[英]pivot html table using php
我有這個PHP代碼:
$query_production = "SELECT uploadedby as name, sum(points) as points,
date_format(uploaddate,'%Y-%m-%d') as date FROM imsexport
WHERE uploaddate BETWEEN '2014-01-01 00:00:00' and '2014-01-20 23:59:59'
GROUP BY uploadedby,date";
$result_production = mysql_query($query_production);
$row_production = mysql_fetch_assoc($result_production);
HTML表輸出為
name points date
John 147 2014-01-01
Bob 79 2014-01-01
Joe 156 2014-01-01
Sue 116 2014-01-01
John 117 2014-01-02
Bob 186 2014-01-02
Sue 74 2014-01-02
Bob 233 2014-01-03
John 159 2014-01-03
Sue 162 2014-01-03
Bob 162 2014-01-04
Sue 38 2014-01-05
如何使用php旋轉此表使其看起來像這樣? 我已經使用mysql完成此操作,但是代碼太長。
Name |2014-01-01|2014-01-02|2014-01-03|2014-01-04|2014-01-05
Bob 79 186 233 162 0
Joe 156 0 0 0 0
John 147 117 159 0 0
Sue 116 74 162 0 38
這將為您提供大部分幫助。 我剩下的是如何正確校正左偏移量(或date / points列對齊方式;提示,您需要將日期與名稱/點對保持一致,並知道它在$cols
的偏移量)。
顯然,您正在處理數據庫中的行,因此會有所不同。
請參閱代碼下方的鍵盤演示鏈接。
<?php
$data = "John 147 2014-01-01
Bob 79 2014-01-01
Joe 156 2014-01-01
Sue 116 2014-01-01
John 117 2014-01-02
Bob 186 2014-01-02
Sue 74 2014-01-02
Bob 233 2014-01-03
John 159 2014-01-03
Sue 162 2014-01-03
Bob 162 2014-01-04
Sue 38 2014-01-05";
$data = explode("\n", $data);
$cols = array();
$pivot = array();
while ($line = array_shift($data)) {
list($name, $points, $date) = explode(' ', trim($line));
if (!$pivot[$name]) $pivot[$name] = array();
array_push($cols, $date);
array_push($pivot[$name], array('date' => $date, 'points' => $points));
}
$cols = array_unique($cols);
print_r($pivot);
echo implode('|', $cols);
echo PHP_EOL;
foreach ($pivot as $name => $row) {
while ($entry = array_shift($row)) {
echo str_pad($name, 7, ' ') . str_pad($entry['points'], 3, ' ', STR_PAD_LEFT) . '|';
}
echo PHP_EOL;
}
?>
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