Find missing entries in one file - regex

Question

I've got two files:
1st: Entries.txt
confirmation.resend
send
confirmation.showResendForm
login.header
login.loginBtn
2nd: Used_Entries.txt
confirmation.showResendForm = some value
login.header = some other value
I want to find all entries from the first file (Entries.txt) that have not been asigned a value in the 2nd file (Used_Entries.txt)
In this example I'd like the following result:
confirmation.resend
send
login.loginBtn
In the result confirmation.showResendForm and login.header do not show up because these exist in the Used_Entries.txt
How do I do this? I've been playing around with regular expressions but haven't been able to solve it. A bash script or sth would be much appreciated!

Answer 1

You can do this with regex. But get your code mood ready, because you can't match both files with regex at once, and we do want to match both contents with regex at once. Well, that means you must have at least some understanding of your language, I would like you to concatenate the contents from the two files with at least a new line in between.
This regex solution expects your string to be matched to be in this format:
text (no equals sign)
text
text
...
key (no equals sign) ␣ (optional whitespace) = (literal equal) whatever (our regex will skip this part.)
key=whatever
key=whatever
Do I have your attention? Yes? Please see the following regex (using techniques accessible to most regex engines):
/(^[^=\n]+$)(?!(?s).*^\1\s*=)/m
Inspired from a recent answer I saw from zx81, you can switch to (?s) flag in the middle to switch to DOTALL mode suddenly, allowing you to start multiline matching with . in the middle of a RegExp. Using this technique and the set syntax above, here's what the regex does, as an explanation:
(^[^=\n]+$) Goes through all the text (no equals sign) elements. Enforces no equals signs or newlines in the capture. This means our regex hits every text element as a line, and tries to match it appropriately.
(?! Opens a negative lookahead group. Asserts that this match will not locate the following:
(?s).* Any number of characters or new lines - As this is a greedy match, will throw our matcher pointer to the very end of the string, skipping to the last parts of the document to backtrack and scoop up quickly.
^\1\s*= The captured key, followed by an equals sign after some optional whitespaces, in its own line.
) Ends our group.
View a Regex Demo!
A regex demo with more test cases
I'm stupid. I could had just put this:
/(^[^=\n]+$)(?!.*^\1\s*=)/sm

Answer 2

I've been going at this a little bit to complex and just solved it with a small script in scala:
import scala.io.Source
object HelloWorld {
def main(args: Array[String]) {
val entries = (for(line <- Source.fromFile("Entries.txt").getLines()) yield {
line
}).toList
val usedEntries = (for(line <- Source.fromFile("Used_Entries.txt").getLines()) yield {
line.dropRight(line.length - line.indexOf(' '))
}).toList
println(entries)
println(usedEntries)
val missingEntries = (for {
entry <- entries
if !usedEntries.exists(_ == entry)
} yield {
entry
}).toList
println(missingEntries)
println("Missing Entries: ")
println()
for {
missingEntry <- missingEntries
} yield {
println(missingEntry)
}
}
}

Answer 3

import re
e=open("Entries.txt",'r')
m=e.readlines()
u=open("Used_Entries.txt",'r')
s=u.read()
y=re.sub(r"= .*","",s)
for i in m:
if i.strip() in [k.strip() for k in y.split("\n")] :
pass
else:
print i.strip()