If condition giving error in shell script when checking two strings - linux

In following shell script I want to perform two different tasks depending on file type,
but it is giving an error: "[==c]: command not found"
echo "enter file name"
read num
var_check= echo $str |awk -F . '{if (NF>1) {print $NF}}'
if ["$var_check"=="c"];then
echo "Some task for c"
elif ["$var_check"=="cpp"];then
echo "Some task for cpp"
else
echo "Wrong file extension"
fi

You wrote:
if ["$var_check"=="c"];then
The [ command is a command; its name must be surrounded by spaces (put simplistically).
if [ "$var_check" == "c" ]; then
The last argument, ], must also be preceded by a space. The operands within must also be space separated; they need to be separate arguments. The rules for the [[ ... ]] operator are a bit different, but using spaces helps people read the code even there. What you wrote is a bit like expecting:
ls"-l"/dev/tty
to work; it won't.
You also need to double check whether your test or [ operator supports ==; the normal form is =.
The line:
var_check= echo $str |awk -F . '{if (NF>1) {print $NF}}'
This runs the echo command with var_check set as an environment variable, which is unlikely to be what you wanted. You almost certainly intended to write:
var_check=$(echo $str |awk -F . '{if (NF>1) {print $NF}}')
This runs the echo and awk commands and captures the output in var_check. Use the $(...) notation in preference to the older but more complex to use `...` notation. In simple cases, they look the same; when you nest them, the $(...) notation is far, far simpler to understand and use.
Also, looking on the larger scale (3 lines instead of just 1 line):
echo "enter file name"
read num
var_check=$(echo $str |awk -F . '{if (NF>1) {print $NF}}')
You read the file name into variable num; you then echo $str instead of $num. If you've already got $str set somewhere earlier in the script (in unshown code), what you've got may be fine. Taken as a standalone fragment, it isn't right.
You could also simplify the awk a little:
var_check=$(echo $str |awk -F . 'NF > 1 {print $NF}')
This would work the same as what you wrote, but uses fewer parentheses and braces.

Related

How to monitor CPU usage automatically and return results when it reaches a threshold

I am new to shell script , i want to write a script to monitor CPU usage and if the CPU usage reaches a threshold it should print the CPU usage by top command ,here is my script , which is giving me error bad number and also not storing any value in the log files
while sleep 1;do if [ "$(top -n1 | grep -i ^cpu | awk '{print $2}')">>sy.log - ge "$Threshold" ]; then echo "$(top -n1)">>sys.log;fi;done
Your script HAS to be indented and stored to a file, especially if you are new to shell !
#!/bin/sh
while sleep 1
do
if [ "$(top -n1 | grep -i ^cpu | awk '{print $2}')">>sy.log - ge "$Threshold" ]
then
echo "$(top -n1)" >> sys.log
fi
done
Your condition looks a bit odd. It may work, but it looks really complex. Store intermediate results in variables, and evaluate them.
Then, you will immediately see the syntax error on the “-ge”.
You HAVE to store logfiles within an absolute path for security reasons. Use variables to simplify the reading.
#!/bin/sh
LOGFILE=/absolute_path/sy.log
WHOLEFILE=/absolute_path/sys.log
Thresold=80
while sleep 1
do
TOP="$(top -n1)"
CPU="$(echo $TOP | grep -i ^cpu | awk '{print $2}')"
echo $CPU >> $LOGFILE
if [ "$CPU" -ge "$Threshold" ] ; then
echo "$TOP" >> $WHOLEFILE
fi
done
You have a couple of errors.
If you write output to sy.log with a redirection then that output is no longer available to the shell. You can work around this with tee.
The dash before -ge must not be followed by a space.
Also, a few stylistic remarks:
grep x | awk '{y}' is a useless use of grep; this can usefully and more economically (as well as more elegantly) be rewritten as awk '/x/{y}'
echo "$(command)" is a useless use of echo -- not a deal-breaker, but you simply want command; there is no need to capture what it prints to standard output just so you can print that text to standard output.
If you are going to capture the output of top -n 1 anyway, there is no need really to run it twice.
Further notes:
If you know the capitalization of the field you want to extract, maybe you don't need to search case-insensitively. (I could not find a version of top which prints a CPU prefix with the load in the second field -- it the expression really correct?)
The shell only supports integer arithmetic. Is this a bug? Maybe you want to use Awk (which has floating-point support) to perform the comparison? This also allows for a moderately tricky refactoring. We make Awk output an exit code of 1 if the comparison fails, and use that as the condition for the if.
#!/bin/sh
while sleep 1
do
if top=$(top -n 1 |
awk -v thres="$Threshold" '1; # print every line
tolower($1) ~ /^cpu/ { print $2 >>"sy.log";
exitcode = ($2 >= thres ? 0 : 1) }
END { exit exitcode }')
then
echo "$top" >>sys.log
fi
done
Do you really mean to have two log files with nearly the same name, or is that a typo? Including a time stamp in the log might be useful both for troubleshooting and for actually using the log files.

