从JdbcTemplate中的ResultSet获取DateTime [英] getting DateTime from ResultSet in JdbcTemplate

问题描述

我的列是类型TIMESTAMP，所以我的类的属性类型为Datetime像这样：

in database my column is of type TIMESTAMP, so my class has properties of type Datetime like this:

public void setDiscoveryDate(final DateTime discoveryDtTm) {
        this.discoveryDtTm = discoveryDtTm;
    }

现在在JdbcTemplate我想得到它，所以一些代码如下： / p>

now in JdbcTemplate I want to get it, so some code like this:

variant.setDiscoveryDate(rs.getTimestamp("discovery_dt_tm"));

哪些不工作，因为列的结果集的获取我找不到返回DateTime的东西，我只看到getDate或getTime。

which does Not work because column the get for resultset I could not find something that returns DateTime, I only saw either getDate or getTime.

推荐答案

这是因为 DateTime 不是标准Java类型。如果您指的是JodaTime类型，请尝试以下方式：

That's because DateTime is not a standard Java type. If you're referring to the JodaTime type, then try this:

variant.setDiscoveryDate(
   new DateTime(rs.getTimestamp("discovery_dt_tm").getTime())
);

如果 rs.getTimestamp 返回 null ，所以你可能想把它分解成较小的语句，并为 null 添加检查。

This will break if rs.getTimestamp returns null, so you may want to break this up into smaller statements and add checks for null.

注意，这可以更容易，因为 DateTime 的构造函数需要一个 java.util.Date ，其中 Timestamp 是以下子类：

Note that this can be made easier, since DateTime's constructor takes a java.util.Date, which Timestamp is a subclass of:

variant.setDiscoveryDate(
   new DateTime(rs.getTimestamp("discovery_dt_tm"))
);

但这也是错误的，因为 Timestamp 类（请参阅 javadoc for解释）

But it's also wrong, due to bad design of the Timestamp class (see javadoc for explanation).

坚持第一个例子（使用 getTime（））

Stick with the first example (with getTime())

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