Postgresql regexp_replace negative lookahead不工作

I am trying to replace street with st only if street isn't followed by any alphabet. Replacement is allowed if after street EITHER there is non-alphabet character OR end of string.

我正试着用st来代替街道，除非街上没有任何字母。如果街道后面有非字母表字符或字符串尾，则允许替换。

I am trying to achieve this in Postgresql 9.5 regex_replace function. Sample query i wrote:

我试图在Postgresql 9.5 regex_replace函数中实现这一点。我写查询样例:

select regexp_replace('super streetcom','street(?!=[a-z])','st');

的选择regexp_replace(“超级streetcom”、“街(? ! =[a - z])”,“圣”);

street here shouldn't have been replaced by st since street is followed by 'c'. So the expected output is 'super streetcom' but the output i am getting is 'super stcom'.

这里的街道不应该被st取代，因为street后面跟着c。所以预期输出是" super streetcom "但我得到的输出是" super stcom "

Any help for why i am getting the unexpected output and what can be the right way to achieve the intended result.

对于为什么我得到了意想不到的输出，以及什么是实现预期结果的正确方法，有什么帮助吗?

2 个解决方案

#1

A lookahead construct looks like (?!...), all what follows ?! is a lookahead pattern that the engine will try to match, and once found, the match will be failed.

一个前瞻结构看起来像(?!)，接下来会发生什么?!是引擎将尝试匹配的前瞻模式，一旦找到，匹配将失败。

It seems you need to match a whole word street. Use \y, a word boundary:

看来你需要把整条街搭配起来。使用\y，一个词的边界:

select regexp_replace('super streetcom street','\ystreet\y','st');

See the online demo

查看在线演示

From the docs:

从文档:

\y matches only at the beginning or end of a word

只在单词的开头或结尾匹配

#2

This looks like a syntax issue. Try: ?! instead of ?!= . e.g.

这看起来像是一个语法问题。试题:? !而不是? !=。如。

select regexp_replace('super street','street(?![a-z])','st');

will return

将返回

super st