I would like to replace a variable inside the the awk command with a bash variable.
For example:
var="one two three"
echo $var | awk "{print $2}"
I want to replace the $2 with the var variable. I have tried awk -v as well as something like awk "{ print ${$wordnum} } to no avail.
Sightly different approach:
$ echo $var
one two three
$ field=3
$ echo $var | awk -v f="$field" '{print $f}'
three
$ field=2
$ echo $var | awk -v f="$field" '{print $f}'
two
You've almost got it...
$ myfield='$3'
$ echo $var | awk "{print $myfield}"
three
The hard quotes on the first line prevent interpretation of $3 by the shell. The soft quotes on the second line allow variable replacement.
You can concatenate parts of awk statements with variables. Maybe this is what you want in your script file:
echo $1|awk '{print($'$2');}'
Here the parts {print($ and the value of local variable $2 and );} are concatenated and given to awk.
EDIT: After some advice rather don't use this. Maybe as a one-time solution. It's better to get accustomed to doing it right right away - see link in first comment.
Related
I have this script that's designed to assign variables to commands that collect information about a system and then echo them back. This works very well for the first few commands, but the last one continues to return the value without "PRETTY_NAME=" stripped out of the output.
Is there some problem with this that I'm not seeing?
I have tried using grep to separate awk:
grep PRETTY_NAME /etc/*-release | awk -F '=' '{print $2}'
Using escaped quotes:
awk -F \"=\" '/PRETTY_NAME/ {print $2}' /etc/*-release
Whole block (edited somewhat for relevance)
declare -A CMDS=(
[primaryMacAddress]="cat /sys/class/net/$(ip route show default | awk '/default/ {print $5}')/address"
[primaryIpAddress]="hostname --ip-address"
[hostname]="hostname"
[osType]="awk -F '=' '/PRETTY_NAME/ {print $2}' /etc/*-release"
)
#This bit is actually nested in another function
for kpair in "${!CMDS[#]}" do
echo "$kpair=\"$( eval ${CMDS[$kpair]} )\""
done
Results when run from .sh file:
osType="PRETTY_NAME="Red Hat Enterprise Linux Server 7.4 (Maipo)""
expected:
osType=""Red Hat Enterprise Linux Server 7.4 (Maipo)""
When this command is run by itself, it seems to work as intended:
$ awk -F '=' '/PRETTY_NAME/ {print $2}' /etc/*-release
"Red Hat Enterprise Linux Server 7.4 (Maipo)"
Because your Awk command is specified in double quotes, interior dollar signs are subject to special treatment: the $2 is treated as a parameter substitution by your shell, and so the array element doesn't store the text $2 but rather its expansion. The Awk interpreter never sees the $2 syntax.
However, you have a second problem in your command dispatcher. Your eval command does not prevent word splitting:
eval ${CMDS[$kpair]}
you want this:
eval "${CMDS[$kpair]}"
without the quotes, your command is arbitrarily chopped into fields on whitespace. Then eval catenates the pieces together, using one space between them, and evaluates the resulting syntax. The difference can be demonstrated with the following example:
$ cmd="awk '/foo/ { print \$1\" \"\$2 }'"
$ echo 'foo a' | eval $cmd
foo a
$ echo 'foo a' | eval "$cmd"
foo a
We can just use echo to understand the issue:
$ echo $cmd
awk '/foo/ { print $1" "$2 }'
$ echo "$cmd"
awk '/foo/ { print $1" "$2 }'
The substitution of $cmd and the subsequent word splitting is done irrespective of any shell syntax that `cmd contains. We can see the pieces like this:
$ for x in $cmd ; do echo "<$x>" ; done
<awk>
<'/foo/>
<{>
<print>
<$1">
<"$2>
<}'>
When we execute eval $cmd, the above pieces are generated and re-combined by eval and evaluated. Needless to say, you don't want your command syntax to be chopped up and re-combined like this; who knows what sort of hidden bug will arise. It may be okay for the commands you have now, but as a generic command dispatch mechanism, it is flawed.
I noticed when running a command that this statement doesn't recognize the delimiter
awk -F',' "{print $4}" wtd.csv
However, this one does.
awk -F',' '{print $4}' wtd.csv
Any reason why? I'm sure this is part of some general bash rule I'm forgetting.
If you're using double quotes, $4 will get replaced by Bash (probably with the empty string). You'd need to escape the $ to use it in double quotes.
Example where this also is happening:
[thom#lethe ~]$ echo '$4'
$4
[thom#lethe ~]$ echo "$4"
[thom#lethe ~]$ echo "\$4"
$4
You are forgetting that double-quotes allow bash variable interpolation. In this case it tries to replace $4 with the fourth argument to the shell which is usually empty.
The single-quotes prevent bash interpolation and passes the literal $4 to awk.
You'll have identical results with:
awk -F',' '{print $4}' wtd.csv
awk -F',' "{print \$4}" wtd.csv
I'm trying to do some manipulation with Wordpress and I'm trying to write a script for it...
# cat /usr/local/uftwf/_wr.sh
#!/bin/sh
# $Id$
#
table_prefix=`grep ^\$table_prefix wp-config.php | awk -F\' '{print $2}'`
echo $table_prefix
#
Yet I'm getting following output
# /usr/local/uftwf/_wr.sh
ABSPATH ABSPATH wp-settings.php_KEY LOGGED_IN_KEY NONCE_KEY AUTH_SALT SECURE_AUTH_SALT LOGGED_IN_SALT NONCE_SALT wp_0zw2h5_ de_DE WPLANG WP_DEBUG s all, stop editing! Happy blogging. */
#
Running from command line, I get the correct output that I'm looking for:
# grep ^\$table_prefix wp-config.php | awk -F\' '{print $2}'
wp_0zw2h5_
#
What is going wrong in the script?