How to use awk to print "hello" if pattern is found

I want to search for a pattern in a tab-separated .txt-file and, if the pattern is found in a line, print the third field of that line.
I only need to find the first occurence in the line, since the pattern appears only once for sure.
Structure of .txt-file:
XXX01 foo target1
XXX02 bar target2
XXX03 foobar target3
My first idea was, to print "hello", if the pattern is found, to control, if my code works. I also included echos of the variables I pass to my bash script.
Command line call and Script:
$ ./script.sh file.txt foo
#!/bin/bash
file=$1
pattern=$2
awk '/"$pattern"/{print "hello"}' "$file"
echo "$file"
echo "$pattern"
As far as I found it for awk, to get the third field printed, I would have to substitute print "hello" with print "\$2".
But printing "hello" already does not work:
Actual output:
file.txt
foo
Desired output:
hello (respectively target1)
file.txt
foo
And I also checked for sure, that "foo" is in the file.txt
Progress (see comments and answer please):
#!/bin/bash
awk -v p="$2"'$2=="$p"{print "hello",$3}' "$1"
echo "$1"
echo "$2"
new output:
awk: 1:unexpected character '.'
file.txt
foo
I believe you want something like:
$ ./script.sh file.txt foo
#!/bin/bash
file=$1
pattern=$2
awk -v pattern=$pattern'$2==pattern{print "hello",$3}' "$file"
echo "$file"
echo "$pattern"
Here we get rid of the loop since awk checks every record when it is fed a file. We also use the -v flag to pass in the $pattern variable into the awk script. Then we check that the second field $2 is pattern and print "hello" as well as the contents of the third field $3.
You could change that awk condition to $2~/pattern/ to truly utilize regex if you want but I suspect it will print the 1st and 3rd line as foo shows up in both.
If you want to check if your pattern exists in anywhere in the line then you can drop the $2~ so it's just '/pattern/{print "hello",$2}.
Look:
$ x="foo"'bar' && echo "$x"
foobar
$ x="foo" 'bar' && echo "$x"
-bash: bar: command not found
Your script is:
awk -v p="$2"'$2=="$p"{print "hello",$3}' "$1"
so guess what not leaving a space between -v p="$2" and '$2=="$p" is doing. Right, it's concatenating them so don't do that - add a space:
awk -v p="$2" '$2=="$p"{print "hello",$3}' "$1"
The unexpected . btw was the . in your file name file.txt when awk was trying to evaluate the string file.txt as its cript due to the concatenation consuming the actual script into the assignment to p.
Now to actually USE the variable p in the comparison you'd have to use it as a variable instead of putting it inside a string:
awk -v p="$2" '$2==p{print "hello",$3}' "$1"
The above simply answers your question about the syntax error. To actually do what you WANT would require one of these, depending on whether you want a string or regexp match and whether you want partial or full matching:
awk -v p="$2" '$2==p{print "hello",$3}' "$1"
awk -v p="$2" '$2~p{print "hello",$3}' "$1"
awk -v p="$2" '$2~"\\<"p"\\>"{print "hello",$3}' "$1"
or some other solution depending on your so far unstated requirements.

