问题描述
我的字符串是这样的
'{ "city": "(not set)","cityId": "9108665","continent": "Americas","country": "United States","latitude": "0.0000","longitude": "0.0000","metro": "(not set)","networkDomain": "(not set)","networkLocation": "(not set)","region": "Pennsylvania","subContinent": "northern America" }'
我如何使用 REGEXP_SUBSTR
获取宾夕法尼亚州的区域
我正在努力
select REGEXP_SUBSTR(string,':[^:]+[[:alpha:]]',1,10) as region
然而,它没有给我想要的。
解决方法
您尝试的模式 :[^:]+[[:alpha:]]
匹配太多,因为否定字符类 [^:]+
匹配除 :
之外的任何字符,该匹配将到达下一个键值对。
如果您只想选择宾夕法尼亚州,您可以使用更具体的模式从第 10 场比赛中获取捕获组值:
"[^"]*":\s*"([^"]*)"
例如:
SELECT regexp_substr('{ "city": "(not set)","cityId": "9108665","continent": "Americas","country": "United States","latitude": "0.0000","longitude": "0.0000","metro": "(not set)","networkDomain": "(not set)","networkLocation": "(not set)","region": "Pennsylvania","subContinent": "Northern America" }','"[^"]*":\s*"([^"]*)"',1,10,NULL,1) as region from dual;
输出
REGION
Pennsylvania
但也许将区域添加到模式中以从示例数据中获取单个匹配项可能更容易。
"region":\s*"([^"]*)"