What command would I use on linux to see if the lines in a file are correctly formatted? So if a text file (names.txt) had the lines;
{"first":joe "last":smith} and "{"first:joe last":smith] it would show me that the first line is correct and the 2nd line is not.
You could use a simple grep like this:
grep -v -E '^{".*?":.*? ".*?":.*?}$' names.txt
Assuming, that your example provided is no real JSON (missing the comma after the first value).
Related
I have output file that can be read by a visualizing tool, the problem is that at the beginning of each line there is one space. Because of this the visualizing tool isn't able to recognize the scripts and, hence crashing. When I manually remove the first column of spaces, it works fine. Is there a way to remove the 1st empty column of spaces using bash command
What I want is to remove the excess column of empty space like shown in this second image using a bash command
At present I use Vim editor to remove the 1st column manually. But I would like to do it using a bash command so that I can automate the process. My file is not just full of columns, it has some independent data line
Using cut or sed would be two simple solutions.
cut
cut -c2- file > file.cut && mv file.cut file
cut cannot modify a file, therefore you need to redirect its output to a different file and then overwrite the old file with the new file.
sed
sed -i 's/^.//' file
-i modifies the file in-place.
I would use
sed -ie 's/^ //' file
to just remove spaces (in case a line does not contain it)
So, I have this assignment which requires me to delete a certain line from a .dat file. Now my file is basically a phone book. I have a Bash script that adds the ID, name, last name, phone number, address, etc., to the .dat file. Now one of the flags is supposed to be -delete and it takes the parameter id. So, basically I need to implement the function where I'd put ./phonebook.sh -delete -id 7 and it would delete the row where the id is 7.
I tried using sed and awk, but nothing is working and it's frustrating. My current code for that short script (delete.sh) is:
id=$1
sed "/$id/d" phonebook.dat
Try this:
On Mac:
sed -i '' -e "/$id/d" phonebook.dat
Otherwise:
sed -i -e "/$id/d" phonebook.dat
By default, sed will output the results to stdout. So, your command was working, but the output wasn't going back into the file. The -i flag says that the file should be replaced with the results. -i is also meant to backup the original file. For example:
sed -i .bk -e "/$id/d" phonebook.dat
The above will create a copy of the original called: phonebook.dat.bk. However, to do in place replacement without a backup, you can specify no value for -i. On the MAC, sed really really really wants a value, so you can pass it an empty string ( making sure there is a space between the -i and the empty quotes ).
I'm making some assumptions because I don't know what the format of your dat file is. I'll assume that the id field is the first field and the file is comma delimited. If I'm wrong, you should be able to modify the below to fit your needs.
I personally like to use grep -v for this problem. From the --help:
-v, --invert-match select non-matching lines
Running this will output every line of a file that does not match your pattern.
id="$1"
grep -v "^${id}," phonebook.dat > phonebook.temp
mv phonebook.temp phonebook.dat
The pattern consists of
^: Beginning of the line
${id}: Your variable
,: Our assumed delimiter
The reason for specifying the beginning of the line to the first delimiter is to avoid deleting entries where the entered id ($1) is a substring of other ids. You wouldn't want to enter 22 and delete id 22 as well as id 122.
I am trying to replace a specific line in a txt file with my shell script, for example;
cat aa.txt:
auditd=0
bladeServerSlot=0
When I run my script I would like to change "bladeServerSlot" to 12 as following;
cat aa.txt:
auditd=0
bladeServerSlot=12
Could you please help me?
Using sed and backreferencing:
sed -r '/bladeServerSlot/ s/(^.*)(=.*)/\1=12/g' inputfile
Using awk , this will search for the line which contains bladeServerSlot and replace the second column of that line.
awk 'BEGIN{FS=OFS="="}/bladeServerSlot/{$2=12}1' inputfile
perl -pe 's/bladeServerSlot=\K\d+/12/' aa.txt > output.txt
The \K is a particular form of the positive lookbehind, which discards all previous matches. So we need to replace only what follows. The s/ is applied by default to $_, which contains the current line. The -p prints $_ for every line, so all other lines are copied. We redirect output to a file.
Is it really necessary to replace the line in your example? As bladeServerSlot is a variable you could reset the value.
bladeServerSlot=`any command`
Or you could just let this variable be filled by a Parameter provided to this script.
bladeServerSlot=$1
With $1being the first parameter of your script. I think this would be the cleaner way do solve your issue than to do fancy regex here. The sed/perl solutions will work, but they are not very clear to other people reading your script.
I have a file contained \n hidden behind each line:
input:
s3741206\n
s2561284\n
s4411364\n
s2516482\n
s2071534\n
s2074633\n
s7856856\n
s11957134\n
s682333\n
s9378200\n
s1862626\n
I want to remove \n behind
desired output:
s3741206
s2561284
s4411364
s2516482
s2071534
s2074633
s7856856
s11957134
s682333
s9378200
s1862626
however, I try this:
tr -d '\n' < file1 > file2
but it goes like below without space and new line
s3741206s2561284s4411364s2516482s2071534s2074633s7856856s11957134s682333s9378200s1862626
I also try sed $'s/\n//g' -i file1 and it doesn't work in mac os.
Thank you.
This is a possible solution using sed:
sed 's/\\n/ /g'
with awk
awk '{sub(/\\n/,"")} 1' < file1 > file2
What you are describing so far in your question+comments doesn't make sense. How can you have a multi-line file with a hidden newline character at the end of each line? What you show as your input file:
s3741206\n
s2561284\n
s4411364\n
etc.
where each "\n" above according to your comment is a single newline character "\n" is impossible. If those "\n"s were newline characters then your file would simply look like:
s3741206
s2561284
s4411364
etc.
There's really only 2 possibilities I can think of:
You are wrongly interpreting what you are seeing in your input file
and/or using the wrong terminology and you actually DO have \r\n
at the end of every line. Run cat -v file to see the \rs as
^Ms and run dos2unix or similar (e.g. sed 's/\r$//' file) to
remove the \rs - you do not want to remove the \ns or you will
no longer have a POSIX text file and so POSIX tools will exhibit
undefined behavior when run on it. If that doesn't work for you then
copy/paste the output of cat -v file into your question so we can
see for sure what is in your file.
Or:
It's also entirely possible that your file is a perfectly fine POSIX
text file as-is and you are incorrectly assuming you will have a
problem for some reason so also include in your question a
description of the actual problem you are having, include an example
of the command you are executing on that input file and the output
you are getting and the output you expected to get.
You could use bash-native string substitution
$ cat /tmp/newline
s3741206\n
s2561284\n
s4411364\n
s2516482\n
s2071534\n
s2074633\n
s7856856\n
s11957134\n
s682333\n
s9378200\n
s1862626\n
$ for LINE in $(cat /tmp/newline); do echo "${LINE%\\n}"; done
s3741206
s2561284
s4411364
s2516482
s2071534
s2074633
s7856856
s11957134
s682333
s9378200
s1862626
I am trying to extract no.s from a file, so I created a script, but grep is giving error:grep: line too long. Can anyone tell me where am I wrong. command is:
echo $(cat filename|grep '\<[0-9]*\>')
Thanks in advance
grep is line-oriented; it will print matching lines to output. Probably you have a huge line in your file, and the resulting line cannot be converted into a string value by shell, as $(...) requires.
First of all, try just cat filename | grep '\<[0-9]*\>' > results and see what is in the results file. Maybe it's enough.
But if you have multiple numbers in a line and you want to extract them all, use -o: grep -o '\<[0-9]*\>'. This will print only matching parts, every match on a new line, even if original matches are on the same line. If you need line numbers, too, add -n.