I'm getting this error
Syntax error: redirection unexpected
in the line:
if grep -q "^127.0.0." <<< "$RESULT"
How I can run this in Ubuntu?
<<< is a bash-specific redirection operator (so it's not specific to Ubuntu). The documentation refers to it as a "Here String", a variant of the "Here Document".
3.6.7 Here Strings
A variant of here documents, the format is:
<<< word
The word is expanded and supplied to the command on its
standard input.
A simple example:
$ cat <<< hello
hello
If you're getting an error, it's likely that you're executing the command using a shell other than bash. If you have #!/bin/sh at the top of your script, try changing it to #!/bin/bash.
If you try to use it with /bin/sh, it probably assumes the << refers to a "here document", and then sees an unexpected < after that, resulting in the "Syntax error: redirection unexpected" message that you're seeing.
zsh and ksh also support the <<< syntax.
if grep -q "^127.0.0." <<< "$RESULT"
then
echo IF-THEN
fi
is a Bash-specific thing. If you are using a different bourne-compatable shell, try:
if echo "$RESULT" | grep -q "^127.0.0."
then
echo IF-THEN
fi
It works for me on Ubuntu, if I complete you IF block:
if grep -q "^127.0.0." <<< "$RESULT"; then echo ""; fi
Related
I was looking at an answer in another thread about which bracket pair to use with if in a bash script. [[ is less surprising and has more features such as pattern matching (=~) whereas [ and test are built-in and POSIX compliant making them portable.
Recently, I was attempting to test the result of a grep command and it was failing with [: too many arguments. I was using [. But, when I switched to [[ it worked. How would I do such a test with [ in order to maintain the portability?
This is the test that failed:
#!/bin/bash
cat > slew_pattern << EOF
g -x"$
EOF
if [ $(grep -E -f slew_pattern /etc/sysconfig/ntpd) ]; then
echo "slew mode"
else
echo "not slew mode"
fi
And the test that succeeded:
#!/bin/bash
cat > slew_pattern << EOF
g -x"$
EOF
if [[ $(grep -E -f slew_pattern /etc/sysconfig/ntpd) ]]; then
echo "slew mode"
else
echo "not slew mode"
fi
if [ $(grep -E -f slew_pattern /etc/sysconfig/ntpd) ]; then
This command will certainly fail for multiple matches. It will throw an error as the grep output is being split on line ending.
Multiple matches of grep are separated by new line and the test command becomes like:
[ match1 match2 match3 ... ]
which doesn't make much of a sense. You will get different error messages as the number of matches returned by grep (i.e the number of arguments for test command [).
For example:
2 matches will give you unary operator expected error
3 matches will give you binary operator expected error and
more than 3 matches will give you too many arguments error or such, in Bash.
You need to quote variables inside [ to prevent word splitting.
On the other hand, the Bash specific [[ prevents word splitting by default. Thus the grep output doesn't get split on new line and remains a single string which is a valid argument for the test command.
So the solution is to look only at the exit status of grep:
if grep -E -f slew_pattern /etc/sysconfig/ntpd; then
Or use quote when capturing output:
if [ "$(grep -E -f slew_pattern /etc/sysconfig/ntpd)" ]; then
Note:
You don't really need to capture the output here, simply looking at the exit status will suffice.
Additionally, you can suppress output of grep command to be printed with -q option and errors with -s option.
I want to echo a string that might contain the same parameters as echo. How can I do it without modifying the string?
For instance:
$ var="-e something"
$ echo $var
something
... didn't print -e
A surprisingly deep question. Since you tagged bash, I'll assume you mean bash's internal echo command, though the GNU coreutils' standalone echo command probably works similarly enough.
The gist of it is: if you really need to use echo (which would be surprising, but that's the way the question is written by now), it all depends on what exactly your string can contain.
The easy case: -e plus non-empty string
In that case, all you need to do is quote the variable before passing it to echo.
$ var="-e something"
$ echo "$var"
-e something
If the string isn't eaxctly an echo option or combination, which includes any non-option suffix, it won't be recognized as such by echo and will be printed out.
Harder: string can be -e only
If your case can reduce to just "-e", it gets trickier. One way to do it would be:
$ echo -e '\055e'
-e
(escaping the dash so it doesn't get interpreted as an option but as on octal sequence)
That's rewriting the string. It can be done automatically and non-destructively, so it feels acceptable:
$ var="-e something"
$ echo -e ${var/#-/\\055}
-e something
You noticed I'm actually using the -e option to interpret an octal sequence, so it won't work if you intended to echo -E. It will work for other options, though.
The right way
Seriously, you're not restricted to echo, are you?
printf '%s\n' "$var"
The proper bash way is to use printf:
printf "%s\n" "$var"
By the way, your echo didn't work because when you run:
var="-e something"
echo $var
(without quoting $var), echo will see two arguments: -e and something. Because when echo meets -e as its first argument, it considers it's an option (this is also true for -n and -E), and so processes it as such. If you had quoted var, as shown in other answers, it would have worked.
Quote it:
$ var="-e something"
$ echo "$var"
-e something
If what you want is to get echo -e's behaviour (enable interpretation of backslash escapes), then you have to leave the $var reference without quotes:
$ var="hi\nho"
$ echo $var
hi
ho
Or use eval:
$ var="hi\nho"
$ eval echo \${var}
hi\nho
$ var="-e hi\nho"
$ eval echo \${var}
hi
ho
Since we're using bash, another alternative to echo is to simply cat a "here string":
$ var="-e something"
$ cat <<< "$var"
-e something
$ var="-e"
$ cat <<< "$var"
-e
$
printf-based solutions will almost certainly be more portable though.
Try the following:
$ env POSIXLY_CORRECT=1 echo -e
-e
Due to shell aliases and built-in echo command, using an unadorned
echo interactively or in a script may get you different functionality
than that described here. Invoke it via env (i.e., env echo ...)
to avoid interference from the shell.
The environment variable POSIXLY_CORRECT was introduced to allow the user to force the standards-compliant behaviour. See: POSIX at Wikipedia.
Or use printf:
$ printf '%s\n' "$var"
Source: Why is bash swallowing -e in the front of an array at stackoverflow SE
Use printf instead:
var="-e bla"
printf "%s\n" "$var"
Using just echo "$var" will still fail if var contains just a -e or similar. If you need to be able to print that as well, use printf.
I am trying to work out how to make bash (force?) expand variables in a string (which was loaded from a file).
I have a file called "something.txt" with the contents:
hello $FOO world
I then run
export FOO=42
echo $(cat something.txt)
this returns:
hello $FOO world
It didn't expand $FOO even though the variable was set. I can't eval or source the file - as it will try and execute it (it isn't executable as it is - I just want the string with the variables interpolated).
Any ideas?
I stumbled on what I think is THE answer to this question: the envsubst command:
echo "hello \$FOO world" > source.txt
export FOO=42
envsubst < source.txt
This outputs: hello 42 world
If you would like to continue work on the data in a file destination.txt, push this back to a file like this:
envsubst < source.txt > destination.txt
In case it's not already available in your distro, it's in the
GNU package gettext.
#Rockallite
I wrote a little wrapper script to take care of the '$' problem.
(BTW, there is a "feature" of envsubst, explained at
https://unix.stackexchange.com/a/294400/7088
for expanding only some of the variables in the input, but I
agree that escaping the exceptions is much more convenient.)
Here's my script:
#! /bin/bash
## -*-Shell-Script-*-
CmdName=${0##*/}
Usage="usage: $CmdName runs envsubst, but allows '\$' to keep variables from
being expanded.
With option -sl '\$' keeps the back-slash.
Default is to replace '\$' with '$'
"
if [[ $1 = -h ]] ;then echo -e >&2 "$Usage" ; exit 1 ;fi
if [[ $1 = -sl ]] ;then sl='\' ; shift ;fi
sed 's/\\\$/\${EnVsUbDolR}/g' | EnVsUbDolR=$sl\$ envsubst "$#"
Many of the answers using eval and echo kind of work, but break on various things, such as multiple lines, attempting to escaping shell meta-characters, escapes inside the template not intended to be expanded by bash, etc.
