awk shell script error

ApacheRoseXbones · 07-17-2008, 12:33 PM

hey I have a small shell script (below), and I'm trying to use it to make a variable out of a number thats in another file. The reason I have the "for" statement is because I want to have it do this for multiple files.

shell script (to pull the number 2 from the 5th line of the example file below)

Code:

#!/bin/bash
for i in `ls *.log`
do
	a="`awk '/The following ModRedundant/ {getline; print $2}'`"
	echo $a
done

example file data being pulled from

Code:

  D15                 177.16243                  
  D16                 -64.62156                  
 
 The following ModRedundant input section has been read:
 A    2   11   13 116.97 S  19 0.0100                                          
 Iteration  1 RMS(Cart)=  0.00295973 RMS(Int)=  0.00108084
 Iteration  2 RMS(Cart)=  0.00001517 RMS(Int)=  0.00108072

My trouble is that it works if i execute it as shellscriptname.sh < filename but not if I execute it as ./shellscriptname. (the command prompt locks up) I'm thinking it has something to do with the fact that it is a for statement but I'm not sure. Any ideas?

forrestt · 07-17-2008, 12:36 PM

You are never specifying a file to awk. Try:

Code:

#!/bin/bash
for i in `ls *.log`
do
	a="`awk '/The following ModRedundant/ {getline; print $2}' $i`"
	echo $a
done

HTH

Forrest

ApacheRoseXbones · 07-17-2008, 01:43 PM

Ah thanks that worked!

Now I have another issue: I want to use awk to do a substitution with $a in file $ANG. With the way my script is now, it echoes what is in $ANG rather than printing to it. Any idea what to do there?

Code:

#!/bin/bash
for i in `ls *.log`
do
    a="`awk '/The following ModRedundant/ {getline; print $2}' $i`"
    ANG=${i/log/sh}
    awk '{sub(/x/,$a);print}' $ANG
done

what is in $ANG:

Code:

scanf=\$1
\awk -v mine=\$mine 'BEGIN {found=0; ene=0; count=0; printf \"%2s %8s %12s\n\", \"#\", \"Ang1\", \"E\";}\ "
/E\(RHF\)/ {ene=\$5; found=0;} \ " 
/Stationary point found/ {found=1; count++} \ 
(found==1 && \$0 ~ /Optimized Parameters/) {found=2} \ 
(found==2 && \$0 ~ /A\(x,y,z\)/) {found=3; Ang1=\$4} \ 
(found==3) {found=4; printf \"%-2d %8.3f %12.6f\n\", count, Ang1, ene }' \$1 \

forrestt · 07-17-2008, 01:55 PM

If you want something to be parsed by the shell, you use double quotes (""). If you do not want it to be parsed, you use single quotes (''). You are using single quotes to wrap your awk script, so the shell isn't parsing it. Replace those with double quotes, and things should work properly. However, you won't be able to do this if you wanted to do something like print $1 as the shell would replace $1 with the first parameter to the shell or NULL if you didn't give one.

Code:

#!/bin/bash
for i in `ls *.log`
do
    a="`awk '/The following ModRedundant/ {getline; print $2}' $i`"
    ANG=${i/log/sh}
    awk "{sub(/x/,$a);print}" $ANG                 <--------------- Note the double quotes
done

Someone with a bit more bash/awk skills than I have may be able to suggest a way to do both (I'd have to look it up).

HTH

Forrest

ApacheRoseXbones · 07-18-2008, 11:26 AM

Thanks for your help, the script works perfectly now

forrestt · 07-18-2008, 12:06 PM

Glad I could help.

Forrest