sed或awk在一行的最后100个字符中查找字符串或删除该行

Question

first question so hopefully I form it well.

I'm looking to match a string, namely "lang":"en" in the last 100 characters of a line and if there's no match, delete the line.

I have tried using sed by doing

sed '/"lang":"en"/!d' file > output

But unfortunately many lines have that string more than once and I only care about the final occurrence of it.

I'm learning sed still, but don't know anything about awk and most of my searches have come up with "first/last instance in a file" rather than "in a line" so any help in learning the best method to do this would be great. thanks.

Answer 1

This should work with any Posix awk:

awk 'match(substr($0,length-99),/"lang":"en"/)' file

You can do it with a simple string find, instead of a regular expression, but the string is more annoying to type:

awk 'index(substr($0,length-99),"\"lang\":\"en\"")' file

Both simply extract the last 100 characters of each line, and if the test pattern is found in the substring, print the line (print is the default action, so the program consists only of the condition.)

Answer 2

For a simple regex-based solution,

grep -E '"lang":"en".{0,89}$' file

I subtracted the length of "lang":"en" from the maximum amount, assuming you mean the string must be found entirely within the last 100 characters.

This looks like you are attempting to process JSON data, so perhaps you can come up with a better, structure-based rule, and use jq instead.

jq 'select(path["to"]["lang"] == "en")' file

to find "en" in the structure "path": { ... "to": { ..., "lang": "en" ...} } } . This will also be robust against newlines in the JSON, spacing variations in "lang": "en" , etc.

Answer 3

sed '/"lang":"en".\{0,89\}$/!d' file > output

在选择的末尾之前添加可能的其他89个字符