Linux Scripting with Spaces in Filenames

I am currently working with a vendor-provided software that is trying to handle sending attachment files to another script that will text-extract from the listed file. The script fails when we receive files from an outside source that contain spaces, as the vendor-supplied software does not surround the filename in quotes - meaning when the text-extraction script is run, it receives a filename that will split apart on the space and cause an error on the extractor script. The vendor-provided software is not editable by us.
This whole process is designed to be an automated transfer, so having this wrench that could be randomly thrown into the gears is an issue.
What we're trying to do, is handle the spaced name in our text extractor script, since that is the piece we have some control over. After a quick Google, it seems like changing the IFS value for the script would be the quick solution, but unfortunately, that script would take effect after the extensions have already mutilated the incoming data.
The script I'm using takes in a -e value, a -i value, and a -o value. These values are sent from the vendor supplied script, which I have no editing control over.
#!/bin/bash
usage() { echo "Usage: $0 -i input -o output -e encoding" 1>&2; exit 1; }
while getopts ":o:i:e:" o; do
case "${o}" in
i)
inputfile=${OPTARG}
;;
o)
outputfile=${OPTARG}
;;
e)
encoding=${OPTARG}
;;
*)
usage
;;
esac
done
shift $((OPTIND-1))
...
...
<Uses the inputfile, outputfile, and encoding variables>
I admit, there may be pieces to this I don't fully understand, and it could be a simple fix, but my end goal is to be able to extract -o, -i, and -e that all contain 1 value, regardless of the spaces within each section. I can handle quoting the script after I can extract the filename value
The script fragment that you have posted does not have any issues with spaces in the arguments.
The following, for example, does not need quoting (since it's an assignment):
inputfile=${OPTARG}
All other uses of $inputfile in the script should be double quoted.
What matters is how this script is called.
This would fail and would assign only hello to the variable inputfile:
$ ./script.sh -i hello world.txt
The string world.txt would prompt the getopts function to stop processing the command line and the script would continue with the shift (world.txt would be left in $1 afterwards).
The following would correctly assign the string hello world.txt to inputfile:
$ ./script.sh -i "hello world.txt"
as would
$ ./script.sh -i hello\ world.txt
The following script uses awk to split the arguments while including spaces in the file names. The arguments can be in any order. It does not handle multiple consecutive spaces in an argument, it collapses them to one.
#!/bin/bash
IFS=' '
str=$(printf "%s" "$*")
istr=$(echo "${str}" | awk 'BEGIN {FS="-i"} {print $2}' | awk 'BEGIN {FS="-o"} {print $1}' | awk 'BEGIN {FS="-e"} {print $1}')
estr=$(echo "${str}" | awk 'BEGIN {FS="-e"} {print $2}' | awk 'BEGIN {FS="-o"} {print $1}' | awk 'BEGIN {FS="-i"} {print $1}')
ostr=$(echo "${str}" | awk 'BEGIN {FS="-o"} {print $2}' | awk 'BEGIN {FS="-e"} {print $1}' | awk 'BEGIN {FS="-i"} {print $1}')
inputfile=""${istr}""
outputfile=""${ostr}""
encoding=""${estr}""
# call the jar
There was an issue when calling the jar where Java threw a MalformedUrlException on a filename with a space.
So after reading through the commentary, we decided that although it may not be the right answer for every scenario, the right answer for this specific scenario was to extract the pieces manually.
Because we are building this for a pre-built script passing to it, and we aren't updating that script any time soon, we can accept with certainty that this script will always receive a -i, -o, and -e flag, and there will be spaces between them, which causes all the pieces passed in to be stored in different variables in $*.
And we can assume that the text after a flag is the response to the flag, until another flag is referenced. This leaves us 3 scenarios:
The variable contains one of the flags
The variable contains the first piece of a parameter immediately after the flag
The variable contains part 2+ of a parameter, and the space in the name was interpreted as a split, and needs to be reinserted.
One of the other issues I kept running into was trying to get string literals to equate to variables in my IF statements. To resolve that issue, I pre-stored all relevant data in array variables, so I could test $variable == $otherVariable.
Although I don't expect it to change, we also handled what to do if the three flags appear in a different order than we anticipate (Our assumption was that they list as i,o,e... but we can't see excatly what is passed). The parameters are dumped into an array in the order they were read in, and a parallel array tracks whether the items in slots 0,1,2 relate to i,o,e.
The final result still has one flaw: if there is more than one consecutive space in the filename, the whitespace is trimmed before processing, and I can only account for one space. But saying as we processed over 4000 files before encountering one with a space, I find it unlikely with the naming conventions that we would encounter something with more than one space.
At that point, we would have to be stepping in for a rare intervention anyways.
Final code change is as follows:
#!/bin/bash
IFS='|'
position=-1
ioeArray=("" "" "")
previous=""
flagArr=("-i" "-o" "-e" " ")
ioePattern=(0 1 2)
#echo "for loop:"
for i in $*; do
#printf "%s\n" "$i"
if [ "$i" == "${flagArr[0]}" ] || [ "$i" == "${flagArr[1]}" ] || [ "$i" == "${flagArr[2]}" ]; then
((position += 1));
previous=$i;
case "$i" in
"${flagArr[0]}")
ioePattern[$position]=0
;;
"${flagArr[1]}")
ioePattern[$position]=1
;;
"${flagArr[2]}")
ioePattern[$position]=2
;;
esac
continue;
fi
if [[ $previous == "-"* ]]; then
ioeArray[$position]=${ioeArray[$position]}$i;
else
ioeArray[$position]=${ioeArray[$position]}" "$i;
fi
previous=$i;
done
echo "extracting (${ioeArray[${ioePattern[0]}]}) to (${ioeArray[${ioePattern[1]}]}) with (${ioeArray[${ioePattern[2]}]}) encoding."
inputfile=""${ioeArray[${ioePattern[0]}]}"";
outputfile=""${ioeArray[${ioePattern[1]}]}"";
encoding=""${ioeArray[${ioePattern[2]}]}"";

