replacing multiple lines in shell script with only one output file

Question

I have one file Length.txt having multiples names (40) line by line. I want to write a small shell script where it will count the character count of each line of the file and if the count is less than 9 replace those lines with adding extra 8 spaces and 1 at the end of each line.

For example, if the name is

XXXXXX

replace as

XXXXXX        1

I tried with the below coding. It is working for me, however whenever it's replacing the line it is displaying all the lines at a time. So suppose I have 40 lines in Length.txt and out of that 4 lines having less than 9 character count then my output has 160 lines. Can anyone help me to display only 40 line output with the 4 changed lines?

#!/usr/bin/sh
#set -x

while read line;
do
count=`echo $line|wc -m`
        if [ $count -lt 9 ]
        then
        Number=`sed -n "/$line/=" Length.txt`;
        sed -e ""$Number"s/$line/$line       1/" Length4.txt
        fi
done < Length.txt

"Can anyone help me to display only 40 line output with the 4 changed lines?" Yes, we can help. — nanofarad
– nanofarad, Commented Sep 5, 2013 at 12:25

konsolebox · Accepted Answer · 2013-09-05 13:07:21Z

2

A single sed command can do that:

sed -E 's/^.{,8}$/&        1/' file

To modify the contents of the file add -i:

sed -iE 's/^.{,8}$/&        1/' file

Partial output:

94605320        1
105018263
2475218231
7728563        1
        1

* Fixed to include only add 8 spaces not 9, and include empty lines. If you don't want to process empty lines, use {1,8}.

edited Sep 5, 2013 at 13:07

answered Sep 5, 2013 at 12:37

konsolebox

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1 Comment

Adrian Frühwirth Over a year ago

That's actually way more elegant, +1.

Adrian Frühwirth · Accepted Answer · 2013-09-05 12:26:16Z

1

$ cat foo.input
I am longer than 9 characters
I am also longer than 9 characters
I am not
Another long line
short

$ while read line; do printf "$line"; (( ${#line} < 9 )) && printf "        1"; echo; done < foo.input
I am longer than 9 characters
I am also longer than 9 characters
I am not        1
Another long line
short        1

answered Sep 5, 2013 at 12:26

Adrian Frühwirth

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Comments

dogbane · Accepted Answer · 2013-09-05 12:29:47Z

1

Let me show you what is wrong with your script. The only thing missing from your script is that you need to use sed -i to edit file and re-save it after making the replacement.

I'm assuming Length4.txt is just a copy of Length.txt?

I added sed -i to your script and it should work now:

cp Length.txt Length4.txt
while read line;
do
count=`echo $line|wc -m`
        if [ $count -lt 9 ]
        then
        Number=`sed -n "/$line/=" Length.txt`
        sed -ie ""$Number"s/$line/$line       1/" Length4.txt
        fi
done < Length.txt

However, you don't need sed or wc. You can simplify your script as follows:

while IFS= read -r line
do
    count=${#line}
    if (( count < 9 ))
    then
        echo "$line        1"
    else 
        echo "$line"
    fi
done < Length.txt > Length4.txt

answered Sep 5, 2013 at 12:29

dogbane

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2 Comments

Adrian Frühwirth Over a year ago

+1 for actually pointing out the problem of the original snippet and not being as lazy as I am :-)

dogbane Over a year ago

@user2642751 Looks like your version of sed doesn't support -i. You can use a temp file instead, like this: sed -e ""$Number"s/$line/$line 1/" Length4.txt > Length4.tmp && mv Length4.tmp Length4.txt. But I strongly suggest you use an alternative approach.

Ed Morton · Accepted Answer · 2013-09-05 17:37:13Z

0

$ awk -v FS= 'NF<9{$0=sprintf("%s%*s1",$0,8,"")} 1' file
XXXXXX        1

Note how simple it would be to check for a number other than 9 characters and to print some sequence of blanks other than 8.

answered Sep 5, 2013 at 17:37

Ed Morton

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