I am stuck on a homework question. The question asks to display the lines, with grep and I can't use -w option, that contain no duplicate vowels.
My teacher said to find the grep command that could display two or more 'a's in a line which would, I think, be grep 'a.*a' file and then find the grep command that would display two or more 'u's which, I think, would be grep 'u.*u' file, combine them and then I should be able to get it. But I don't know how I would combine the grep commands.
You can combine different regular expressions with |:
grep 'a.*a|e.*e|i.*i|o.*o|u.*u' file
Related
real beginner in Unix commands so not sure if the following is actually possible but here goes.
Is it possible to highlight just one item in a ls output?
I.e.: in a directory I use the following
ls -l --color=auto
this lists 4 items in green
file1.xls
file2.xls
file3.xls
file4.xls
But I want to highlight a specific item, in this case file2.
Is this possible?
The ls program will not do this for you. But you could filter the results from ls through a custom script which modifies the text to highlight just one item. It would be simpler if no color was originally given; then you could match on the given filename (for example as the pattern in an awk script, or in a sed script) and modify just that one item, adding colors.
That is, certainly it is possible. Writing a sample script is a different question.
How you approach the problem depends on what you want from the output. If that is (literally) the output from ls with a single filename in color, then a script would be the normal approach. You could use grep as suggested in the other answer, which raises a few issues:
commenting on ls -l --color=auto makes it sound as if you are using GNU ls, hence likely using Linux. An appropriate tag for the question would be linux rather than unix. If you ask for unix, the answers should differ.
supposing that you are using Linux. Then likely you have GNU grep, which can do colors. That would let you do something like this:
ls -l | grep --color=always file2 |less -R
however, there is a known bug in GNU grep's use of color (see xterm FAQ "grep --color" does not show the right output).
using grep like this shows only the matching lines. For ls that might be a good choice. For matches in a manual page -- definitely not.
Alternatively, less (which is found more often on Unix systems than GNU grep) also can highlight matches (not in color) and would show the file you are looking for in context. You could do this:
ls -l | less -p file2
(Both grep and less use patterns aka regular expressions, but I left the example simple — read the documentation to learn more).
If you're a beginner I would strongly suggest you learn the grep command if you want to filter results - A Unix users best friend (mine anyway)
Use grep to only display the list items you want to see...
ls- l | grep "file2"
NOTE: This is no different to typing ls -l file2 by the way but your pattern could be expanded based on what you actually want displayed on the screen.
So if you had a directory full of files ".txt", ".xls", ".doc" and you wanted to only see ".doc" with the word "work" in the name (work1.doc) you could write:
ls -ls | grep "work" | grep "txt"
This would list work1.txt, work2.txt, work3.txt and so on.
This is a very basic example but I use grep extensively whilst in the unix shell and would advise using this to filter all results instead of colours.
A little side note using grep -v will show you everything but the pattern you give it
ls -l | grep -v ".txt" will show everything BUT .txt files.
I need a unix command that would search multiple patterns (basically an AND), however those patterns need not be on the same line (otherwise I could use grep AND command). For e.g. suppose I have a file like following:
This is first line.
This is second line.
This is last line.
If I search for words 'first' and 'last', above file should be included in the result.
Try this question, seems to be the same as yours with plenty of solutions: How to find patterns across multiple lines using grep?
I think instead of AND you actually mean OR:
grep 'first\|last' file.txt
Results:
This is first line.
This is last line.
If you have a large number of patterns, add them to a file; for example if patterns.txt contains:
first
last
Run:
grep -f patterns.txt file.txt
Q1. I want to grep something like that:
grep -Ir --exclude-dir="some*dirs" "my-text" ~/somewhere
but I don't want to show the whole strings containing "my-text", I want to see only list of files.
Q2. I want to see list of files containing "my-text" but not containing "another-text". How to do that?
Sorry, but I could not find the answer in man grep, neither in google.
Q1. You mustn't have googled very hard on that one.
man grep
-l, --files-with-matches
Suppress normal output; instead print the name of each input
file from which output would normally have been printed. The
scanning will stop on the first match.
Q2. Unless you expect both patterns to be on the same line, you'll need multiple invocations of grep. Something like:
$ grep -l my-text | xargs grep -vl another-text
I want to grep files that contain text "wp_" but do not contain text "wp3_". E.g. I've got a file with two strings:
wp_123
wp3_123
I try $ grep -lr wp_ ~/tmp | xargs grep -vl wp3_
It outputs this file name! But if I remove the linebreak, it's working like I want, i.e. handles string "wp_123 wp3_123" correctly.
How to make it work with search conditions on different strings?
P.S. Sorry for kind of duplicate, but seems that nobody noticed my comment during last hour...
This should work
$ grep -lr 'wp_' ~/tmp | xargs grep -L 'wp3_'
The single quotes are not necessary in this case, but are a good habit to prevent pattern characters from being interpreted by the shell. In your original attempt, -vl means "print each file with at least one line that does not match". Here, -L means "print each file with no lines that match".
I have a file and i want to sort it according to a word and to remove the special characters.
The grep command is used to search for the characters
-b Display the block number at the beginning of each line.
-c Display the number of matched lines.
-h Display the matched lines, but do not display the filenames.
-i Ignore case sensitivity.
-l Display the filenames, but do not display the matched lines.
-n Display the matched lines and their line numbers.
-s Silent mode.
-v Display all lines that do NOT match.
-w Match whole word
but
How to use the grep command to do the file sort and remove the special character and number.
grep searches inside all the files to find matching text. It doesn't really sort and it doesn't really chop and change output. What you want is probably to use the sort command
sort <filename>
and the output sent to either the awk command or the sed command, which are common tools for manipulating text.
sort <filename> | sed 's/REPLACE/NEW_TEXT/g'
something like above I'd imagine.
The following command would do it.
sort FILE | tr -d 'LIST OF SPECIAL CHARS' > NEW_FILE