在Hive尋找選擇列中具有特殊字符的行

Question

我期待在列中具有特殊字符[a-zA-Z0-9]的Hive中選擇行。

我不太清楚如何構建WHERE子句，但是基於其他線程我想這應該是這個樣子：

SELECT DISTINCT user_name 
FROM user_info 
WHERE user_name like regexp_extract('%[^a-zA-Z\d\s:]%') 

但蜂巢提供了這個錯誤：

Error while compiling statement: FAILED: SemanticException [Error 10014]: Line 3:21 Wrong arguments ''%[^a-zA-Z\d\s:]%'': No matching method for class org.apache.hadoop.hive.ql.udf.UDFRegExpExtract with (string). Possible choices: FUNC(string, string) FUNC(string, string, int)

如何我應該設置regexp_extract子句嗎？

感謝

Answer 1

SELECT DISTINCT user_name 
FROM user_info 
WHERE user_name rlike '[^a-zA-Z\\d\\s:]' 

https://cwiki.apache.org/confluence/display/Hive/LanguageManual+UDF#LanguageManualUDF-StringOperators

hive> select 'Dudu Markovitz: 123' rlike '[^a-zA-Z\\d\\s:]'; 
OK 
false 
hive> select 'Dudu Markovitz: @123' rlike '[^a-zA-Z\\d\\s:]'; 
OK 
true