SED, using variables and in with an array - bash

What I am trying to do is run the sed on multiple files in the directory Server_Upload, using variables:
AB${count}
Corresponds, to some variables I made that look like:
echo " AB1 = 2010-10-09Three "
echo " AB2 = 2009-3-09Foo "
echo " AB3 = Bar "
And these correspond to each line which contains a word in master.ta, that needs changing in all the text files in Server_Upload.
If you get what I mean... great, I have tried to explain it the best I can, but if you are still miffed I'll give it another go as I found it really hard to convey what I mean.
cd Server_Upload
for fl in *.UP; do
mv $fl $fl.old
done
count=1
saveIFS="$IFS"
IFS=$'\n'
array=($(<master.ta))
IFS="$saveIFS"
for i in "${array[#]}"
do
sed "s/$i/AB${count}/g" $fl.old > $fl
(( count++ ))
done
It runs, doesn't give me any errors, but it doesn't do what I want, so any ideas?

Your loop should look like this:
while read i
do
sed "s/$i/AB${count}/g" $fl.old > $fl
(( count ++ ))
done < master.ta
I don't see a reason to use an array or something similar. Does this work for you?

It's not exactly clear to me what you are trying to do, but I believe you want something like:
(untested)
do
eval repl=\$AB${count}
...
If you have a variable $AB3, and a variable $count, $AB${count} is the concatenation of $AB and $count (so if $AB is empty, it is the same as $count). You need to use eval to get the value of $AB3.

It looks like your sed command is dependent on $fl from inside the first for loop, even though the sed line is outside the for loop. If you're on a system where sed does in-place editing (the -i option), you might do:
count=1
while read i
do
sed -i'.old' -e "s/$i/AB${count}/g" Server_Upload/*.UP
(( count ++ ))
done < master.ta
(This is the entire script, which incorporates Patrick's answer, as well.) This should substitute the text ABn for every occurrence of the text of the nth line of master.ta in any *.UP file.

Does it help if you move the first done statement from where it is to after the second done?

Related

sed not parsing variable

I'm stuck with a bash script here...
I have variables:
hostname1="sxxx" hostname2="vbbbb" hostname3="sggg" hostname4="aaa" ...
I'm trying to change the 12th line of every files in a folder with the host-name variables.
The files are server1.txt server2.txt server3.txt server4.txt ...
I'm trying to do this with a while loop:
i=1
imax=1
while [[ $i -le 20 ]]
do
sed -i "12s/.*/$hostname$imax/" server$((imax)).txt
(( i++ ))
(( imax++ ))
if [[ imax -eq 21 ]]
then
imax=1
fi
done
what I want to do with sed is to concatenate the word host-name with imax and then use it as variable.
Maybe with this I'm clear enough:
$hostname=hostname$imax; //for exammple
sed -i "12s/.*/$hostname/" server$((imax)).txt // i need here the variable $hostname to have the content "sxxx"
You achieve this by using indirect parameter expansion,
${!var}
Here, var - is a dynamically generated variable name.
Try this:
sed -i "12s/.*/${!hostname}/" server$((imax)).txt
Example:
$ hostname1="sat"
$ i=1
$ hostval="hostname$i"
$ echo ${!hostval}
sat
I'd use the following. Use change, instead of switch in sed. Then strong quote my sed, and unquote my variables.
for i in {1..20}; do
eval h='$'hostname$i
sed -i '12c'$h'' server$i.txt
done
Bash 3 and over supports number ranges in a for looop, easier than your while loop. I also have no idea what you are doing with imax instead of just i, b/c you exit at 20, but change imax value to 1... which it will never use.
edit: b/c I misread
Basically, your problem comes from a variable interpretation.
Try using sed >=4.2.2, should work with your code.

How do you name output files using an increment after a bash file loop?

