Question

我们可以编写一个Hive UDF，它采用＆＃39; String-value＆＃39;作为输入。任何示例或链接都会有所帮助。我访问了以下链接： http://blog.matthewrathbone.com/2013/08/10/guide-to-writing-hive-udfs.html

需要更多帮助。

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:context="http://www.springframework.org/schema/context"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:mvc="http://www.springframework.org/schema/mvc"
    xsi:schemaLocation="
        http://www.springframework.org/schema/beans     
        http://www.springframework.org/schema/beans/spring-beans.xsd
        http://www.springframework.org/schema/context 
        http://www.springframework.org/schema/context/spring-context.xsd
        http://www.springframework.org/schema/mvc
        http://www.springframework.org/schema/mvc/spring-mvc.xsd">

    <bean id="dataSource"
        class="org.springframework.jdbc.datasource.DriverManagerDataSource">
        <property name="driverClassName" value="org.postgresql.Driver" />
        <property name="url" value="jdbc:postgresql://localhost:5432/test" />
        <property name="username" value="root" />
        <property name="password" value="root" />
    </bean>

    <bean id="jdbcTemplate" class="org.springframework.jdbc.core.JdbcTemplate">
        <property name="dataSource" ref="dataSource" />
    </bean>

</beans>

Hive UDF获取参数作为输入

0 个答案: