I want to read all the files in a particular directory, and I want to read it one by one.
Here's what i've done so far. ls successfully get all the files from a specified directory, but could not give the file names to me one by one. It Echos the files one time only. I want to get it one by one because I need to do some parsing and use it somehwere.
#!/bin/sh
echo Content-type: application/json
echo ""
for output in "ls /home/myComputer/Desktop/*";
do
echo $output
done
You can do
for output in /home/myComputer/Desktop/*
do
echo $output
done
If there's any risk that /home/myComputer/Desktop is empty, please follow #JIDs advice in the comments below.
The next solution also works when the dir is empty.
ls /home/myComputer/Desktop/ 2>/dev/null |while read -r output; do
echo ${output}
done
This construction is nice to know, for when you want to split the input lines in some fields:
cat someFile | while read -r field1 field2 remainingfields; do
Related
I would like to know how to access the contents of a variety of txt files by passing arguments into shell scripts. I'll have different files and I'm expecting to execute with this command:
./script.sh FileA.txt
What should I put into my shell script so that I can access and manipulate the contents of the files?
I tried this but it outputs 0:
echo "$#"
I also tried these, but both output nothing:
for i in $1
do
echo "$i"
done
echo "$1"
To sum up the contents see this link to understand bash arguments more https://tecadmin.net/tutorial/bash-scripting/bash-command-arguments/ . Also as #Barmar said, to iterate through a list of arguments of unknown quantity use for i in "$#" .
edit
and as #Barmar said, $1 is simply the name of the argument. So echoing $1 will just echo the name.
I don't understand your question fully. Lets assume you have list of file names in a text file "FileA.txt".
And you wanted to run some commands for each file in the "FileA.txt" file.
Can you try below:
for i in `cat $1`
do
echo $i
done
I'm sorry for my poor English, first.
I want to read a file (tel.txt) that contains many tel numbers (a number per line) and use that line to grep command to search about the specific number in the source file (another file)!
I wrote this code :
dir="/home/mujan/Desktop/data/ADSL_CDR_Text_Parts_A"
file="$dir/tel.txt"
datafile="$dir/ADSL_CDR_Like_Tct4_From_960501_to_97501_Part0.txt"
while IFS= read -r line
do
current="$line"
echo `grep -F $current "$datafile" >> output.txt`
done < $file
the tel file sample :
44001547
44001478
55421487
but that code returns nothing!
when I declare 'current' variable with literals it works correctly!
what happened?!
Your grep command is redirected to write its output to a file, so you don't see it on the terminal.
Anyway, you should probably be using the much simpler and faster
grep -Ff "$file" "$datafile"
Add | tee -a output.txt if you want to save the output to a file and see it at the same time.
echo `command` is a buggy and inefficient way to write command. (echo "`command`" would merely be inefficient.) There is no reason to capture standard output into a string just so that you can echo that string to standard output.
Why don't you search for the line var directly? I've done some tests, this script works on my linux (CentOS 7.x) with bash shell:
#!/bin/bash
file="/home/mujan/Desktop/data/ADSL_CDR_Text_Parts_A/tel.txt"
while IFS= read -r line
do
echo `grep "$line" /home/mujan/Desktop/data/ADSL_CDR_Text_Parts_A/ADSL_CDR_Like_Tct4_From_960501_to_97501_Part0.tx >> output.txt`
done < $file
Give it a try... It shows nothing on the screen since you're redirecting the output to the file output.txt so the matching results are saved there.
You should use file descriptors when reading with while loop.instead use for loop to avoid false re-directions
dir="/home/mujan/Desktop/data/ADSL_CDR_Text_Parts_A"
file="$dir/tel.txt"
datafile="$dir/ADSL_CDR_Like_Tct4_From_960501_to_97501_Part0.txt"
for line in `cat $file`
do
current="$line"
echo `grep -F $current "$datafile" >> output.txt`
done
I need help copying content from various files to others (same name and format, different path).
For example, $HOME/initial/baby.desktop has text which I need to write into $HOME/scripts/baby.desktop. This is very simple for a single file, but I have 2500 files in $HOME/initial/ and the same number in $HOME/scripts/ with corresponding names (same names and format). I want append (copy) the content of file in path A to path B (which have the same name and format), to the end of file in path B without erase the content of file in path B.
