Extracting value from a flat file using shell script - shell

I'm trying to extract the value present between brackets in the last row of a flat file e.g. " last_line (4) ". This is the last line and I want to extract 4 and store it in a variable. I have extracted the last row using tail command but now I am unable to extract the value between the brackets.
Kindly help.

Using awk:
$ cat input
first line
2nd line
last line (4) with some data
$ awk -F'[()]' 'END{print $2}' input
4

l=$(tail -n1 filename); tmp=${l##*(}; tmp=${tmp%)*}; printf "tmp: %s\n" $tmp
Output
tmp: 4
Written in script format, you are using substring removal to trim everything up to the first ( and everything after the last ) from the last line, leaving only 4:
l=$(tail -n1 filename) ## get the last line
tmp=${l##*(} ## trim to ( from left
tmp=${tmp%)*} ## trim to ) from right
printf "tmp: %s\n" $tmp

sed:
sed -n '${s/.*(//;s/).*//;p}' file

U can use this script.
In this script i saved the last line in a tmp file and at last removed it.
the number between the brackets() is in variable WORD
#!/bin/ksh
if test "${DEBUG}" = "Y"
then
set -vx
fi
tail -1 input>>tmp
WORD=`sed -n 's/.*(//;s/).*//;p' tmp`
echo $WORD
rm tmp

Related

Using sed in order to change a specific character in a specific line

I'm a beginner in bash and here is my problem. I have a file just like this one:
Azzzezzzezzzezzz...
Bzzzezzzezzzezzz...
Czzzezzzezzzezzz...
I try in a script to edit this file.ABC letters are unique in all this file and there is only one per line.
I want to replace the first e of each line by a number who can be :
1 in line beginning with an A,
2 in line beginning with a B,
3 in line beginning with a C,
and I'd like to loop this in order to have this type of result
Azzz1zzz5zzz1zzz...
Bzzz2zzz4zzz5zzz...
Czzz3zzz6zzz3zzz...
All the numbers here are random int variables between 0 and 9. I really need to start by replacing 1,2,3 in first exec of my loop, then 5,4,6 then 1,5,3 and so on.
I tried this
sed "0,/e/s/e/$1/;0,/e/s/e/$2/;0,/e/s/e/$3/" /tmp/myfile
But the result was this (because I didn't specify the line)
Azzz1zzz2zzz3zzz...
Bzzzezzzezzzezzz...
Czzzezzzezzzezzz...
I noticed that doing sed -i "/A/ s/$/ezzz/" /tmp/myfile will add ezzz at the end of A line so I tried this
sed -i "/A/ 0,/e/s/e/$1/;/B/ 0,/e/s/e/$2/;/C/ 0,/e/s/e/$3/" /tmp/myfile
but it failed
sed: -e expression #1, char 5: unknown command: `0'
Here I'm lost.
I have in a variable (let's call it number_of_e_per_line) the number of e in either A, B or C line.
Thank you for the time you take for me.
Just apply s command on the line that matches A.
sed '
/^A/{ s/e/$1/; }
/^B/{ s/e/$2/; }
# or shorter
/^C/s/e/$3/
'
s command by default replaces the first occurrence. You can do for example s/s/$1/2 to replace the second occurrence, s/e/$1/g (like "Global") replaces all occurrences.
0,/e/ specifies a range of lines - it filters lines from the first up until a line that matches /e/.
sed is not part of Bash. It is a separate (crude) programming language and is a very standard command. See https://www.grymoire.com/Unix/Sed.html .
Continuing from the comment. sed is a poor choice here unless all your files can only have 3 lines. The reason is sed processes each line and has no way to keep a separate count for the occurrences of 'e'.
Instead, wrapping sed in a script and keeping track of the replacements allows you to handle any file no matter the number of lines. You just loop and handle the lines one at a time, e.g.
#!/bin/bash
[ -z "$1" ] && { ## valiate one argument for filename provided
printf "error: filename argument required.\nusage: %s filename\n" "./$1" >&2
exit 1
}
[ -s "$1" ] || { ## validate file exists and non-empty
printf "error: file not found or empty '%s'.\n" "$1"
exit 1
}
declare -i n=1 ## occurrence counter initialized 1
## loop reading each line
while read -r line || [ -n "$line" ]; do
[[ $line =~ ^.*e.*$ ]] || continue ## line has 'e' or get next
sed "s/e/1/$n" <<< "$line" ## substitute the 'n' occurence of 'e'
((n++)) ## increment counter
done < "$1"
Your data file having "..." at the end of each line suggests your files is larger than the snippet posted. If you have lines beginning 'A' - 'Z', you don't want to have to write 26 separate /match/s/find/replace/ substitutions. And if you have somewhere between 3 and 26 (or more), you don't want to have to rewrite a different sed expression for every new file you are faced with.
That's why I say sed is a poor choice. You really have no way to make the task a generic task with sed. The downside to using a script is it will become a poor choice as the number of records you need to process increase (over 100000 or so just due to efficiency)
Example Use/Output
With the script in replace-e-incremental.sh and your data in file, you would do:
$ bash replace-e-incremental.sh file
Azzz1zzzezzzezzz...
Bzzzezzz1zzzezzz...
Czzzezzzezzz1zzz...
To Modify file In-Place
Since you make multiple calls to sed here, you need to redirect the output of the file to a temporary file and then replace the original by overwriting it with the temp file, e.g.
$ bash replace-e-incremental.sh file > mytempfile && mv -f mytempfile file
$ cat file
Azzz1zzzezzzezzz...
Bzzzezzz1zzzezzz...
Czzzezzzezzz1zzz...

How to get a number with variable number of digits from a string in a file using bash script?

I have the following file:
APP_VERSION.ts
export const APP_VERSION = 1;
This is the only content of that file, and the APP_VERSION variable will be incremented as needed.
So, the APP_VERSION could be a single digit number or multiple digit number, like 15 or 999, etc.
I need to use that value in one of my bash scripts.
use-app-version.sh
APP_VERSION=`cat src/constants/APP_VERSION.ts`
echo $APP_VERSION
I know I can read it with cat. But how can I parse that string so I can get exactly the APP_VERSION value, whether it's 1 or 999, for example.
sed -En 's/(^.*APP_VERSION.*)([[:digit:]]+.*)(\;.*$)/\2/p' src/constants/APP_VERSION
Using sed, split the line into three sections defined by opening and closing brackets. Substitute the line for second section on ( the version value) and print.
You may use this awk:
app_ver=$(awk -F '[[:blank:];=]+' '$(NF-2) == "APP_VERSION" {print $(NF-1)}' src/constants/APP_VERSION.ts)
echo "$app_ver"
1
You can concat some commands to remove everything else:
APP_VERSION=`cat src/constants/APP_VERSION.ts | awk -F '=' '{print $2}' | tr -d ' ' | tr -d ';'`
1 - Cat get all file content
2 - AWK gets all content after '='
3 - Remove space
4 - Remove ;
A simple
APP_VERSION=$(grep --text -Eo '[0-9]+' src/constants/APP_VERSION.ts)
should be enough
With bash only:
APP_VERSION=$(cat src/constants/APP_VERSION.ts)
APP_VERSION=${APP_VERSION%;}
APP_VERSION=${APP_VERSION/*= }
Line 2 removes the trailing ';', line 3 removes everything before "= ".
Alternatively, you could set APP_VERSION as an array, take 5th element, and remove trailing ';'.
Or, another solution, using IFS:
IFS='=;' read a APP_VERSION < src/constants/APP_VERSION.ts
In this version, the space will remain before version number.
Assuming that the task can be rephrased to "extract the digits from a file", there are a few options:
Delete all characters that aren't digits with tr:
version=$(tr -cd '[:digit:]' < infile)
Use grep to match all digits and retain nothing but the match:
version=$(grep -Eo '[[:digit:]]+' infile)
Read file into string and delete all non-digits with just Bash:
contents=$(< infile)
version=${contents//[![:digit:]]}

sed insert line after a match only once [duplicate]

UPDATED:
Using sed, how can I insert (NOT SUBSTITUTE) a new line on only the first match of keyword for each file.
Currently I have the following but this inserts for every line containing Matched Keyword and I want it to only insert the New Inserted Line for only the first match found in the file:
sed -ie '/Matched Keyword/ i\New Inserted Line' *.*
For example:
Myfile.txt:
Line 1
Line 2
Line 3
This line contains the Matched Keyword and other stuff
Line 4
This line contains the Matched Keyword and other stuff
Line 6
changed to:
Line 1
Line 2
Line 3
New Inserted Line
This line contains the Matched Keyword and other stuff
Line 4
This line contains the Matched Keyword and other stuff
Line 6
You can sort of do this in GNU sed:
sed '0,/Matched Keyword/s//New Inserted Line\n&/'
But it's not portable. Since portability is good, here it is in awk:
awk '/Matched Keyword/ && !x {print "Text line to insert"; x=1} 1' inputFile
Or, if you want to pass a variable to print:
awk -v "var=$var" '/Matched Keyword/ && !x {print var; x=1} 1' inputFile
These both insert the text line before the first occurrence of the keyword, on a line by itself, per your example.
Remember that with both sed and awk, the matched keyword is a regular expression, not just a keyword.
UPDATE:
Since this question is also tagged bash, here's a simple solution that is pure bash and doesn't required sed:
#!/bin/bash
n=0
while read line; do
if [[ "$line" =~ 'Matched Keyword' && $n = 0 ]]; then
echo "New Inserted Line"
n=1
fi
echo "$line"
done
As it stands, this as a pipe. You can easily wrap it in something that acts on files instead.
If you want one with sed*:
sed '0,/Matched Keyword/s//Matched Keyword\nNew Inserted Line/' myfile.txt
*only works with GNU sed
This might work for you:
sed -i -e '/Matched Keyword/{i\New Inserted Line' -e ':a;n;ba}' file
You're nearly there! Just create a loop to read from the Matched Keyword to the end of the file.
After inserting a line, the remainder of the file can be printed out by:
Introducing a loop place holder :a (here a is an arbitrary name).
Print the current line and fetch the next into the pattern space with the ncommand.
Redirect control back using the ba command which is essentially a goto to the a place holder. The end-of-file condition is naturally taken care of by the n command which terminates any further sed commands if it tries to read passed the end-of-file.
With a little help from bash, a true one liner can be achieved:
sed $'/Matched Keyword/{iNew Inserted Line\n:a;n;ba}' file
Alternative:
sed 'x;/./{x;b};x;/Matched Keyword/h;//iNew Inserted Line' file
This uses the Matched Keyword as a flag in the hold space and once it has been set any processing is curtailed by bailing out immediately.
If you want to append a line after first match only, use AWK instead of SED as below
awk '{print} /Matched Keyword/ && !n {print "New Inserted Line"; n++}' myfile.txt
Output:
Line 1
Line 2
Line 3
This line contains the Matched Keyword and other stuff
New Inserted Line
Line 4
This line contains the Matched Keyword and other stuff
Line 6

shell: how to read a certain column in a certain line into a variable

I want to extract the first column of the last line of a text file. Instead of output the content of interest in another file and read it in again, can I just use some command to read it into a variable directly?
For exampole, if my file is like this:
...
123 456 789(this is the last line)
What I want is to read 123 into a variable in my shell script. How can I do that?
One approach is to extract the line you want, read its columns into an array, and emit the array element you want.
For the last line:
#!/bin/bash
# ^^^^- not /bin/sh, to enable arrays and process substitution
read -r -a columns < <(tail -n 1 "$filename") # put last line's columns into an array
echo "${columns[0]}" # emit the first column
Alternately, awk is an appropriate tool for the job:
line=2
column=1
var=$(awk -v line="$line" -v col="$column" 'NR == line { print $col }' <"$filename")
echo "Extracted the value: $var"
That said, if you're looking for a line close to the start of a file, it's often faster (in a runtime-performance sense) and easier to stick to shell builtins. For instance, to take the third column of the second line of a file:
{
read -r _ # throw away first line
read -r _ _ value _ # extract third value of second line
} <"$filename"
This works by using _s as placeholders for values you don't want to read.
I guess with "first column", you mean "first word", do you?
If it is guaranteed, that the last line doesn't start with a space, you can do
tail -n 1 YOUR_FILE | cut -d ' ' -f 1
You could also use sed:
$> var=$(sed -nr '$s/(^[^ ]*).*/\1/p' "file.txt")
The -nr tells sed to not output data by default (-n) and use extended regular expressions (-r to avoid needing to escape the paranthesis otherwise you have to write \( \))). The $ is an address that specifies the last line. The regular expression anchors the beginning of the line with the first ^, then matches everything that is not a space [^ ]* and puts that the result into a capture group ( ) and then gets rid of the rest of the line .* by replacing the line with the capture group \1, then print p to print the line.

appending text to specific line in file bash

So I have a file that contains some lines of text separated by ','. I want to create a script that counts how much parts a line has and if the line contains 16 parts i want to add a new one. So far its working great. The only thing that is not working is appending the ',' at the end. See my example below:
Original file:
a,a,a,a,a,a,a,a,a,a,a,a,a,a,a,a
b,b,b,b,b,b
a,a,a,a,a,a,a,a,a,a,a,a,a,a,a,a,a
b,b,b,b,b,b
a,a,a,a,a,a,a,a,a,a,a,a,a,a,a,a
Expected result:
a,a,a,a,a,a,a,a,a,a,a,a,a,a,a,a,xx
b,b,b,b,b,b
a,a,a,a,a,a,a,a,a,a,a,a,a,a,a,a,a
b,b,b,b,b,b
a,a,a,a,a,a,a,a,a,a,a,a,a,a,a,a,xx
This is my code:
while read p; do
if [[ $p == "HEA"* ]]
then
IFS=',' read -ra ADDR <<< "$p"
echo ${#ADDR[#]}
arrayCount=${#ADDR[#]}
if [ "${arrayCount}" -eq 16 ];
then
sed -i "/$p/ s/\$/,xx/g" $f
fi
fi
done <$f
Result:
a,a,a,a,a,a,a,a,a,a,a,a,a,a,a,a
,xx
b,b,b,b,b,b
a,a,a,a,a,a,a,a,a,a,a,a,a,a,a,a,a
b,b,b,b,b,b
a,a,a,a,a,a,a,a,a,a,a,a,a,a,a,a
,xx
What im doing wrong? I'm sure its something small but i cant find it..
It can be done using awk:
awk -F, 'NF==16{$0 = $0 FS "xx"} 1' file
a,a,a,a,a,a,a,a,a,a,a,a,a,a,a,a,xx
b,b,b,b,b,b
a,a,a,a,a,a,a,a,a,a,a,a,a,a,a,a,a
b,b,b,b,b,b
a,a,a,a,a,a,a,a,a,a,a,a,a,a,a,a,xx
-F, sets input field separator as comma
NF==16 is the condition that says execute block inside { and } if # of fields is 16
$0 = $0 FS "xx" appends xx at end of line
1 is the default awk action that means print the output
For using sed answer should be in the following:
Use ${line_number} s/..../..../ format - to target a specific line, you need to find out the line number first.
Use the special char & to denote the matched string
The sed statement should look like the following:
sed -i "${line_number}s/.*/&xx/"
I would prefer to leave it to you to play around with it but if you would prefer i can give you a full working sample.

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