I'm trying to treat a bunch of files (five) with an awk command and name the output files using an incrementation.
The input files have complicated names. I know how to reuse the files' basenames to rename the outputs but I want to simplify the file names.
this is my code:
for f in *.txt; do
for i in {1..5}; do
echo processing file $f
awk '
{ if ($1=="something" && ($5=="60" || $5=="61"|| $5=="62"|| $5=="63"|| $5=="64"|| $5=="65"|| $5=="66"|| $5=="67"|| $5=="68"|| $5=="69"|| $5=="70"))
print }' $b.txt>"file_treated"$i.txt
echo processing file $f over
done
done
I understand that the error is in the second line because what I wrote runs the second loop for each value of the first one. I want each value of the first loop to correspond to one value of the second one.
Hope this was clear enough
How about:
i=0
for f in *.txt; do
let i++;
awk '$1=="something" && ($5 >= 60 && $5 <=70)' "$f" > file_treated"${i}".txt
done
I simplified your awk command and straightened out your various quoting issues. I also removed the $b.txt since you were simply recreating $f. I left the echo $b etc in case you actually wanted that but it could just as easily be replaced with echo "$f".
Use a counter:
i=1
for f in *.txt
do
echo "$f is number $((i++))"
done

Replace part of lines of a file that match a string in Bash

Here is what I'd like to do:
I have a file which is space delimited, and I have an array of integers. I need to search each line of the file (inv.txt) for each element in the array. If the pattern is found, I need to retain the first 7 characters of the line and replace the rest with " 0; 0;"
I'm very new to scripting and what I've tried so far is this: (I think I've gotten a little off track)
removing=(3199 26543)
for line in inv.txt
do
for i in "${#removing[#]}"
do
sed -i "s/${removing[$1]/c\ ${line:0,7} 0; 0;/g inv.txt
done
done
This gave me a sed -e expression #1 char 0: no previous regular expression
I doubt this is the best way to accomplish the task (if it even will)...Any help is appreciated, thanks in advance!
Edit: I'm using an array because it will actually contain a couple thousand entries! The two I put in it for this scenario is just for testing.
You could actually do that with one sed (if it supports -r):
sed -rie '/3199|26543/s/^(.{7}).*/\1\t0;\t0;/' inv.txt
With the arrays:
#!/bin/bash
removing=(3199 26543)
( IFS='|'; sed -rie "/${removing[*]}/s/^(.{7}).*/\\1\\t0;\\t0;/" inv.txt; )
You can try something like this:
removing=(3199 26543)
>output
while read line
do
for i in "${removing[#]}"
do
if [[ "$line" =~ ^.*\ ${i}\ ]]; then
echo "${line:0:7} 0; 0;" >> output
else
echo "$line" >> output
fi
done
done < inv.txt
# you can uncomment below mv command if you want to overwrite inv.txt
# mv output inv.txt
sed -i '/3199\|26543/s/^\(.\{7\}\).*/\1\t0\t0/' inv.txt

