replace a string with "/" in a variable

[Billa-User]

hello,

i use often sed to replace a string, but
here it is not a good way, other options ?

what's the best way to replace a string with "/"

input for example: /dev/vgold_fs/config
new output : /dev/vgneu

it work's good with
echo /dev/vgold_fs/config |sed "sX/dev/vgold_fs/configX/dev/vgneuXg"

But when you have a X in the string, then you get with:

echo "/dev/vgold_fs/configX" | sed "sX/dev/vgold_fsX/dev/vgneuXg"

sed: Function sX/dev/vgold_fs/configXXX/dev/vgneuXg cannot be parsed.

i have to prepare the input string like
"\\\\\/dev/vgold_fs/configX" ?

not so good

regards

[James R. Ferguson · Accepted solution]

Hi:

Change your delimiters:

# X=/dev/vgold_fs/config
# echo $X|sed -e 's|/dev/vgold_fs/config|/dev/vgneu|'
/dev/vgneu

Regards!

...JRF...

[Billa-User]

great, great.

maybe you have also an improvement for "awk" and escape-ing like this:

it is only an example (but used for parse commands like "bdf", "lvdisplay" )

OK (because i escape )

X="/dev/vgold_fs/config"
X_AWK=$( echo ${X} | sed "s|/|\\\\\/|g" )
echo "/dev/vgold_fs/config feld2 feld3" | awk '$1 ~ /^'${X_AWK}'$/ {print $1}'
/dev/vgold_fs/config

NOT-OK:

X="/dev/vgold_fs/config"
echo "/dev/vgold_fs/config feld2 feld3" | awk '$1 ~ /^'${X}'$/ {print $1}'

syntax error The source line is 1.
The error context is
$1 ~ /^/dev/vgold_fs/config$/ {print >>> $1} <<<
awk: Quitting
The source line is 1.

[Dennis Handly]

>maybe you have also an improvement for "awk" and escape-ing like this:

What are you really trying to do? It seems you are looking for an exact match for the first field. Why not:
echo "/dev/vgold_fs/config feld2 feld3" |
awk -v select="$X" '$1 == select {print $1}'

>X_AWK=$( echo ${X} | sed "s|/|\\\\\/|g" )

Better to use single quotes and then the number of "\"s are predictable:
X_AWK=$( echo "${X}" | sed -e 's|/|\\/|g' )

>echo "/dev/vgold_fs/config feld2 feld3" | awk '$1 ~ /^'${X_AWK}'$/ {print $1}'

If you really want a pattern match, better go with variables than constant regular expression that has "//":
echo "/dev/vgold_fs/config feld2 feld3" |
awk -v pat="$X" '$1 ~ ("^" pat "$") {print $1}'

(The "^" and "$" anchors could be included in the original $X.)

[Billa-User]

>What are you really trying to do? It seems you are looking for an exact match for the > first field. Why not:

if want to search with a shell variable , which include "/" or "\" with "awk" in field 1 for example.

example: fs=/tmp

then i want search for "${fs}" with command "bdf" and work with "awk"

like:
bdf | awk "$NF ~ ... { print ....

regards.

sorry, i explain it not very good :-((

[Dennis Handly]

>if want to search with a shell variable, which include "/" or "\" with "awk" in field 1 for example.

I as asking if you wanted an exact match (first example: ==) by a pattern match (ERE) (the last example: ~).

>then I want search for "${fs}" with command "bdf" and work with "awk"

awk "$NF ~ ...

It is much easier to put the pattern in a variable. Then you don't have to worry about the "/" delimiters.

You can also use the match function.

[James R. Ferguson]

Hi (again):

> if want to search with a shell variable , which include "/" or "\" with "awk" in field 1 for example.

> example: fs=/tmp

First, I would avoid the temptation to use a variable name, that if upper-cased would become an 'awk' reserved variable name -- FS.

Consider these two ways of matching:

# |awk -v FSYS=/ho '$NF~FSYS'
/home

...which is probably *not* what you want.

# bdf | awk -v FSYS=/ho '$NF==FSYS {print $NF}'

...returns nothing since the comparison for equality fails.

# bdf | awk -v FSYS=/home '$NF==FSYS {print $NF}'
/home

...which is what you appear to want.

Regards!

....JRF...

...and:

# bdf |
~/Sandbox jrf $ df -h | awk -v FSYS=/home '$NF==FSYS'
map auto_home 0Bi 0Bi 0Bi 100% /home
~/Sandbox jrf $ df -h | awk -v FSYS=/ho '$NF==FSYS'
~/Sandbox jrf $

[Billa-User]

for a example,
i have more filesystems like /home1, /home2
then :

exact match for "/home" , right ?
bdf | awk -v FSYS=/home '$NF==FSYS {print $NF}'
matches all filesystems
bdf | awk -v FSYS=/home '$NF ~ FSYS {print $NF}'

i can use "~" with ERE like FSYS[0-9]
bdf | awk -v FSYS=/home '$NF ~ FSYS[0-9] {print $NF}'

in another usage, when i have to search in a file where
FS is or then i use following command:

awk -v text=search ' BEGIN { FS = "[ \t]*|[ \t]+" }
$NF ~ search {print $NF}' /tmp/text

[Dennis Handly]

>I have more filesystems like /home1, /home2 then:
>exact match for "/home", right?
>... | awk -v FSYS=/home '$NF==FSYS

Yes.

>matches all filesystems
... | awk -v FSYS=/home '$NF ~ FSYS

Yes, those /home* ones.

>I can use "~" with ERE like FSYS[0-9]
... | awk -v FSYS=/home '$NF ~ FSYS[0-9]

Not like that. Either:
... | awk -v FSYS='/home[0-9]' '$NF ~ FSYS
Or:
... | awk -v FSYS=/home '$NF ~ (FSYS "[0-9]")

>when I have to search in a file where FS is or then I use following command:
awk -v text=search 'BEGIN { FS = "[ \t]*|[ \t]+" }
$NF ~ search {print $NF}' /tmp/text

Don't do that. The default is already tab or blank but ignoring leading tab/blanks.
(Not sure why you are using both "*" and "+" in your ERE? Only "+" should be used for one or more.)
And you could use -F on the command line to set FS.

[Billa-User]

great, great, great.

one more example:
i have more filesystems like /home1, /home2, /home33

then i match /home1, /home2 like:
... awk -v FSYS="/home[0-9]$" '$NF ~ FSYS { print $0 }'
or
... awk -v FSYS=/home '$NF ~ (FSYS"[0-9]""$") { print $0 }'
/dev/vg00/lvol4 1048576 757648 288720 72% /home1
/dev/vg00/lvol4 1048576 757648 288720 72% /home2

then i match /home1, /home2, /home33 like:
... awk -v FSYS="/home[0-9]+$" '$NF ~ FSYS { print $0 }'
or
....awk -v FSYS=/home '$NF ~ (FSYS"[0-9]+""$") { print $0 }'
/dev/vg00/lvol4 1048576 757648 288720 72% /home1
/dev/vg00/lvol4 1048576 757648 288720 72% /home2
/dev/vg00/lvol4 1048576 757648 288720 72% /home33

OK , my syntax ? it works, like i want and you show me :-)