The problem is the grep command:
table_prefix=`grep ^\$table_prefix wp-config.php | awk -F\' '{print $2}'`
It either needs three backslashes - not one - or you need to use single quotes (which is much simpler):
table_prefix=$(grep '^$table_prefix' wp-config.php | awk -F"'" '{print $2}')
It's also worth using the $( ... ) notation in general.
The trouble is that the backquotes removes the backslash, so the shell variable is evaluated, and what's passed to grep is, most likely, just ^, and each line starts with a beginning of line.
This has all the appearance as though the grep is not omitting all the lines that are not matching, when you issue the echo $table_prefix without quotes it collapses all the white space into a single output line, if you issue an: echo "$table_prefix", you would see the match with all the other white-space that was output.
I'd recommend the following sed expression instead:
table_prefix=$(sed -n "s/^\$table_prefix.*'\([^']*\)'.*/\1/p" wp-config.php)
You should try
#!/bin/sh
table_prefix=$(awk -F"'" '/^\$table_prefix/{print $2}' wp-config.php)
echo $table_prefix
Does this one work for you?
awk -F\' '/^\$table_prefix/ {print $2}' wp-config.php
Update
If you are using shell scripting, there is no need to call up awk, grep:
#!/bin/sh
while read varName op varValue theRest
do
if [ "_$varName" = "_\$table_prefix" ]
then
table_prefix=${varValue//\'/} # Remove the single quotes
table_prefix=${table_prefix/;/} # Remove the semicolon
break
fi
done < wp-config.php
echo "Found: $table_prefix"
I tried the following snippet in a shell script but awk didn't find $REF
REF=SEARCH_TEXT
echo "some text" | awk '/$REF/{print $2}'
Instead of quoting games in the shell, use the -v option to pass the shell variable as an awk variable:
awk -v ref="$REF" 'match($0, ref) {print $2}'
If $REF is just text and not a regular expression, use the index() function instead of match().
You question is worded really poor...
Anyway, I think you want this:
REF=SEARCH_TEXT
echo "some text" | awk "/$REF/{print \$2}"
Note the escaping of $2 and the double quotes.
or this:
REF=SEARCH_TEXT
echo "some text" | awk "/$REF/"'{print $2}'
Note the judicious use of double and single quotes and no escaping on $2.
You have to use shell expansion, as otherwise it would encompass exporting a shell variable and using it from the environment with awk - which is overkill in this situation:
export REF=SEARCH_TEXT
echo "some text" | awk '{if (match($0, ENVIRON["REF"])) print $2}'
I think awk does not support variables in /.../ guards. Please correct me if I'm wrong.
In gawk, you have the ENVIRON array, e.g. awk 'END{print ENVIRON["REF"]}' /dev/null will print your variable if you've exported it out from the shell to sub-processes.
I have a shell script that constructs an awk program as a string then pass that string to awk. This is because I want to use values of shell variables in the awk program.
My code looks like this:
awk_prog="'{if (\$4~/$shell_var/) print \$1,\$2}'"
echo $awk_prog
awk $awk_prog $FILENAME
However, when I pass the string to awk, I always get the error:
'{if ($4~/regex/) print $1,$2}'
awk: '{if
awk: ^ invalid char ''' in expression
What does that error message mean? I tried the -F: switch but it does not help. How can I settle this issue?
Thank you.
This is caused by shell quoting. The following will work:
awk_prog="{ if (\$4 ~ /$shell_var/) print \$1, \$2 }"
echo "$awk_prog"
awk "$awk_prog" $FILENAME
When you run awk '{ print }' foo from the command line, the shell interprets and removes the quotes around the program so awk receives two arguments - the first is the program text and the second is the filename foo. Your example was sending awk the program text '{if ...}' which is invalid syntax as far as awk is concerned. The outer quotes should not be present.
In the snippet that I gave above, the shell uses the quotes in the awk_prog= line to group the contents of the string into a single value and then assigns it to the variable awk_prog. When it executes the awk "$awk_prog"... line, you have to quote the expansion of $awk_prog so awk receives the program text as a single argument.
There's another way to get your shell variable into awk -- use awk's -v option:
awk -v pattern="$shell_var" '$4 ~ pattern {print $1, $2}' "$FILENAME"
Use -v multiple times if you have several variables to pass to awk.
If you truly want to hold your awk program in a shell variable, build it up using printf:
awk_script="$( printf '$4 ~ /%s/ {print $1, $2}' "$shell_var" )"
awk "$awk_script" "$FILENAME"
Note the use of quotes in the printf command: single quotes around the template to protect the dollar signs you want awk to interpret, double quotes for shell variables.
Another (IMO simpler) solution which (I think) addresses what you are intuitively trying to do is simply to use eval. You want the shell to behave as if you had literally typed:
awk '{if ($4~/foo/) print $1,$2}' path
(where foo and path are the literal contents of $shell_var and $FILENAME). To make that happen, just slap an eval on the front of your last line (and perhaps quotes for good measure, but they aren't necessary in this case) so that your last line is:
eval "awk $awk_prog $FILENAME"