Replace a substring in Lua with a pattern

Question

I have a string like this

     str = '["username"] = "user";
     ["deepscan"] = "true";
     ["token"] = true;
     ["password"] = "krghfkghkfghf";
     ["uploadMethod"] = "JSON";
     ["serviceIsRunning"] = {};
     ["host"] = "sample.com";
     ["instance_ID"] = 405454058;'

I would like the pattern match ["password"] = and have it replace only the string in between the ";' that would be '"krghfkghkfghf" in this instance.

str = str:gsub('(%["password"%]%s*=%s*").-"','%1'..new_pwd:gsub('%%','%%%%')..'"') — Egor Skriptunoff
– Egor Skriptunoff, Commented May 19, 2013 at 20:05
@EgorSkriptunoff - why do you always post answers as comments :)? — W.B.
– W.B., Commented May 19, 2013 at 20:08

W.B. · Accepted Answer · 2013-05-19 20:19:49Z

1

local function replacePass(configstr, newpass)
    return configstr:gsub("(%[\"password\"%]%s*=%s*)%b\"\"", "%1\"" .. newpass .. "\"")
end

That won't work if your password contains a double quote inside.

edited May 19, 2013 at 20:19

answered May 19, 2013 at 20:04

W.B.

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wesleylim1993 · Accepted Answer · 2020-05-25 05:19:06Z

-1

i have same question as well, how about the following password replace ?

"password" : "krghfkghkfghf"

answered May 25, 2020 at 5:19

wesleylim1993

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