Help with passing multiple variables into SED

Shell Programming and Scripting
1

Hello I am hoping you can help.
I use ksh in Solaris9

I am trying to pass user imputed variables into SED but for some reason can only get SED to recognize one variable.
I have experimented with te below command with putting ' ' and " " in different places but I cant seem to get it to recognize the second variable.
Examples I can find dont seem to cover this in this form.

cat filename | sed -e '/./{H;$!d;}' -e "x;/$AAAA/!d;" -e '/./{H;$!d;}' -e "x;/$BBBB/!d"

I am using this SED to specifically strip out paragraphs containing text that the user inputs.
Normal I would simply use the below command if I were not using variables and that works fine - for example I have known values of 12345 and 8888.

cat filename | sed -e '/./{H;$!d;}' -e 'x;/12345/!d; -e 'x;/8888/!d'

If you are able to help or spot what I am doing wrong here I would appreciate it.
Regards

2

I don't see anything obviously wrong with it -- your variables are in properly double-quoted strings -- but surely it depends on what the values of AAAA and BBBB are...

3

Hello.
Yes, the strings are user input, just numbers or straight text no special characters or symbols.
Regards.
:frowning:

---------- Post updated at 02:55 PM ---------- Previous update was at 12:47 PM ----------

I made a cut and paste error earlier. Here is the more expanded version of what I am doing.

printf "enter search string: "
read string1
printf "Another search string [yes|no] "
read ans
if [ "$ans" == "yes" ]
then
printf "enter 2nd search string: "
read string2
cat filename | sed -e '/./{H;$!d;}' "-e 'x;/$string1/!d; -e 'x;/$string2/!d"

If that shows any noticeable problems.
I have tried double quotes around both strings "$string" and further out

4

You don't need to cat the file and pipe it to sed, sed can read the file directly. The problem seems to me to be the quotes. I believe this is what you need; double quotes round the expressions that you wish to have shell variables expanded in.

sed -e '/./{H;$!d;}'  -e "x;/$string1/!d;"  -e "x;/$string2/!d" filename >output-file

---------- Post updated at 23:22 ---------- Previous update was at 23:13 ----------

You might also consider logic like the sample below, so the sed is executed with the correct number of expressions when the user does not enter 'yes' at the prompt:

printf "enter search string: "
read string1
printf "Another search string [yes|no] "
read ans
if [ "$ans" == "yes" ]
then
    printf "enter 2nd search string: "
    read string2
    p2="-e x;/$string2/!d"
fi

/usr/bin/sed -e '/./{H;$!d;}' -e "x;/$string1/!d;"  $p2  filename >output-file

Note that $p2 is purposely not in quotes on the sed command.

---------- Post updated at 23:37 ---------- Previous update was at 23:22 ----------

And one last thing....
I'm not quite sure what your sed is trying to accomplish. If I had to guess, I would say that you are trying to use sed to print all lines in the file that contain either string1 or string2. At least when I run it, with that assumption, it isn't doing that.

If that is what you are trying to accomplish, I'd suggest the following:

printf "enter search string: "
read string1

printf "Another search string [yes|no] "
read ans
if [ "$ans" == "yes" ]
then
    printf "enter 2nd search string: "
    read string2
    p2="/$string2/!d;"
else
    p2="d"
fi

sed -e "/$string1/b" -e "$p2" input-file >output-file
5

Thanks yes I did try that but it did not print out. when I changed the command from

-e "x;/$string1/!d;"  to   -e "x;/$string/d"

then I get a limited print out.
It closes with

6

This seems like a semicolon in one of your expressions is not quoted and the shell is seeing that as the end of the command and is trying to execute the -e as a command.

7

Thanks yes. I tried with a number of senarios with " " around the

-e "x;/12345/!d; -e 'x;/8888/!d" and have tried   -e "x;/12345/!d"; -e "x;/8888/!d" and tried  -e "x;/12345/!d;" -e "x;/8888/!d"

None of which seems to work taking the second variable.
I have also tried

-e "x;/12345/!d; -e 'x;/8888/!d"

but that dos not work either?

So I am a little stuck. I was wondering if there is a way using ( ) brackets or something I am missing. I am using Solaris 9 ksh

Regards

8

I think you're overthinking it a bit. The quotes join things together. If you don't want them joined, don't put them in quotes.

Put your expressions in double quotes.

sed -e '/./{H;$!d;}' "-e "x;/$string1/!d;" -e "x;/$string2/!d" filename
9

Thanks Agama
I will re-write the script with what you showed me there and let you know.

---------- Post updated at 05:12 PM ---------- Previous update was at 12:05 PM ----------

Hi I tried the modified script

#!/usr/bin/env ksh
cd /usr/logs/appserver/
ls -ltr XSLog*

printf "enter filename: "
read filename
printf "enter search string: "
read string1
printf "Another search string [yes|no] "
read ans
if [ "$ans" == "yes" ]
then
    printf "enter 2nd search string: "
    read string2
    p2="-e x;/$string2/!d"
fi

sed -e '/./{H;$!d;}' -e "x;/$string1/!d"  $p2  $filename 

bash-2.05$

but I could not get the second variable to be recognized.
Trying the first variable works but two, it seems to just hang and not print out.
I tried with the ";" before and after the first

 -e "x;/$string1/!d"

.
sorry to take up your time with this.

---------- Post updated 04-05-12 at 03:27 PM ---------- Previous update was 04-04-12 at 05:12 PM ----------

agama, Corona688 many thanks for your suggestions and help. I managed to get it working. I noticed I had an extra

 -e "x;/$string2/!d"

on the second string when actually it only required

  "/$string2/!d"

 sed -e '/./{H;$!d;}' -e "x;/$string1/!d; $p2" filename

Many thanks again
Regards:)