This is just a simple problem but I don't understand why I got an error here. This is just a for loop inside an if statement.
This is my code:
#!/bin/bash
if (!( -f $argv[1])) then
echo "Argv must be text file";
else if ($#argv != 1) then
echo "Max argument is 1";
else if (-f $argv[1]) then
for i in `cut -d ',' -f2 $argv[1]`
do
ping -c 3 $i;
echo "finish pinging host $i"
done
fi
Error is in line 16, which is the line after fi, that is a blank line .....
Can someone please explain why i have this error ????
many, many errors.
If I try to stay close to your example code:
#!/bin/sh
if [ ! -f "${1}" ]
then
echo "Argv must be text file";
else if [ "${#}" -ne 1 ]
then
echo "Max argument is 1";
else if [ -f "${1}" ]
then
for i in $(cat "${1}" | cut -d',' -f2 )
do
ping -c 3 "${i}";
echo "finish pinging host ${i}"
done
fi
fi
fi
another way, exiting each time the condition is not met :
#!/bin/sh
[ "${#}" -ne 1 ] && { echo "There should be 1 (and only 1) argument" ; exit 1 ; }
[ ! -f "${1}" ] && { echo "Argv must be a file." ; exit 1 ; }
[ -f "${1}" ] && {
for i in $(cat "${1}" | cut -d',' -f2 )
do
ping -c 3 "${i}";
echo "finish pinging host ${i}"
done
}
#!/usr/local/bin/bash -x
if [ ! -f "${1}" ]
then
echo "Argument must be a text file."
else
while-loop-script "${1}"
fi
I have broken this up, because I personally consider it extremely bad form to nest one function inside another; or truthfully to even have more than one function in the same file. I don't care about file size, either; I've got several scripts which are 300-500 bytes long. I'm learning FORTH; fractalism in that sense is a virtue.
# while-loop-script
while read line
do
IFS="#"
ping -c 3 "${line}"
IFS=" "
done < "${1}"
Don't use cat in order to feed individual file lines to a script; it will always fail, and bash will try and execute the output as a literal command. I thought that sed printing would work, and it often does, but for some reason it very often substitutes spaces for newlines, which is extremely annoying as well.
The only absolutely bulletproof method of feeding a line to a script that I know of, which will preserve all space and formatting, is to use while-read loops, rather than substituted for cat or for sed loops, as mentioned.
Something else which you will need to do, in order to be sure about preserving whitespace, is to set the internal field seperator (IFS) to something that you know your file will not contain, and then resetting it back to whitespace at the end of the loop.
For every opening if, you must have a corresponding closing fi. This is also true for else if. Better use elif instead
if test ! -f "$1"; then
echo "Argv must be text file";
elif test $# != 1; then
echo "Max argument is 1";
elif test -f "$1"; then
for i in `cut -d ',' -f2 "$1"`
do
ping -c 3 $i;
echo "finish pinging host $i"
done
fi
There's also no argv variable. If you want to access the command line arguments, you must use $1, $2, ...
Next point is $#argv, this evaluates to $# (number of command line args) and argv. This looks a lot like perl.
Furthermore, testing is done with either test ... or [ ... ], not ( ... )
And finally, you should enclose at least your command line arguments in double quotes "$1". If you don't and there is no command line argument, you have for example
test ! -f
instead of
test ! -f ""
This lets the test fail and go on to the second if, instead of echoing the proper message.
Related
I have a following list.txt file with the content
cat list.txt
one
two
zero
three
four
I have a shell script (check.sh) like below,
for i in $(cat list.txt)
do
if [ $i != zero ]; then
echo "the number is $i"
else
exit 1
fi
done
it gives output like below,
./check.sh
the number is one
the number is two
I want to have script which continue with the rest of the items in the list.txt, but it should not process zero and continue with the rest of item.
eg.
the number is one
the number is two
the number is three
the number is four
I tried using "return" but it did not work, gave error.
./check.sh: line 6: return: can only `return' from a function or sourced script
About exit (and return)
The command exit will quit running script. There is no way to continue.
As well, return command will quit function. There in no more way to continue.
About reading input file
For processing line based input file, you'd better to use while read instead of for i in $(cat...:
Simply try:
while read -r i;do
if [ "$i" != "zero" ] ;then
echo number $i
fi
done <list.txt
Alternatively, you could drop unwanted entries before loop:
while read -r i;do
echo number $i
done < <( grep -v ^zero$ <list.txt)
Note: In this specific case, ^zero$ don't need to be quoted. Consider quoting if your string do contain special characters or spaces.
If you have more than one entries to drop, you could use
while read -r i;do echo number $i ;done < <(grep -v '^\(zero\|null\)$' <list.txt)
Alternatively, once input file filtered, use xargs:
If your process is only one single command, you could avoid bash loop by using xargs:
xargs -n 1 echo number < <(grep -v '^\(zero\|null\)$' <list.txt)
How to use continue in bash script
Maybe you are thinking about something like:
while read -r i;do
if [ "$i" = "zero" ] ;then
continue
fi
echo number $i
done <list.txt
Argument of continue is a number representing number of loop to shortcut.
