVisitTime;
        $time2 = $row['timestamp'];
        $sec_diff = strtotime($time1) - strtotime($time2);
        $hourdiff = (int)($sec_diff/3600);
        $diff = $hourdiff .' hours ';*/

        $lastVisitTime = $row['timestamp'];
    }
    else
    {
        $diff = 'last page';
        $lastDate = $thisDate;
        $lastVisitTime = $row['timestamp'];
    }

    echo $row['page']." ___ ".$row['timestamp']." ___ ". $diff.'<br>';
}

如何根据mySQL中的日期拆分行并使用PHP显示它？

3 个答案: