如何在 GROUP BY 和第一个空格之间匹配字符串，前面没有逗号

Question

有字符串：

GROUP BY id,asd, asddd, adfasd LIMIT 4;
=> id,asd, asddd, adfasd
GROUP BY id,asd, asddd, adfasd order id asc;
=> id,asd, asddd, adfasd
GROUP BY id limit 4;
=> id

如何在不带逗号的情况下匹配 GROUP BY 和第一个空格之间的值...？

Answer 1

试试这个正则表达式：

GROUP BY (.*[^s]+s*,s*[^s]+)

在这里演示：

正则表达式101

Answer 2

这个正则表达式完成了这项工作： (?<=GROUP BYs+)(w+(?:s*,s*w+)*)

解释：

(?<=GROUP BYs+): lookbehind, make sure we have "GROUP BY"
(               : start group 1    
  w+           : 1 or more word character
  (?:           : start non capture group
    s*         : 0 or more spaces
    ,           : a comma
    s*         : 0 or more spaces
    w+         : 1 or more word character
  )*            : group may occurs 0 or more times
)               : end group 1

输入字符串：

GROUP BY id,asd, asddd, adfasd LIMIT 4;
GROUP BY id,asd, asddd, adfasd order id asc;
GROUP BY id , asd limit 4;

输出：

'id,asd, asddd, adfasd'
'id,asd, asddd, adfasd'
'id , asd'