I had the same issue, and wrote this shell function, which as far as I can tell, handles everything correctly. This will still strip only trailing newlines from the template, because of bash's command substitution rules, but I've never found that to be an issue as long as everything else remains intact.
apply_shell_expansion() {
declare file="$1"
declare data=$(< "$file")
declare delimiter="__apply_shell_expansion_delimiter__"
declare command="cat <<$delimiter"$'\n'"$data"$'\n'"$delimiter"
eval "$command"
}
For example, you can use it like this with a parameters.cfg which is really a shell script that just sets variables, and a template.txt which is a template that uses those variables:
. parameters.cfg
printf "%s\n" "$(apply_shell_expansion template.txt)" > result.txt
In practice, I use this as a sort of lightweight template system.
you can try
echo $(eval echo $(cat something.txt))
You don't want to print each line, you want to evaluate it so that Bash can perform variable substitutions.
FOO=42
while read; do
eval echo "$REPLY"
done < something.txt
See help eval or the Bash manual for more information.
Another approach (which seems icky, but I am putting it here anyway):
Write the contents of something.txt to a temp file, with an echo statement wrapped around it:
something=$(cat something.txt)
echo "echo \"" > temp.out
echo "$something" >> temp.out
echo "\"" >> temp.out
then source it back in to a variable:
RESULT=$(source temp.out)
and the $RESULT will have it all expanded. But it seems so wrong !
Single line solution that doesn't need temporary file :
RESULT=$(source <(echo "echo \"$(cat something.txt)\""))
#or
RESULT=$(source <(echo "echo \"$(<something.txt)\""))
If you only want the variable references to be expanded (an objective that I had for myself) you could do the below.
contents="$(cat something.txt)"
echo $(eval echo \"$contents\")
(The escaped quotes around $contents is key here)
If something.txt has only one line, a bash method, (a shorter version of Michael Neale's "icky" answer),
using process & command substitution:
FOO=42 . <(echo -e echo $(<something.txt))
Output:
hello 42 world
Note that export isn't needed.
If something.txt has one or more lines, a GNU sed evaluate method:
FOO=42 sed 's/"/\\\"/g;s/.*/echo "&"/e' something.txt
Following solution:
allows replacing of variables which are defined
leaves unchanged variables placeholders which are not defined. This is especially useful during automated deployments.
supports replacement of variables in following formats:
${var_NAME}
$var_NAME
reports which variables are not defined in environment and returns error code for such cases
TARGET_FILE=someFile.txt;
ERR_CNT=0;
for VARNAME in $(grep -P -o -e '\$[\{]?(\w+)*[\}]?' ${TARGET_FILE} | sort -u); do
VAR_VALUE=${!VARNAME};
VARNAME2=$(echo $VARNAME| sed -e 's|^\${||g' -e 's|}$||g' -e 's|^\$||g' );
VAR_VALUE2=${!VARNAME2};
if [ "xxx" = "xxx$VAR_VALUE2" ]; then
echo "$VARNAME is undefined ";
ERR_CNT=$((ERR_CNT+1));
else
echo "replacing $VARNAME with $VAR_VALUE2" ;
sed -i "s|$VARNAME|$VAR_VALUE2|g" ${TARGET_FILE};
fi
done
if [ ${ERR_CNT} -gt 0 ]; then
echo "Found $ERR_CNT undefined environment variables";
exit 1
fi
foo=45
file=something.txt # in a file is written: Hello $foo world!
eval echo $(cat $file)
$ eval echo $(cat something.txt)
hello 42 world
$ bash --version
GNU bash, version 3.2.57(1)-release (x86_64-apple-darwin17)
Copyright (C) 2007 Free Software Foundation, Inc.
envsubst is a great solution (see LenW's answer) if the content you're substituting is of "reasonable" length.
In my case, I needed to substitute in a file's content to replace the variable name. envsubst requires that the content be exported as environment variables and bash has a problem when exporting environment variables that are more than a megabyte or so.
awk solution
Using cuonglm's solution from a different question:
needle="doc1_base64" # The "variable name" in the file. (A $ is not needed.)
needle_file="doc1_base64.txt" # Will be substituted for the needle
haystack=$requestfile1 # File containing the needle
out=$requestfile2
awk "BEGIN{getline l < \"${needle_file}\"}/${needle}/{gsub(\"${needle}\",l)}1" $haystack > $out
This solution works for even large files.
expenv () {
LF=$'\n'
echo "cat <<END_OF_TEXT${LF}$(< "$1")${LF}END_OF_TEXT" | bash
return $?
