[SOLVED] awk

webhope · 04-23-2010, 04:23 AM

Hi,
I tried to get a block of lines in awk, but unfortunately it returns output of one line only. I don't state the code here, because it's too short and too poor. What exactly I wanted to do: from file "/boot/grub/menu.lst" get blocks of lines, starting by title and ending by \n\n

Now I have just

Code:

script="/boot/grub/menu.lst";
OLDIFS=$IFS; IFS=$'\n';
content=($(cat $script |
    awk '' ));
echo ${content[2]}
read

I want to get every block as one row of array. Can you help me with this? Thanks

colucix · 04-23-2010, 05:17 AM

Quote:

Originally Posted by webhope

I want to get every block as one row of array

Hi. What do you mean for 'as one row'? Arrays in bash are not multi-dimensional (you can only mimic them using indirect variable reference). Maybe do you want to assign one block of lines to a single element of the array?

webhope · 04-23-2010, 05:35 AM

Quote:

Originally Posted by colucix

Maybe do you want to assign one block of lines to a single element of the array?

I think so. If the "single element" means array[0], array[1], etc. Other words I mean output like this:

Code:

array[0]='title           Ubuntu 9.04 - boot_ide 160GB uuid
uuid            40af8b13-eb6d-4257-bace-50ba2e5ce544
kernel          /boot/vmlinuz-2.6.28-11-generic root=UUID=40af8b13-eb6d-4257-bace-50ba2e5ce544b ro xforcevesa quiet splash
initrd          /boot/initrd.img-2.6.28-11-generic
quiet
' 
array[1]='Next block
asd
asd
'

catkin · 04-23-2010, 06:11 AM

This should do it

Code:

#!/bin/bash

readonly false=
readonly true=true

in_stanza_flag=$false
i=-1
while read line
do
    case $line in
        title* )
            let i++
            in_stanza_flag=$true
            ;;
        * )
    esac
    [[ $in_stanza_flag ]] && array[i]=${array[i]}$line$'\n'
done < /boot/grub/menu.lst

for (( i=0; i<${#array[*]}; i++ ))
do
    echo "Stanza $i"$'\n'"${array[i]}"
done

If there is stuff after the end of the stanzas list you can replace the case statement's wild (*) case with a string match for it and code in_stanza_flag=$false to avoid all the remaining guff ending up in the last array element.

webhope · 04-23-2010, 06:18 AM

catkin:
Uff. This looks complicated. I hoped in some simple solution on one or two lines by awk. I think awk is much more better because it uses strong regular expressions. But thanks for your efforts.

grail · 04-23-2010, 07:14 AM

Code:

awk 'BEGIN{RS=""}/^title/' /boot/grub/menu.lst

catkin · 04-23-2010, 08:01 AM

Quote:

Originally Posted by grail

Code:

awk 'BEGIN{RS=""}/^title/' /boot/grub/menu.lst

That's cunning. How does it work?

webhope · 04-23-2010, 08:06 AM

Quote:

Originally Posted by grail

Code:

awk 'BEGIN{RS=""}/^title/' /boot/grub/menu.lst

What's wrong? I have problem with it even with my variation.

menu.lst

Code:

timeout 5                                                                
color black/cyan yellow/cyan                                             
gfxmenu (hd0,2)/boot/gfxmenu                                             
default 0                                                                

title Sata Mandriva
kernel (hd0,2)/boot/vmlinuz BOOT_IMAGE=linux root=UUID=eab515e9-bc3e-4024-9f01-55fddaa0fb1c  resume=UUID=e12487ff-6d6f-44c4-9e03-33f545b3b798 splash=silent vga=788                                                           
initrd (hd0,2)/boot/initrd.img                                                                                 


title sata XP
 unhide (hd0,0)                 
 hide (hd0,1)                   
rootnoverify (hd0,0)            
chainloader +1                  
makeactive                      
savedefault

Output:

Code:

# clear; script="/boot/grub/menu.lst";
# content=($( awk 'BEGIN{RS="\n"}/^title/' $script ));
# echo ${content[0]}
title
# echo ${content[1]}
Sata
# echo ${content[2]}
Mandriva
# echo ${content[3]}
title
# echo ${content[4]}
sata
[root@localhost grub]# echo ${content[5]}
XP1

(Missing IFS=$'\n')

My code:

Code:

clear; script="/boot/grub/menu.lst";
content=($( awk 'BEGIN{RS=""}/^title/' $script ));
echo ${content[1]}

Code 2

Code:

IFS=$'\n';
clear; script="/boot/grub/menu.lst";
content=($( awk 'BEGIN{RS="\n"}/^title/' $script ));
echo ${content[0]}

Outputs:

Code:

# echo ${content[0]}
title Sata Mandriva
# echo ${content[1]}
title sata XP

In 1st case it is words of 1st line of block 1, then words of 1st line of next block.

In 2nd case it is 1st line of next block.

I need to edit 2nd code (having IFS=$'\n'

to get next lines of the same block to content[n]. So output should be:

Code:

# echo ${content[0]}
title Sata Mandriva
kernel (hd0,2)/boot/vmlinuz BOOT_IMAGE=linux root=UUID=eab515e9-bc3e-4024-9f01-55fddaa0fb1c  resume=UUID=e12487ff-6d6f-44c4-9e03-33f545b3b798 splash=silent vga=788                                                           
initrd (hd0,2)/boot/initrd.img

grail · 04-23-2010, 08:33 AM

Quote:

That's cunning. How does it work?

Setting the record separator to blank allows you to look through the file and keep all consecutive lines that contain
something together, so now in your normal grub file a $0 may look like:

Code:

title		Ubuntu 9.10, kernel 2.6.31-21-generic
root		(hd0,1)
kernel		/boot/vmlinuz-2.6.31-21-generic root=/dev/mapper/isw_cehbidjbfj_RaidME2 ro quiet splash 
initrd		/boot/initrd.img-2.6.31-21-generic

@OP - The solution I presented was able to get the necessary lines, but I would hazard a guess that no amount of playing with IFS will work
as it does not take a regular expression as an option (that I know of).
Basically you can't have your cake and eat it too, either you use something like catkin had or if using something like the awk you will need to change
how you deal with the data, ie into an array may not be the best option.
You could perhaps also try converting the awk to output each field using a different separator, maybe pipe, and then use your IFS='\n',
but of course then you need a way to break it back up at the other end.

Maybe if you would enlighten us a little further as to what you need to do with each stanza from menu.lst?

webhope · 04-23-2010, 08:52 AM

I am going to write a script where I can do this actions:
1) change title
2) change hdx to uuid or uuid to hdx
3) increase hdx+1 or descrease hdx-1
4) save

Similar to script I did yesterday.

catkin · 04-23-2010, 09:10 AM

Quote:

Originally Posted by grail

Setting the record separator to blank allows you to look through the file and keep all consecutive lines that contain
something together, so now in your normal grub file a $0 may look like:

Code:

title		Ubuntu 9.10, kernel 2.6.31-21-generic
root		(hd0,1)
kernel		/boot/vmlinuz-2.6.31-21-generic root=/dev/mapper/isw_cehbidjbfj_RaidME2 ro quiet splash 
initrd		/boot/initrd.img-2.6.31-21-generic

Thanks, grail. Neat!

colucix · 04-23-2010, 09:25 AM

@webhope: if you want each element of the array contain multiple lines, you cannot use "\n" as IFS. Following grail's suggestion you can choose a character that does not appear in menu.lst stanzas, e.g. @, then simply:

Code:

#!/bin/bash
OLD_IFS=$IFS
IFS="@"
content=( $(awk 'BEGIN{RS=""; ORS="@"}/^title/' /boot/grub/menu.lst) )
IFS=$OLD_IFS
echo ${content[0]}

webhope · 04-23-2010, 09:25 AM

I don't understand how you get the block of lines. When I run your code

Code:

awk 'BEGIN{RS=""}/^title/' /boot/grub/menu.lst

I have something like this:

Code:

title           Ubuntu 9.04, kernel 2.6.28-11-generic                                                          
uuid            40af8b13-eb6d-4257-bace-50ba2e5ce544                                                           
kernel          /boot/vmlinuz-2.6.28-11-generic root=UUID=40af8b13-eb6d-4257-bace-50ba2e5ce544 ro xforcevesa quiet splash                                                                                                     
initrd          /boot/initrd.img-2.6.28-11-generic                                                             
quiet                                                                                                          
title           Ubuntu 9.04 - boot_sata 500GB                                                                  
root (hd0,2)                                                                                                   
kernel          /boot/boot_sata/vmlinuz-2.6.28-11-generic root=/dev/sda3 ro xforcevesa access=v2 quiet splash  
# uuid          40af8b13-eb6d-4257-bace-50ba2e5ce544                                                           
# kernel                /boot/boot_sata/vmlinuz-2.6.28-11-generic root=UUID=40af8b13-eb6d-4257-bace-50ba2e5ce544 ro xforcevesa access=v2 quiet splash                                                                         
initrd          /boot/boot_sata/initrd.img-2.6.28-11-generic
quiet
title           Ubuntu 9.04 - boot_ide 160GB - root hd1,4
root (hd0,4)
kernel          /boot/vmlinuz-2.6.28-11-generic root=/dev/sdb1 ro xforcevesa quiet splash
initrd          /boot/initrd.img-2.6.28-11-generic
quiet

grail · 04-23-2010, 10:08 AM

Quote:

title Ubuntu 9.04, kernel 2.6.28-11-generic
uuid 40af8b13-eb6d-4257-bace-50ba2e5ce544
kernel /boot/vmlinuz-2.6.28-11-generic root=UUID=40af8b13-eb6d-4257-bace-50ba2e5ce544 ro xforcevesa quiet splash
initrd /boot/initrd.img-2.6.28-11-generic
quiet

So this is one of the stanzas from your menu.lst.

If you know follow colucix advice, you can even use select:

Code:

#!/bin/bash
OLD_IFS=$IFS
IFS="@"
content=( $(awk 'BEGIN{RS=""; ORS="@"}/^title/' /etc/grub/menu.lst) )

select x in ${content[@]}
do
    echo "$x"
done
IFS=$OLD_IFS

webhope · 04-23-2010, 10:24 AM

Sorry I didn't see colucix response. I will try