Escaping ~! in awk (bash command), backslash not last character on line

I am trying to run a bash command in the following format:
declare "test${nb}"="$(cat file.txt | awk '{if($3>0.5 && $3 !~ "ddf") $2="NA"; print $1,$2}')"
where $nb is an int (e.g. 2) and file.txt contains a table with various numeric and string values (I can provide more details if needed, but it should not be relevant here)
when running this, the shell substitutes !~ for the name of a file that I have (not sure why). I tried escaping this using the backslash like this:
declare "test${nb}"="$(cat file.txt | awk '{if($3>0.5 && $3 \!~ "ddf") $2="NA"; print $1,$2}')"
but then I get this error:
awk: {if($3>0.5 && $3 \!~ "ddf") $2="NA"; print $1,$2}
awk: ^ backslash not last character on line
I also tried having the table contained in the variable "var" and writing it this way:
declare test[$nb]=$(echo "$var" | awk '{if($3>0.5 && $3 !~ "ddf") $2="NA"; print $1,$2}')
Then there is no error, but the output is just the first field of first column of the table, which is not the case when I don't expand the variable name. For example, if I do this:
declare test2=$(echo "$var" | awk '{if($3>0.5 && $3 !~ "ddf") $2="NA"; print $1,$2}')
then it works perfectly and test2 has the expected value. But I need to be able to use any number instead of 2 (something like test[$nb]).
any idea how I could fix this? Any help will be very appreciated!
thanks
Lose the quotes:
$ declare x="$( echo '!' )"
-bash: !': event not found
$ declare x=$( echo '!' )
$ echo "$x"
!
You have a lot of other issues with your statement, though, including UUOC, using a scalar to emulate an array, non idiomatic awk syntax, etc. Try this instead:
declare test[$nb]=$( awk '{print $1, (($3 > 0.5) && ($3 !~ /ddf/) ? "NA" : $2)}' file.txt )
Are you talking about interactively entering a command, or running the command from a file (i.e. script)? The history expansion of !~ occurs only in interactive use. BTW, the bash manpage says: If enabled, history expansion will be performed unless an ! appearing in double quotes is escaped using a backslash. The backslash preceding the ! is not removed.
This means that, as long as you do scripting (or, more precisley, as long as you do not have an interactive shell), you don't have to worry about this special meaning of '!'.
If you have an interactive shell and history expansion bothers you, you can turn it off by
set +H

Unable to delete a line in file in shell script

I have to delete a line in a file from inside a shell script.
I am trying this:
linenumber=0
##CHeck If server IP exists
if grep -wq $serverip $FILE; then
echo "IP exists"
linenumber=$(awk -v serverip="$serverip" '$0 ~ serverip {print NR}' $FILE)
echo "$linenumber"
sed -e '${$linenumber}d' $FILE
fi
Basically I extract the line number and then want to delete it.
sed -e '1d' $FILE --> WOrks on CLI but inside script does not work
Why? How to get it working ?
This is simply a case of using the incorrect quotes around your sed command, so the variable isn't being used. Ignoring the fact that you're unnecessarily using 3 tools when 1 would suffice, the fix is this:
sed -e "${linenumber}d" "$FILE"
Perhaps your requirement is more complex than it appears but I would suggest changing your entire script to this:
awk -v serverip="$serverip" '!($0 ~ serverip)' "$FILE"
This prints every line that doesn't contain the shell variable $serverip. It is assumed that you have escaped any regex meta-characters present in the variable.
Alternatively (and more succinctly):
sed "/$serverip/d" "$FILE"
If you actually want the messages to be printed out (I assumed that they were for debugging), then that's easy enough to achieve:
awk -v serverip="$serverip" '$0 ~ serverip { print "IP exists"; print NR; next } 1' "$FILE"
If you're not familiar with the 1 at the end, it's just a common shorthand which causes awk to print each line (1 is always true and the default action is { print }).

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