Hadoop Pig-如何一步连接和定义模式

我采取以下措施：

A = LOAD 'a.txt' USING PigStorage('\\u001') AS (
    foo:int
    ,bar:chararray
);
B = LOAD 'b.txt' USING PigStorage('\\u001') AS (
    foo:int
    ,baz:long
);
C = JOIN A BY foo, B BY foo;
D = FOREACH C GENERATE
    A::foo AS foo
    ,A::bar AS bar
    ,B::baz AS baz
;

如何在一个步骤中加入并定义架构？

根据连接关系时不能定义架构。
注: 从语法上讲，您可以嵌套命令，让人感觉您保存了一些步骤，如：

D = foreach
    (join (LOAD 'a.txt' USING PigStorage('\\u001') AS (foo:int ,bar:chararray)) by foo,
          (LOAD 'b.txt' USING PigStorage('\\u001') AS (foo:int ,baz:long)) by foo
    ) generate $0 as foo, $1 as bar, $3 as baz;

但我会避免这样做。它是混乱的，但它生成的解释计划与原始计划相同。

根据规则，在连接关系时不能定义模式。
注: 从语法上讲，您可以嵌套命令，让人感觉您保存了一些步骤，如：

D = foreach
    (join (LOAD 'a.txt' USING PigStorage('\\u001') AS (foo:int ,bar:chararray)) by foo,
          (LOAD 'b.txt' USING PigStorage('\\u001') AS (foo:int ,baz:long)) by foo
    ) generate $0 as foo, $1 as bar, $3 as baz;

但我会避免这样做。它是混乱的，但它生成的解释计划与原始计划相同。

根据规则，在连接关系时不能定义模式。
注: 从语法上讲，您可以嵌套命令，让人感觉您保存了一些步骤，如：

D = foreach
    (join (LOAD 'a.txt' USING PigStorage('\\u001') AS (foo:int ,bar:chararray)) by foo,
          (LOAD 'b.txt' USING PigStorage('\\u001') AS (foo:int ,baz:long)) by foo
    ) generate $0 as foo, $1 as bar, $3 as baz;

但我会避免这样做。它是混乱的，但它生成的解释计划与原始计划相同。

根据规则，在连接关系时不能定义模式。
注: 从语法上讲，您可以嵌套命令，让人感觉您保存了一些步骤，如：

D = foreach
    (join (LOAD 'a.txt' USING PigStorage('\\u001') AS (foo:int ,bar:chararray)) by foo,
          (LOAD 'b.txt' USING PigStorage('\\u001') AS (foo:int ,baz:long)) by foo
    ) generate $0 as foo, $1 as bar, $3 as baz;

但我会避免这样做。它是混乱的，尽管如此，它还是产生了与原始计划相同的解释计划