Try this:
for i in {1..5};do
for l in {a..d};do
if [ "$i" -eq 3 ] && [ "$l" = "b" ] ;then
continue 2
fi
echo $i.$l
done
done
(This print 3.a and stop 3 serie at 3.b, breaking 2 loop level)
Then compare with
for i in {1..5};do
for l in {a..d};do
if [ "$i" -eq 3 ] && [ "$l" = "b" ] ;then
continue 1
fi
echo $i.$l
done
done
(This print 3.a , 3.c and 3.d. Only 3.b are skipped, breaking only 1 loop level)
I was doing this little script in which the first argument must be a path to an existing directory and the second any other thing.
Each object in the path indicated in the first argument must be renamed so that the new
name is the original that was added as a prefix to the character string passed as the second argument. Example, for the string "hello", the object OBJECT1 is renamed hello.OBJECT1 and so on
Additionally, if an object with the new name is already present, a message is shown by a standard error output and the operation is not carried out continuing with the next object.
I have the following done:
#! /bin/bash
if [ "$#" != 2 ]; then
exit 1
else
echo "$2"
if [ -d "$1" ]; then
echo "directory"
for i in $(ls "$1")
do
for j in $(ls "$1")
do
echo "$i"
if [ "$j" = "$2"."$i" ]; then
exit 1
else
mv -v "$i" "$2"."$i"
echo "$2"."$i"
fi
done
done
else
echo "no"
fi
fi
I am having problems if I run the script from another file other than the one I want to do it, for example if I am in /home/pp and I want the changes to be made in /home/pp/rr, since that is the only way It does in the current.
I tried to change the ls to catch the whole route with
ls -R | sed "s;^;pwd;" but the route catches me badly.
Using find you can't because it puts me in front of the path and doesn't leave the file
Then another question, to verify that that object that is going to create new is not inside, when doing it with two for I get bash errors for all files and not just for coincidences
I'm starting with this scripting, so it has to be a very simple solution thing
An obvious answer to your question would be to put a cd "$2 in the script to make it work. However, there are some opportunities in this script for improvement.
#! /bin/bash
if [ "$#" != 2 ]; then
You might put an error message here, for example, echo "Usage: $0 dir prefix" or even a more elaborate help text.
exit 1
else
echo $2
Please quote, as in echo "$2".
if [ -d $1 ]; then
Here, the quotes are important. Suppose that your directory name has a space in it; then this if would fail with bash: [: a: binary operator expected. So, put quotes around the $1: if [ -d "$1" ]; then
echo "directory"
This is where you could insert the cd "$1".
for i in $(ls $1)
do
It is almost always a bad idea to parse the output of ls. Once again, this for-loop will fail if a file name has a space in it. A possible improvement would be for i in "$1"/* ; do.
for j in $(ls $1)
do
echo $i
if [ $j = $2.$i ]; then
exit 1
else
The logic of this section seems to be: if a file with the prefix exists, then exit instead of overwriting. It is always a good idea to tell why the script fails; an echo before the exit 1 will be helpful.
The question is why you use the second loop? a simple if [ -f "$2.$i ] ; then would do the same, but without the second loop. And it will therefore be faster.
mv -v $i $2.$i
echo $2.$i
Once again: use quotes!
fi
done
done
else
echo "no"
fi
fi
So, with all the remarks, you should be able to improve your script. As tripleee said in his comment, running shellcheck would have provided you with most of the comment above. But he also mentioned basename, which would be useful here.
With all that, this is how I would do it. Some changes you will probably only appreciate in a few months time when you need some changes to the script and try to remember what the logic was that you had in the past.
#!/bin/bash
if [ "$#" != 2 ]; then
echo "Usage: $0 directory prefix" >&2
echo "Put a prefix to all the files in a directory." >&2
exit 1
else
directory="$1"
prefix="$2"
if [ -d "$directory" ]; then
for f in "$directory"/* ; do
base=$(basename "$f")
if [ -f "Sdirectory/$prefix.$base" ] ; then
echo "This would overwrite $prefix.$base; exiting" >&2
exit 1
else
mv -v "$directory/$base" "$directory/$prefix.$base"
fi
done
else
echo "$directory is not a directory" >&2
fi
fi
Can anybody tell me what's wrong in this script, it's not working. When I run it, there is no output/error on the screen.
The script is to monitor a log file to check the value of one of the columns, and if it is more than 20 it will echo a message.
#!/bin/bash
while true ; do
COUNT=`tail -f /monitoring/log.20160121|cut -d" " -f39`
echo $COUNT
if [ $COUNT -gt 20 ] ;then
echo "Count is high"
break
fi
sleep 10
done
tail -f does not exit, so your script gets stuck there. I assume you are just interested in the last line of the log; tail -n 1 does that.
