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根据值从MySQL填充表列

[英]Fill table column from mysql depending on value

我有一个如下所示的mysql表:

+------+-------+-------+----------+------------+
| id   | name  | RowSZ | TimeSZ   | DateSZ     | 
+------+-------+-------+----------+------------+
| 1    | namex | 2     | 12:00:00 | 2018-06-29 | 
+------+-------+-------+----------+------------+
| 2    | namey | 1     | 10:00:00 | 2018-06-30 |

现在,我想填充一个如下所示的html表:

+-------+----------+----------+------------+
| name  | Timerow1 | Timerow2 | DateSZ     | 
+-------+----------+----------+------------+
| namex |          | 12:00:00 | 2018-06-29 | 
+-------+----------+----------+------------+
| namey | 10:00:00 |          | 2018-06-30 |

根据值“ RowSZ”填充Timerow1和Timerow2。

我现在的SQL查询:

$sql = "SELECT *, DATE_FORMAT(DateSZ, '%d-%m-%y') AS DateSZ FROM table1 ORDER BY DateSZ ASC";

如何正确填充Timerow1和Timerow2两行?

使用UNION处理这两种情况。

SELECT name, NULL AS TimeRow1, TimeSZ AS TimeRow2, DateSZ
FROM yourTable
WHERE RowSZ = 2

UNION

SELECT name, TimeSZ AS TimeRow1, NULL AS TimeRow2, DateSZ
FROM yourTable
WHERE RowSZ = 1

ORDER BY DateSZ

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