Increment with bash

I'm stuck trying to increment a variable in an .xml file. The tag may be in a file 100 times or just twice. I am trying to add a value that will increment the amount several times. I have included some sample code I am working on, but when I run the script it will only place a one and not increment further. Advice would be great on what I'm doing wrong.
for xmlfile in $(find $DIRECTORY -type f -name \*.xml); do
TFILE="/tmp/$directoryname.$$"
FROM='><process>'
TO=' value\=""><process>'
i=0
while [ $i -lt 10 ]; do
i=`expr $i + 1`
FROM='value\=""'
TO='value\="'$i'"'
done
sed "s/$FROM/$TO/g" "$xmlfile" > $TFILE && mv $TFILE "$xmlfile"
done
The while loop was something I just placed to test the code. It will insert the <process> but it will not insert the increment.
My end goal:
<process>value="1"</process>
<process>value="2"</process>
<process>value="3"</process>
<process>value="4"</process>
And so on as long as <process> is present in the file it needs to increment.
I just tested your code and it seems to correctly increment i.
You could try changing your increment syntax from:
i=`expr $i + 1`
To
i=$((i+1))
For a proper increment in bash, use a for loop (C style) :
n=10
for ((i=1; i<=n; i++)) {
printf '<process>value="%d"</process>\n' $i
}
OUTPUT
<process>value="1"</process>
<process>value="2"</process>
<process>value="3"</process>
<process>value="4"</process>
<process>value="5"</process>
<process>value="6"</process>
<process>value="7"</process>
<process>value="8"</process>
<process>value="9"</process>
<process>value="10"</process>
NOTE
expr is a program used in ancient shell code to do math. In Posix shells like bash, use $(( expression )). In bash and ksh93, you can also use (( expression )) or let expression if you don't need to use the result in an expansion.
EDIT
If I misunderstood your needs and you have a file with blank values like this :
<process>value=""</process>
try this :
$ perl -i -pe '$c++; s/<process>value=""/<process>value"$c"/g' file.xml
<process>value"1"</process>
<process>value"2"</process>
<process>value"3"</process>
<process>value"4"</process>
<process>value"5"</process>
<process>value"6"</process>
<process>value"7"</process>
-i switch edit the file for real, so take care.
This is the simplest way to increment a variable in bash:
i=0
((i++))
This also works.
Declaring the variable as an integer.
declare -i i=0
Then later you can increment like so:
i+=1
Use awk:
awk '{gsub( "value=\"\"", "value=" i++ ); print }' i=1 input-file
This will replace the string value="" with value="1", value="2", etc. You can easily change the start value and the increment ( eg ..."value=" i ); i+=5; print )

What is the problem with my code for multiplying two numbers in a different way using Bash?

This is my first time using Bash, so bear with me. I'm using Git Bash in Windows for a college project, trying to rewrite some C code that provides an alternate way of multiplying two numbers "a" and "b" to produce "c". This is what I came up with:
#!/bin/bash
declare -i a
declare -i b
declare -i c=0
declare -i i=0 # not sure if i have to initialize this as 0?
echo "Please enter a number: "
read a
echo "Please enter a number: "
read b
for i in {1..b}
do
let "c += a"
done
echo "$a x $b = $c"
I think part of the problem is in the for loop, that it only executes once. This is my first time using Bash, and if anyone could find the fault in my knowledge, that would be all I need.
There are problems with your loop:
You can't use {1..b}. Even if you had {1..$b} it wouldn't work because you would need an eval. It's easiest to use the seq command instead.
Your let syntax is incorrect.
Try this:
for i in $(seq 1 $b)
do
let c+=$a
done
Also, it's not necessary to declare or initialise i.
for i in {1..b}
won't work, because 'b' isn't interpreted as a variable but a character to iterate to.
For instance {a..c} expands to a b c.
To make the brace expansion work:
for i in $(eval echo "{1..$b}")
The let "c += a" won't work either.
let c+=$a might work, but I like ((c+=a)) better.
Another way is this:
for ((i = 1; i <= b; i++))
do
((c += a))
done
(might need to put #!/bin/bash at the top of your script, because sh does less than bash.)
Of course, bash already has multiplication, but I guess you knew that ...
If the absence of "seq" is your issue, you can replace it with something more portable, like
c=0
# Print an endless sequence of lines
yes |
# Only take the first $b lines
head -n "$b" |
# Add line number as prefix for each line
nl |
# Read the numbers into i, and the rest of the line into a dummy variable
while read i dummy; do
# Update the value of c: add line number
c=`expr "$c" + "$i"`
echo "$c"
done |
# Read the last number from the while loop
tail -n 1
This should be portable to any Bourne-compatible shell. The while ... echo ... done | tail -n 1 trick is necessary only if the value of c is not exported outside the while loop, as is the case in some, but not all, Bource shells.
You can implement a seq replacement with a Perl one-liner, but then you might as well write all of this in Perl (or awk, or Python, or what have you).

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