I have this command sed -i 's/[A-Za-z]//g' file.txt that gets rid of any letters in my file but now have come to the realization that I need to be a little more steep with these errors.
How can I alter this command to Delete the line completely if there is letters in it?
000000asd000,12
000000000000,123
Would go to this
000000000000,123
sed -i '/[a-zA-Z]/d' file.txt
The /.../ command to match lines containing a letter, then d to delete the line.
Related
I have a UTF-8/no BOM file (converted from ISO-8859-1) that has 31214 lines. I have already run dos2unix on the file. When I open it in notepad++, I see a blank line underneath. When I remove this blank line, the line count reduces by one. I save it under a different name and when I tail the file, the prompt displays on the same line. From bash, how do I delete the blank line in the 1st file to produce the result displayed below in the 2nd file?
The goal is to do this from bash w/o manually deleting the line in notepad++
1st file:
[user#server]$ cat file1.txt | wc -l
31214
[user#server]$ tail file1.txt
T 31212 Data 20170517
[user#server]$
2nd file (edited with notepad++)
[user#server]$ cat file2.txt | wc -l
31213
[user#server]$ tail file2.txt
T 31212 Data 20170517[user#server]$
That's the trailing newline of the last line. Some editors allow you to go to the nonexisting "empty" line at the end, some don't show it. Again, some programs may allow you to remove the final newline, but note that e.g. POSIX in effect requires it to be there, and some standard utilities act oddly if it isn't present.
E.g. wc -l counts the number of newlines in the input file (printf "foo\nbar" | wc -l shows 1) so removing the final newline does decrease the line count.
Also, Bash prints the prompt wherever it was that the cursor was left on the screen, so if you print something that doesn't have the trailing newline, the prompt will be placed where the final incomplete line ended, as you saw.
There's no need to remove that final newline, just leave it there.
To remove the final newline character it is possible, as explained here, to use
sed -i '$ s/.$//' your.file
which will substitute nothing for the last character in the last line of the file (if you want to delete smth else from the end of the file you can replace the regex .$ with smth-else$). -i means ‘substitute in-place’ (in FreeBSD/MacOS you need to add an empty string as an argument: sed -i "" '$ s/.$//' your.file)
The file2.txt is missing a trailing newline.
Yes, a text file should end on a newline character.
Given that you do know that a trailing newline is missing, this command should be enough to correct the problem:
$ echo >> file2.txt
My question is probably rather simple. I'm trying to replace sequences of strings that are at the beginning of lines in a file. For example, I would like to replace any instance of the pattern "GN" with "N" or "WR" with "R", but only if they are the first 2 characters of that line. For example, if I had a file with the following content:
WRONG
RIGHT
GNOME
I would like to transform this file to give
RONG
RIGHT
NOME
I know i can use the following to replace any instance of the above example;
sed -i 's/GN/N/g' file.txt
sed -i 's/WR/R/g' file.txt
The issue is that I want this to happen only if the above patterns are the first 2 characters in any given line. Possibly an IF statement, although i'm not sure what the condition would look like. Any pointers in the right direction would be much appreciated, thanks.
just add the circumflex, remove g suffix (unnecessary, since you want at most one replacement), you can also combine them in one script.
sed -i 's/^GN/N/;s/^WR/R/' file.txt
Use the start-of-string regexp anchor ^:
sed -i 's/^GN/N/' file.txt
sed -i 's/^WR/R/' file.txt
Since sed is line-oriented, start-of-string == start-of-line.
I want to insert word after nth line after pattern using sed.
I tied to modify this command but it inserts only in first line after pattern.
sed -i '/myPattern/a \ LineIWantToinser ' myFile
What command should I use to insert for example in third line after pattern?
Easiest way to do it with GNU sed is.. (maybe some direct solution exists!?)
sed -n '/pattern/=' file
to see line where pattern is (grep also can be used here with -n)
then if linenumber+ numoflines is for example 123
sed '123aSOME INSERTED TEXT AFTER THAT LINE' file
where little a is append command (after that line, if i is used will be pre pattern line)
ps. I'm eager to see if #neronlevelu (or other sed Lover) will find some better sed solution.
Edit: i've found it, it seems a for append or i for insert must? be on first position on line when using { with ; inside } like
sed '/pattern/{N;N;N;
a SOME TEXT FOR INSERTING
}' file
sed '/pattern/{N;N;N;i \
Line to add after 3 lines with patterne as starting counter
' YourFile
number of N to add line between pattern and inserted line.
there is no check for end of file or pattern in the 3 lines. (not specified in PO)
A version with bash and ed:
ed -s myFile <<<$'/myPattern/+3a\n LineIWantToinser \n.\nwq'
ed enables us to use the line addressing /myPattern/+3.
In looking for a command to delete a line (or lines) from a text file that contain a certain string.
For example
I have a text file as follows
Sat 21-12-2014,10.21,78%
Sat 21-12-2014,11.21,60%
Sun 22-12-2014,09.09,21%
I want to delete all lines that have "21-12-2014" in them.
I'm not able to find a solution that works.
According to #twalberg there is more three alternate solution for this question, which I'm explaining is as follows for future reader of this question for more versatile solutions:
With grep command
grep -v 21-12-2014 filename.txt
explanations:
-v is used to find non-matching lines
With awk command
awk '! /21-12-2014/' filename.txt
explanations:
! is denoting it will print all other lines that contain match of the string. It is not operator signify ignorance.
With sed command
sed -e '/21-12-2014/d' < filename.txt
explanations:
-e is signify scripted regex to be executed
d is denoting delete any match
< is redirecting the input file content to command
Try doing this :
sed -i.bak '/21-12-2014/d' *
A bit of explanations :
sed : the main command line, put the mouse pointer on sed
-i.bak : replace the file in place and make a backup in a .bak file
// is the regex
d means: delete
I have a lot of files that end in the classical ^M, an artifact from my Windows times. As this is all source code, git actually thinks those files changed, so I want to remove those nasty lines once and for all.
Here is what I created:
sed -i 's/^M//g' file
But that does not work. Of course I did not type a literal ^M but rather ^V^M (ctrl V, ctrl M). In vim it works (:%s/s/^M//g) and if I modify it like this:
sed -i 's/^M/a/g' file
It also works, i.e. it ends every line with an 'a'. It also works to do this:
sed -i 's/random_string//g' file
Where random_string exists in the file. So I can replace ^M by any character and I can remove lines but I cannot remove ^M. Why?
Note: It is important that it is just removed, no replacing by another invisible char or something. I would also like to avoid double execution and adding an arbitrary string and removing it afterwards. I want to understand why this fails (but it does not report an error).
That character is matched with \r by sed. Use:
sed -e "s/\r//g" input-file
For my case, I had to do
sed -e "s/\r/\n/g" filename.csv
After that wc -l filename Showed correct output instead of 0 lines.