}
expenv "file name"
The following works: bash -c "echo \"$(cat something.txt)"\"
I am trying to do an if/then statement, where if there is non-empty output from a ls | grep something command then I want to execute some statements. I am do not know the syntax I should be using. I have tried several variations of this:
if [[ `ls | grep log ` ]]; then echo "there are files of type log";
Well, that's close, but you need to finish the if with fi.
Also, if just runs a command and executes the conditional code if the command succeeds (exits with status code 0), which grep does only if it finds at least one match. So you don't need to check the output:
if ls | grep -q log; then echo "there are files of type log"; fi
If you're on a system with an older or non-GNU version of grep that doesn't support the -q ("quiet") option, you can achieve the same result by redirecting its output to /dev/null:
if ls | grep log >/dev/null; then echo "there are files of type log"; fi
But since ls also returns nonzero if it doesn't find a specified file, you can do the same thing without the grep at all, as in D.Shawley's answer:
if ls *log* >&/dev/null; then echo "there are files of type log"; fi
You also can do it using only the shell, without even ls, though it's a bit wordier:
for f in *log*; do
# even if there are no matching files, the body of this loop will run once
# with $f set to the literal string "*log*", so make sure there's really
# a file there:
if [ -e "$f" ]; then
echo "there are files of type log"
break
fi
done
As long as you're using bash specifically, you can set the nullglob option to simplify that somewhat:
shopt -s nullglob
for f in *log*; do
echo "There are files of type log"
break
done
Or without if; then; fi:
ls | grep -q log && echo 'there are files of type log'
Or even:
ls *log* &>/dev/null && echo 'there are files of type log'
The if built-in executes a shell command and selects the block based on the return value of the command. ls returns a distinct status code if it does not find the requested files so there is no need for the grep part. The [[ utility is actually a built-in command from bash, IIRC, that performs arithmetic operations. I could be wrong on that part since I rarely stray far from Bourne shell syntax.
Anyway, if you put all of this together, then you end up with the following command:
if ls *log* > /dev/null 2>&1
then
echo "there are files of type log"
fi
I'm writing a shell script to parse through log file and pull out all instances where sudo succeeded and/or failed. I'm realizing now that this probably would've been easier with shell's equivalent of regex, but I didn't want to take the time to dig around (and now I'm paying the price). Anyway:
sudobool=0
sudoCount=0
for i in `cat /var/log/auth.log`;
do
for word in $i;
do
if $word == "sudo:"
then
echo "sudo found"
sudobool=1;
sudoCount=`expr $sudoCount + 1`;
fi
done
sudobool=0;
done
echo "There were " $sudoCount " attempts to use sudo, " $sudoFailCount " of which failed."
So, my understanding of the code I've written: read auth.log and split it up line by line, which are stored in i. Each word in i is checked to see if it is sudo:, if it is, we flip the bool and increment. Once we've finished parsing the line, reset the bool and move to the next line.
However, judging by my output, the shell is trying to execute the individual words of the log file, typically returning '$word : not found'.
why don't you use grep for this?
grep sudo /var/log/auth.log
if you want a count pipe it to wc -l
grep sudo /var/log/auth.log | wc -l
or still better use -c option to grep, which prints how many lines were found containing sudo
grep -c sudo /var/log/auth.log
or maybe I am missing something simple here?
EDIT: I saw $sudoFailCount after scrolling, do you want to count how many failed attempts were made to use sudo ?? You have not defined any value for $sudoFailCount in your script, so it will print nothing. Also you are missing the test brackets [[ ]] around your if condition checking
Expanding on Sudhi's answer, here's a one-liner:
$ echo "There were $(grep -c ' sudo: ' /var/log/auth.log) attempts to use sudo, $(grep -c ' sudo: .*authentication failure' /var/log/auth.log) of which failed."
There were 17 attempts to use sudo, 1 of which failed.
Your error message arises from a lack of syntax in your if statement: you need to put the condition in [[brackets]]
Using the pattern matching in bash:
#!/bin/bash
sudoCount=0
while read line; do
sudoBool=0
if [[ "$line" = *sudo:* ]]; then
sudoBool=1
(( sudoCount++ ))
# do something with sudobool ?
fi
done < /var/log/auth.log
echo "There were $sudoCount attempts to use sudo."
I'm not initimately familiar with the auth.log -- what is the pattern to determine success or failure?