Question

我的输入中有两个urls格式，我需要解析

URL＆＃39; S

abc.com/abcd?id=123

xyz.com/abcd/id123

解析后，我需要在数据库中存储id=123和id123

这是我的hive查询来解析网址和

insert into table table2
select  
CASE
WHEN parse_url(url_domain,'HOST')="abc.com"
THEN 
    parse_url(url_domain,'HOST') as host,
    split(url_domain,'\\?id=')[1] as id,
    count(*) 
    from table1
    GROUP BY parse_url(url_domain, 'HOST'), split(url_domain,'\\?id=')[1]    
WHEN parse_url(url_domain,'HOST')="xyz.com"
THEN
    parse_url(url_domain,'HOST') as host,
    split(url_domain,'\\/id')[1] as id,
    count(*) 
    from table1
    GROUP BY parse_url(url_domain, 'HOST'), split(url_domain,'\\/id=')[1]
END
ORDER BY host, id DESC limit 100;

但是当我执行查询时，它会出现以下错误

FAILED: ParseException line 6:33 missing KW_END at 'as' near ']'

我尝试删除id别名和其他组合，但没有工作

的信息： 我的hive我不能在Group by子句中使用别名作为其hive限制

split(url_domain,'\\?id=')[1] as id

如果我使用GROUP BY id，则会出错，但效果正常GROUP BY parse_url(url_domain, 'HOST')

因此我无法在CASE声明之外移动GROUP BY

更新

insert into table table2
select  
CASE
WHEN parse_url(url_domain,'HOST')="abc.com"
THEN 
    parse_url(url_domain,'HOST') as host, split(url_domain,'\\?id=')[1] as id,
WHEN parse_url(url_domain,'HOST')="xyz.com"
THEN
    parse_url(url_domain,'HOST') as host, split(url_domain,'\\/id')[1] as id,
END
count(*) 
from table1
GROUP BY parse_url(url_domain, 'HOST')
ORDER BY host, id DESC limit 100;

同样的错误:(

错误记录

NoViableAltException(262@[146:1: selectExpression : ( expression | tableAllColumns );])
        at org.antlr.runtime.DFA.noViableAlt(DFA.java:158)
        at org.antlr.runtime.DFA.predict(DFA.java:116)
        at org.apache.hadoop.hive.ql.parse.HiveParser_SelectClauseParser.selectExpression(HiveParser_SelectClauseParser.java:2882)
        at org.apache.hadoop.hive.ql.parse.HiveParser_SelectClauseParser.selectItem(HiveParser_SelectClauseParser.java:2266)
        at org.apache.hadoop.hive.ql.parse.HiveParser_SelectClauseParser.selectList(HiveParser_SelectClauseParser.java:1052)
        at org.apache.hadoop.hive.ql.parse.HiveParser_SelectClauseParser.selectClause(HiveParser_SelectClauseParser.java:789)
        at org.apache.hadoop.hive.ql.parse.HiveParser.selectClause(HiveParser.java:31425)
        at org.apache.hadoop.hive.ql.parse.HiveParser.regular_body(HiveParser.java:29083)
        at org.apache.hadoop.hive.ql.parse.HiveParser.queryStatement(HiveParser.java:28968)
        at org.apache.hadoop.hive.ql.parse.HiveParser.queryStatementExpression(HiveParser.java:28762)
        at org.apache.hadoop.hive.ql.parse.HiveParser.execStatement(HiveParser.java:1238)
        at org.apache.hadoop.hive.ql.parse.HiveParser.statement(HiveParser.java:938)
        at org.apache.hadoop.hive.ql.parse.ParseDriver.parse(ParseDriver.java:190)
        at org.apache.hadoop.hive.ql.Driver.compile(Driver.java:424)
        at org.apache.hadoop.hive.ql.Driver.compile(Driver.java:342)
        at org.apache.hadoop.hive.ql.Driver.runInternal(Driver.java:1000)
        at org.apache.hadoop.hive.ql.Driver.run(Driver.java:911)
        at org.apache.hadoop.hive.cli.CliDriver.processLocalCmd(CliDriver.java:259)
        at org.apache.hadoop.hive.cli.CliDriver.processCmd(CliDriver.java:216)
        at org.apache.hadoop.hive.cli.CliDriver.processLine(CliDriver.java:413)
        at org.apache.hadoop.hive.cli.CliDriver.executeDriver(CliDriver.java:781)
        at org.apache.hadoop.hive.cli.CliDriver.run(CliDriver.java:675)
        at org.apache.hadoop.hive.cli.CliDriver.main(CliDriver.java:614)
        at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
        at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:94)
        at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:55)
        at java.lang.reflect.Method.invoke(Method.java:619)
        at org.apache.hadoop.util.RunJar.main(RunJar.java:212)
FAILED: ParseException line 6:33 missing KW_END at 'as' near ',' in select expression
line 7:0 cannot recognize input near 'WHEN' 'parse_url' '(' in select expression

Answer 1

我从未使用过hive，但是你是否允许在那么条件下看起来像是子查询？我知道在transact sql中必须有一个值。

Answer 2

最后通过为CASE创建两个单独的parse_url语句，为split_url创建一个语句来解决它。由于hive在GROUP BY子句中使用别名有问题所以这是唯一的方法

最终Hive SQL语句

insert into table table2 select parse_url(url_domain,'HOST') as host,
CASE
WHEN parse_url(url_domain,'HOST')="abc.com" 
THEN 
    parse_url(url_domain,'QUERY','id')
WHEN parse_url(url_domain,'HOST')="def.com"
THEN 
    parse_url(url_domain,'QUERY','packageName')
WHEN parse_url(url_domain,'HOST')="xyz.com"
THEN
    split(split(url_domain,'\\/id')[1],'\\?')[0]
ELSE
    "NULL"
END as appid,
count(*)
from table1
group by parse_url(url_domain,'HOST'),
CASE
WHEN parse_url(url_domain,'HOST')="abc.com" 
THEN 
    parse_url(url_domain,'QUERY','id')
WHEN parse_url(url_domain,'HOST')="def.com"
THEN 
    parse_url(url_domain,'QUERY','packageName')
WHEN parse_url(url_domain,'HOST')="xyz.com"
THEN
    split(split(url_domain,'\\/id')[1],'\\?')[0]
ELSE
"NULL"
END
order by host,appid DESC LIMIT 100;

使用GROUP BY子句解析CASE语句中的配置单元查询中的错误

2 个答案: