awk regexp for uuid

grail · 04-26-2010, 09:36 PM

Assuming the format remains the same for all stanzas you could use:

Code:

echo "$block_new" | awk '/UUID/{$4="root=(hd0,2)"}1

webhope · 04-27-2010, 01:58 AM

Tinkster

Quote:

Originally Posted by Tinkster

Did you see the changes to your match in my post above?

Code:

{ x=gensub(/.*([0-9a-f-]{36,36}).*/,"\\1","g"); print x}

Yes:

Code:

uuid=$( echo $block_new | 
awk --re-interval '/.*UUID=/{ x=gensub(/([0-9a-f-]{36,36}).*/,"\\1","g");  print x}'
echo $uuid; read

Gives:

Code:

title Sata Mandriva
e12487ff-6d6f-44c4-9e03-33f545b3b798
initrd (hd0,2)/boot/initrd.img

Instead:
e12487ff-6d6f-44c4-9e03-33f545b3b798

But How can you be sure it gives correct number? I need to specify what (or what type of device) is there. So "UUID=" before hexadecimal digit expression.

Next thing - in this case I need to get uuid, not whole block

Grail

Quote:

Originally Posted by grail

Assuming the format remains the same for all stanzas you could use:

Code:

echo "$block_new" | awk '/UUID/{$4="root=(hd0,2)"}1

What is sense? It doesn't do what I need. I repeat, one of these expressions:

Code:

    uuid=$( echo $block_new | awk --re-interval '/.*UUID=/{ x=gensub(/([0-9a-f-]{36,36}).*/,"\\1","g");  print x}')
    echo $uuid; read

    uuid=$( echo $block_new |  awk '/.*UUID=/{ x=gensub(/UUID=([^ ]+).*/,"\\1","g");  print x}')
    echo $uuid; read

Outputs:

Code:

kernel (hd0,2)/boot/vmlinuz BOOT_IMAGE=linux root=UUID=eab515e9-bc3e-4024-9f01-55fddaa0fb1c
# // Pressed Enter //
kernel (hd0,2)/boot/vmlinuz BOOT_IMAGE=linux root=eab515e9-bc3e-4024-9f01-55fddaa0fb1c

So I need to remove the preceding part: "kernel (hd0,2)/boot/vmlinuz BOOT_IMAGE=linux root="

grail · 04-27-2010, 02:14 AM

I am sorry, I thought the point was to replace the current root=UUID=<blah> with root=(hd0,2):

Quote:

I wanted to change
root=UUID=eab515e9-bc3e-4024-9f01-55fddaa0fb1c
to
root=(hd0,2)

webhope · 04-27-2010, 02:30 AM

Quote:

Originally Posted by grail

I am sorry, I thought the point was to replace the current root=UUID=<blah> with root=(hd0,2):

I tryied to explain situation. But first I need to get correct uuid. I don't know location of system at this point. I will replace it later. Can you explain me where is the fault my awk pattern doesn't cut out (remove) beginning?

webhope · 04-27-2010, 06:01 AM

I forgot to put there .* before UUID= in the gensub() fnc

Code:

uuid=$( echo $block_new | awk --re-interval '/.*UUID=.*/{ x=gensub(/.*UUID=([0-9a-f-]{36,36}).*/,"\\1","g");  print x}')
echo $uuid; read

uuid=$( echo $block_new |  awk '/.*UUID=/{ x=gensub(/.*UUID=([^ ]+).*/,"\\1","g");  print x}')
echo $uuid; read

Thanx to Tinkster and Pixellany

jschiwal · 04-27-2010, 06:06 AM

Quote:

Originally Posted by pixellany

PS:
Does anyone know if a uuid always has the same number of characters? If so, that could be incorporated into the regex.

Look at the uuid of a fat32 filesystem. It will be short.

Here are examples for uuids from the /dev/disk/by-uuid/ directory on my laptop. The first one is from an NTFS filesystem. The second is for a fat32 filesystem. A Linux filesystem or swap partition follows the last pattern.

Code:

145266D35266B95E
2CBD-5986
5783d30e-8dc4-447a-80f4-8952129e9d9e

webhope · 04-27-2010, 07:08 AM

I found one more problem (or two):

Code:

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

In these code are two uuids. It shows the last one (which is not on my system) - e12487ff-6d6f-44c4-9e03-33f545b3b798 ....

