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 "$correctFilePathAndName" | xargs touch
The '<' redirects via stdin whereas touch needs the filename as an argument. xargs transforms stdin in an argument for touch.
Related
I'm sorry for my poor English, first.
I want to read a file (tel.txt) that contains many tel numbers (a number per line) and use that line to grep command to search about the specific number in the source file (another file)!
I wrote this code :
dir="/home/mujan/Desktop/data/ADSL_CDR_Text_Parts_A"
file="$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 some help...
I'm creating a script of unit tests using shellscripts. That script, stores all the beeline calls from all files inside a directory.
The script is doing it's purpose, but I don't wanna append the file name if grep does not return results.
That's my code:
for file in $(ls)
do
cat $file | egrep -on '^( +)?\bbeeline.*password=;"?' >> testa_scripts.sh
echo $file >> testa_scripts.sh
done
How can I do that?
Thanks
grep returns a falsy exit status (1) if it doesn't find any matching lines, so you can put in an if statement to test if it matched anything. Inverted with ! here:
for file in ./*; do
if ! egrep -on '...' "$file" >> somefile; then
echo 'grep did not match anything'
fi
done
(I don't think there's any need for the ls instead of just a shell glob here.)
I have an executable that takes a file and outputs a line.
I am running a loop over a directory:
for file in $DIRECTORY/*.png
do
./eval $file >> out.txt
done
The output of executable does not contain the name of the file.
I want to append the file name with each output.
EDIT1
Perhaps, I could not explain it correctly
I want the name of the file and the output of the program as well, which is processing the same file, Now I am doing following
for file in $DIRECTORY/*.png
do
echo -n $file >> out.txt
or
printf "%s" "$file" >> out.txt
./eval $file >> out.txt
done
For both new line is inserted
If I understood your question, what you want is:
get the name of the file,
...and the output or the program processing the file (in your case, eval),
...on the same line. And this last part is your problem.
Then I'd suggest composing a single line of text (using echo), comprising:
the name of the file, this is the $file part,
...followed by a separator, you may not need that but it may help further processing of the result. I used ":". You can skip this part if this is not interesting for you,
...followed by the output of the program processing the file: this is the $(...) construct
echo $file ":" $(./eval $file) >> out.txt
...and finally appending this line of text to a file, you got that part right.
please use like this
echo -n `echo ${file}|tr -d '\n'` >> out.txt
OR
newname=`echo ${file}|tr -d '\n'`
echo -n $newname >> out.txt
I'm really new to Linux scripting. I am sure this is simple, but I cannot figure it out.
As part of a script, I am trying to pass the content of a file as arguments of a command in a script:
while read i
do $COMMAND $i
done < file.lst
I want to pass every line of the file.lst as the argument of the command except the very first line of the file. How to I do this?
EDIT:
Here is the section of the script:
while read i
do cp --recursive --preserve=all $i $DIR
done < $DIR/file.lst
while read -r i
do
"$COMMAND" "$i"
done < <(sed -n '2,$p' file.lst)
This solutions does not use a while so I am not entirely sure if it solves your problem, but based on your code sample. you can do the following
tail -n +2 input | xargs echo
This will read all lines from input starting at line 2 and execute echo using the value of the line
the file input contains:
skip
1
2
3
executing that command gives
1
2
3
Just substitute input for the file you want and echo for the command you want
Add an extra read to consume the first line before the while loop begins.
{
read -r;
while read -r i; do
"$COMMAND" "$i"
done
} < file.lst
I'm trying to execute a command for each line coming from a cat command. I'm basing this on sample code I got from a vendor.
Here's the script:
for tbl in 'cat /tmp/tables'
do
echo $tbl
done
So I was expecting the output to be each line in the file. Instead I'm getting this:
cat
/tmp/tables
That's obviously not what I wanted.
I'm going to replace the echo with an actual command that interfaces with a database.
Any help in straightening this out would be greatly appreciated.
You are using the wrong type of quotes.
You need to use the back-quotes rather than the single quote to make the argument being a program running and piping out the content to the forloop.
for tbl in `cat /tmp/tables`
do
echo "$tbl"
done
Also for better readability (if you are using bash), you can write it as
for tbl in $(cat /tmp/tables)
do
echo "$tbl"
done
If your expectations are to get each line (The for-loops above will give you each word), then you may be better off using xargs, like this
cat /tmp/tables | xargs -L1 echo
or as a loop
cat /tmp/tables | while read line; do echo "$line"; done
The single quotes should be backticks:
for tbl in `cat /etc/tables`
Although, this will not get you output/input by line, but by word. To process line by line, you should try something like:
cat /etc/tables | while read line
echo $line
done
With while loop:
while read line
do
echo "$line"
done < "file"
while IFS= read -r tbl; do echo "$tbl" ; done < /etc/tables
read this.
You can do a lot of parsing in bash by redefining the IFS (Input Field Seperator), for example
IFS="\t\n" # You must use double quotes for escape sequences.
for tbl in `cat /tmp/tables`
do
echo "$tbl"
done