从 postgresql 中的结束分隔符与其前一个主题之间的字符串中提取 substring

Question

我真的在这方面苦苦挣扎：

我在 postgresql 数据库中有一些额外数据，其中一个数字很重要。

示例：

• {"code_partenaire":"welcome","sous_statut_abc":"1","statut_alpha":"blabla"}
• {"code_partenaire":"code33","sous_statut_def":"1782","statut_alpha":"blablabla"}
• {"code_partenaire":"222","sous_statut_ghei":"17","statut_alpha":"blablaa"}

我想提取","statut_alpha"和它之前的":"之间的字符

因为有我要找的号码。

你有想法吗？

Output 应该是

• 1
• 1782
• 17

谢谢你的帮助：）

Answer 1

对于您给出的 output，substring 是一个繁琐的过程，条件 (1) before: ","statut_alpha" 为此使用了反向字符串并找到最后一个 position 之后的 ":" ","statut_alpha"

create table substdata (datac varchar(300));
insert into substdata values ('{"code_partenaire":"welcome","sous_statut_abc":"1","statut_alpha":"blabla"}');
insert into substdata values ('{"code_partenaire":"code33","sous_statut_def":"1782","statut_alpha":"blablabla"}');
insert into substdata values ('{"code_partenaire":"222","sous_statut_ghei":"17","statut_alpha":"blablaa"}');



select    
  substring 
  (datac,0,
  POSITION('","statut_alpha"' IN datac)) 
,
  length( substring 
  (datac,0,
  POSITION('","statut_alpha"' IN datac)) ), (
  length( substring 
  (datac,0,
  POSITION('","statut_alpha"' IN datac)) )
   -  
 pOSITION('":"' in 
 reverse(
  substring 
  (datac,0,
  POSITION('","statut_alpha"' IN datac))
  ))) as minus 
  , 
substring (substring 
  (datac,0,
  POSITION('","statut_alpha"' IN datac)) 
  ,
  
  (
  length( substring 
  (datac,0,
  POSITION('","statut_alpha"' IN datac)) )
   -  
 pOSITION('":"' in 
 reverse(
  substring 
  (datac,0,
  POSITION('","statut_alpha"' IN datac))
  )))+2,( length( substring 
  (datac,0,
  POSITION('","statut_alpha"' IN datac)) ))) as finresult 
 from substdata

OUTPUT：

substring   length  minus   finresult
{"code_partenaire":"welcome","sous_statut_abc":"1   49  47  1
{"code_partenaire":"code33","sous_statut_def":"1782     51  46  1782
{"code_partenaire":"222","sous_statut_ghei":"27     47  44  17

如果第一个条件总是在字符串中的第三个 position

你可以使用 unest 通常给出行号并用它过滤

select row_number() over(),t from (
  select
 row_number() over() as id ,t
  from substdata ,
  unnest(string_to_array(
  substring 
  (datac,0,
  POSITION('","statut_alpha"' IN datac)), '":"'))  
  WITH ORDINALITY As T ) as drt 

WHERE (id % 3) = 0
  

OUTPUT 2：

row_number  t
1   1
2   1782
3   17

Answer 2

发现了:)

split_part(extradata,'"',8)