Code:

    uuid=$( echo $block_new |  awk '/.*UUID=/{ x=gensub(/.*UUID=([^ ]+).*/,"\\1","g");  print x}')
echo $uuid

Therefor I need to give all uuid to one string separeted by space.

Do you know how to do it?

This is little more complicated.

grail · 04-27-2010, 07:37 AM

So I will ask again, is there any reason to think the start will be different from:

Quote:

kernel (hd0,2)/boot/vmlinuz BOOT_IMAGE=linux root=UUID=eab515e9-bc3e-4024-9f01-55fddaa0fb1c

ie it might not be the same words exactly but will there be 4+ fields in this type of order?

I do use grub, but am not sure of all the possible combinations?

If it is always to be 4+ with root in specified position then do a sub on $4 and return it.

webhope · 04-27-2010, 07:52 AM

Grail,
I don't know what you mean "4+ fields"?
Do you speak about the resume argument?

"I do use grub, but am not sure of all the possible combinations?"
I use configuration of Mandriva Linux auto generated.

Originaly I wanted awk pattern to return uuid.
But now situation is such that I would reather return all uuids to one string. Then I will check if the uuids is in blkid output. I will replace the existing UUID=uuid to hd(?,?). I will reinstall my system and change target partitions. So this script will be usefull after reinstalling system.

grail · 04-27-2010, 08:04 AM

Quote:

I don't know what you mean "4+ fields"?

Code:

field_1=kernel 
field_2=(hd0,2)/boot/vmlinuz 
field_3=BOOT_IMAGE=linux 
field_4=root=UUID=eab515e9-bc3e-4024-9f01-55fddaa0fb1c
field_+=...

So if we know we always will have at a minimum, 4 fields in this order then we can simply deal with field $4 or if you want resume as well and it is always
there then field $5.

I have found the simplest way to get the first uuid to be 2 simple subs:

Code:

block_new=$( echo "$block_new" | awk '/UUID/{sub(/.*root=UUID=/,"");sub(/ .*/,"");print}')

webhope · 04-27-2010, 08:30 AM

Quote:

Originally Posted by grail

So if we know we always will have at a minimum, 4 fields in this order ...

No. This is changable. In menu.lst are more systems and configurations, depending on installer of distribution. So this is not sure. Therefor I want all uuids to one string. Then I will do some type of replace with help of loop.

Quote:

Originally Posted by grail

I have found the simplest way to get the first uuid to be 2 simple subs:

Code:

block_new=$( echo "$block_new" | awk '/UUID/{sub(/.*root=UUID=/,"");sub(/ .*/,"");print}')

This is interesting (clever), but as I say. There can be situation, that there will be more uuids on one line and in block, as installer creates such configuration. So putting all uuids to one string "... ... ... ..." will be better to fix all possibilities.

grail · 04-27-2010, 09:02 AM

Code:

#!/bin/bash

block_new="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"

block_new=($(echo "$block_new" | awk 'BEGIN{RS="UUID="}/-/{gsub(/ .*/,"");print}'))

for x in ${block_new[@]}
do
	echo "|$x|"
done

webhope · 04-27-2010, 10:12 AM

Thank you, I'll write later.

webhope · 04-28-2010, 02:52 AM

Quote:

Originally Posted by grail

Code:

#!/bin/bash

block_new="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"

I continue. I have this code. In case that 1 of uuids is not found in your system it should echo warning. But the if cond. is not right. What is in the d var if awk nothing found?

Code:

uuid=($(echo "$block_new" | awk 'BEGIN{RS="UUID="}/-/{gsub(/ .*/,"");print}'))
for d in ${uuid[@]}; do
  dev=$(blkid | grep $d |  awk 'BEGIN{FS=":"} {print $1}');
  if [[ "$dev" != "" ]] && [[ "$dev" != "\n" ]]; then
  echo $dev
  else
  echo "uuid $uuid not found in list of devices"
  fi;
done

grail · 04-28-2010, 03:16 AM

Quote:

What is in the d var if awk nothing found?

The for loop is immediately exited as the array has no items to step through.
If you put an echo after done, this will be the only thing on screen.