Find word in a string using a regex pattern lua

Question

I have a rest URL as a string --> rest/dashboard/person/hari/categrory/savingaccount/type/withdraw

In this I have to get the value between person and category && value between categrory and type. Because those value will dynamically change

rest/dashboard/person/{{}}/categrory/{{}}/type/withdraw

I tried with string.gsub(mystring, "([%w]+%/)([%w%d]+)"). But it seems it is not the correct wast to do it

Please help

local t = string.match("rest/dashboard/person/hari/categrory/savingaccount/type/withdraw" ,"/(%w+)/category/(%w+)/") print(t) It is printing nil. Anything i am missing — Cedric
– Cedric, Commented Aug 13, 2018 at 16:05

lhf · Accepted Answer · 2018-08-14 19:00:37Z

4

string.match with captures is the right tool for the job. Try this:

s="rest/dashboard/person/hari/category/savingaccount/type/withdraw" 
print(s:match("/person/(.-)/category/(.-)/type/"))

edited Aug 14, 2018 at 19:00

answered Aug 13, 2018 at 16:02

lhf

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2 Comments

Cedric Over a year ago

thanks @ihf marking it as accepted answer. Any idea on what is wrong with thiss="rest/dashboard/person/hari/category/savingaccount/type/withdraw" local s1,s2,s3 = (s:match("/person/(.-)/category/(.-)/type/")) print (s1) print (s2) print (s3)

lhf Over a year ago

@Hari, your s:match returns two values but the parentheses around it reduce them to one.

wp78de · Accepted Answer · 2018-08-13 16:09:30Z

3

You can use a lazy dot .- as suggested or a negated character class [^/]+

s="rest/dashboard/person/hari/category/savingaccount/type/withdraw" 
print(s:match("person/([^/]+)/category/"))
print(s:match("category/([^/]+)/type/"))

Demo

answered Aug 13, 2018 at 16:09

wp78de

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1 Comment

greenage Over a year ago

Fantastic, this instance was extremely helpful for my use case. Thanks!

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