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If I have a file with rows like this
/some/random/file.csv:some string
/some/random/file2.csv:some string2
Is there some way to get a file that only has the first part before the colon, e.g.
/some/random/file.csv
/some/random/file2.csv
I would prefer to just use a bash one liner, but perl or python is also ok.
896
Using the cut Command Specifying the character index isn't the only way to extract a substring. You can also use the -d and -f flags to extract a string by specifying characters to split on. The -d flag lets you specify the delimiter to split on while -f lets you choose which substring of the split to choose.
Open the file “input.sh” and write the appended code in the file. Here, we have declared an echo statement with the string “foo-bar-123” using the “awk” keyword. The print term is followed by the “-F-“ keyword. This will create a substring after the next special character, which is “123,” and print it.
cut -d: -f1 or
awk -F: '{print $1}' or
sed 's/:.*//' Another pure BASH way:
> s='/some/random/file.csv:some string'
> echo "${s%%:*}"
/some/random/file.csv
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