PHP MySQL select 问题

Question

我有一个名为 emplyees 的数据库，结构如下：

id name   entrydate    totalhours
1  test   11/11/2015      8
2  test2  11/11/2015      7
3  test   11/12/2015      8
4  test2  11/12/2015      9

我想把它变成这样

id name 11/11/2015  11/12/2015 
1  test      8          8
2  test2     7          9

我试了好几种方法都没有结果我该怎么办？提前致谢！

Answer 1

一旦你确定了你的日期，这样的事情应该在 PHP...

<?php

require('path/to/connection/stateme.nts');

$query = "
SELECT name
     , entrydate
     , totalhours
  FROM my_table
 ORDER
    BY name;
";

$result = mysqli_query($db,$query);

$array = array();

while($row = mysqli_fetch_assoc($result)){
$array[] = $row;
}


foreach ($array as $element)
    $newArray[$element["name"]][] = array($element["entrydate"]=>$element["totalhours"]);

print_r($newArray);

?>

输出：

Array
(
    [test] => Array
        (
            [0] => Array
                (
                    [2015-11-11] => 8
                )

            [1] => Array
                (
                    [2015-11-12] => 8
                )

        )

    [test2] => Array
        (
            [0] => Array
                (
                    [2015-11-11] => 7
                )

            [1] => Array
                (
                    [2015-11-12] => 9
                )

        )

)

...可以轻松吐出到 table 或 json

PHP MySQL select 问题

PHP MySQL select issue

php

mysql

database

reporting