Example content of $HOME/initial/*.desktop to final $HOME/scripts/*.desktop. I tried the following, but it don't work:
cd $HOME/initial/
for i in $( ls *.desktop ); do egrep "Icon" $i >> $HOME/scripts/$i; done
Firstly, I would backup $HOME/initial and $HOME/scripts, because there is lots of scope for people misunderstanding your question. Like this:
cd $HOME
tar -cvf initial.tar initial
tar -cvf scripts.tar scripts
That will put all the files in $HOME/initial into a single tarfile called initial.tar and all the files in $HOME/scripts into a single tarfile called scripts.tar.
Now for your question... in general, if you want to put the contents of FileB onto the end of FileA, the command is
cat FileB >> FileA
Note the DOUBLE ">>" which means "append" rather than single ">" which means overwrite.
So, I think you want to do this:
cd $HOME/initial/baby.desktop
cat SomeFile >> $HOME/scripts/baby.desktop/SomeFile
where SomeFile is the name of any file you choose to test with. I would test that has worked and then, if you are happy with that, go ahead and run the same command inside a loop:
cd $HOME/initial/baby.desktop
for SOURCE in *
do
DESTINATION="$HOME/scripts/baby.desktop/$SOURCE"
echo Appending "$SOURCE" to "$DESTINATION"
#cat "$SOURCE" >> "$DESTINATION"
done
When the output looks correct, remove the "#" at the start of the penultimate line and run it again.
I solved it, if some people want learn how to resolve is very simple:
using Sed
I need only the match (or pattern) line "Icon=/usr/share/some_picture.png into $HOME/initial/example.desktop to other with same name and format $HOME/scripts/example.desktop, but I had a lot of .desktop files (2500 files)
cd $HOME/initial
STRING_LINE=`grep -l -R "Icon=" *.desktop`
for i in $STRING_LINE; do sed -ne '/Icon=/ p' $i >> $HOME/scripts/$i ; done
_________
If you need only copy all to other file with same name and format
using cat
cd $HOME/initial
STRING_LINE=`grep -l -R "Icon=" *.desktop`
for i in $STRING_LINE; do cat $i >> $HOME/scripts/$i ; done
My code is given below. Echo works fine. But, the moment I redirect output of echo to touch, I get an error "no such file or directory". Why ? How do i fix it ?
If I copy paste the output of only echo, then the file is created, but not with touch.
while read line
do
#touch < echo -e "$correctFilePathAndName"
echo -e "$correctFilePathAndName"
done < $file.txt
If you have file names in each line of your input file file.txt then you don't need to do any loop. You can just do:
touch $(<file.txt)
to create all the files in one single touch command.
You need to provide the file name as argument and not via standard input. You can use command substitution via $(…) or `…`:
while read line
do
touch "$(echo -e "$correctFilePathAndName")"
done < $file.txt
Ehm, lose the echo part... and use the correct variable name.
while read line; do
touch "$line"
done < $file.txt
try :
echo -e "$correctFilePathAndName" | touch
EDIT : Sorry correct piping is :
echo -e "$correctFilePathAndName" | xargs touch
The '<' redirects via stdin whereas touch needs the filename as an argument. xargs transforms stdin in an argument for touch.
I'm encountering many files with the same content and the same name on some of my servers. I need to quarantine these files for analysis so I can't just remove the duplicates. The OS is Linux (centos and ubuntu).
I enumerate the file names and locations and put them into a text file.
Then I do a for statement to move the files to quarantine.
for file in $(cat bad-stuff.txt); do mv $file /quarantine ;done
The problem is that they have the same file name and I just need to add something unique to the filename to get it to save properly. I'm sure it's something simple but I'm not good with regex. Thanks for the help.
Since you're using Linux, you can take advantage of GNU mv's --backup.
while read -r file
do
mv --backup=numbered "$file" "/quarantine"
done < "bad-stuff.txt"
Here's an example that shows how it works:
$ cat bad-stuff.txt
./c/foo
./d/foo
./a/foo
./b/foo
$ while read -r file; do mv --backup=numbered "$file" "./quarantine"; done < "bad-stuff.txt"
$ ls quarantine/
foo foo.~1~ foo.~2~ foo.~3~
$
I'd use this
for file in $(cat bad-stuff.txt); do mv $file /quarantine/$file.`date -u +%s%N`; done
You'll get everyfile with a timestamp appended (in nanoseconds).
You can create a new file name composed by the directory and the filename. Thus you can add one more argument in your original code:
for ...; do mv $file /quarantine/$(echo $file | sed 's:/:_:g') ; done
Please note that you should replace the _ with a proper character which is special enough.