mysql显示日期而不是长选择结果

Question

我正在尝试在select * from table_name的结果上显示日期而不是长按日期。 SO上的所有答案都是指在where子句上转换日期。但是我不想在日期格式中使用过滤器，我希望以日期格式显示结果，而不是很长。

修改

--------------------------+----------------------+----------+----------+--------------+--------------+-----------------+-----------+----------------+
| connectionUniqueIdentity | deviceUniqueIdentity | simCard1 | simCard2 | phoneNumber1 | phoneNumber2 | lastUpdatedDate | rowStatus | activationDate |
+--------------------------+----------------------+----------+----------+--------------+--------------+-----------------+-----------+----------------+
| -1166116395              | 823167013            | simCard1 | simCard2 | phoneNumber1 | phoneNumber2 |   1413371737879 |         0 |  1413371737879 |
| -1301899739              | 823167013            | simCard1 | simCard2 | phoneNumber1 | phoneNumber2 |   1413371395494 |         1 |  1413371395494 |

我需要它像

--------------------------+----------------------+----------+----------+--------------+--------------+-----------------+-----------+----------------+
| connectionUniqueIdentity | deviceUniqueIdentity | simCard1 | simCard2 | phoneNumber1 | phoneNumber2 | lastUpdatedDate | rowStatus | activationDate |
+--------------------------+----------------------+----------+----------+--------------+--------------+-----------------+-----------+----------------+
| -1166116395              | 823167013            | simCard1 | simCard2 | phoneNumber1 | phoneNumber2 |   2nd Oct 2014 |         0 |  15th Oct 2014 |
| -1301899739              | 823167013            | simCard1 | simCard2 | phoneNumber1 | phoneNumber2 |   3rd Oct 2014 |         1 |  18th Oct 2014 |

Answer 1

SELECT DATE(FROM_UNIXTIME(1413439433));

Answer 2

SELECT DATE_FORMAT(FROM_UNIXTIME(yourDate), '%y-%m-%d %h:%i%p') AS FormattedDate
FROM TableName

• DATE_FORMAT
• FROM_UNIXTIME

Answer 3

您可以使用DATE_FORMAT和FROM_UNIXTIME函数：

SELECT DATE_FORMAT(FROM_UNIXTIME(time_value), '%d/%m/%y %h:%i%p') 
FROM table_name