Other points:
Indentation: not sure how much got lost while copy pasting, but proper indentation massively increases readability of your code
Variable names: all uppercase variable names are discouraged as they might clash with reserved (environment) variable names
Command substitution with backticks (` `) is discouraged and the form $( ) is preferred; makes for example nesting easier
Since you're using Bash, you can use the (( )) conditional construct, which is better suited for comparing numbers than [ ]
Together:
#!/bin/bash
while true; do
count=$(tail -n 1 /monitoring/log.20160121 | cut -d " " -f 39)
echo $count
if (( count > 20 )); then
echo "Count is high"
break
fi
sleep 10
done
I've spent 2 hours with an if statement, that never works like I want:
#should return true
if [ "$1" == "355258054414904" ]; then
Here is the whole script:
#!/bin/bash
param=$1
INPUT=simu_900_imei_user_pass.csv
OLDIFS=$IFS
IFS=,
[ ! -f $INPUT ] && { echo "$INPUT ime not found"; exit 99; }
while read imei email pass
do
echo "First Parameter-IMEI: $1"
if [ "$1" == "355258054414904" ]; then
echo "GOOD"
fi
done < $INPUT
IFS=$OLDIFS
This is the output of the script:
First Parameter-IMEI: 355258054414904
First Parameter-IMEI: 355258054414904
First Parameter-IMEI: 355258054414904
I have seen a lot of pages about the subject, but I can't make it work :(
EDIT: I Join the content of csv for better understanding ! Tx for your help !
4790057be1803096,user1,pass1
355258054414904,juju,capp
4790057be1803096,user2,pass2
358854053154579,user3,pass3
The reason $1 does not match is because $1 means the first parameter given to the script on the command line, while you want it to match the first field read from the file. That value is in $imei.
You probably meant:
if [ "$imei" == "355258054414904" ]; then
echo "GOOD"
fi
Since it is inside the loop where you read input file line by line.
To check content of $1 use:
cat -vet <<< "$1"
UPDATE: To strip \r from $1 have this at top:
param=$(tr -d '\r' <<< "$1")
And then use "$param" in rest of your script.
To test string equality with [ you want to use a single '=' sign.
I need to validate my log files:
-All new log lines shall start with date.
-This date will respect the ISO 8601 standard. Example:
2011-02-03 12:51:45,220Z -
Using shell script, I can validate it looping on each line and verifying the date pattern.
The code is below:
#!/bin/bash
processLine(){
# get all args
line="$#"
result=`echo $line | egrep "[0-9]{4}-[0-9]{2}-[0-9]{2} [012][0-9]:[0-9]{2}:[0-9]{2},[0-9]{3}Z" -a -c`
if [ "$result" == "0" ]; then
echo "The log is not with correct date format: "
echo $line
exit 1
fi
}
# Make sure we get file name as command line argument
if [ "$1" == "" ]; then
echo "You must enter a logfile"
exit 0
else
file="$1"
# make sure file exist and readable
if [ ! -f $file ]; then
echo "$file : does not exists"
exit 1
elif [ ! -r $file ]; then
echo "$file: can not read"
exit 2
fi
fi
# Set loop separator to end of line
BAKIFS=$IFS
IFS=$(echo -en "\n\b")
exec 3<&0
exec 0<"$file"
while read -r line
do
# use $line variable to process line in processLine() function
processLine $line
done
exec 0<&3
# restore $IFS which was used to determine what the field separators are
IFS=$BAKIFS
echo SUCCESS
But, there is a problem. Some logs contains stacktraces or something that uses more than one line, in other words, stacktrace is an example, it can be anything. Stacktrace example:
2011-02-03 12:51:45,220Z [ERROR] - File not found
java.io.FileNotFoundException: fred.txt
at java.io.FileInputStream.<init>(FileInputStream.java)
at java.io.FileInputStream.<init>(FileInputStream.java)
at ExTest.readMyFile(ExTest.java:19)
at ExTest.main(ExTest.java:7)
...
will not pass with my script, but is valid!
Then, if I run my script passing a log file with stacktraces for example, my script will failed, because it loops line by line.
I have the correct pattern and I need to validade the logger date format, but I don't have wrong date format pattern to skip lines.
I don't know how I can solve this problem. Does somebody can help me?
Thanks
You need to anchor your search for the date to the start of the line (otherwise the date could appear anywhere in the line - not just at the beginning).
The following snippet will loop over all lines that do not begin with a valid date. You still have to determine if the lines constitute errors or not.
DATEFMT='^[0-9]{4}-[0-9]{2}-[0-9]{2} [012][0-9]:[0-9]{2}:[0-9]{2},[0-9]{3}Z'
egrep -v ${DATEFMT} /path/to/log | while read LINE; do
echo ${LINE} # did not begin with date.
done
So just (silently) discard a single stack trace. In somewhat verbose bash:
STATE=idle
while read -r line; do
case $STATE in
idle)
if [[ $line =~ ^java\..*Exception ]]; then
STATE=readingexception
else
processLine "$line"
fi
;;
readingexception)
if ! [[ $line =~ ^' '*'at ' ]]; then
STATE=idle
processLine "$line"
fi
;;
*)
echo "Urk! internal error [$STATE]" >&2
exit 1
;;
esac
done <logfile
This relies on processLine not continuing on error, else you will need to track a tad more state to avoid two consecutive stack traces.
This makes 2 assumptions.
lines that begin with whitespace are continuations of previous lines. we're matching a leading space, or a leading tab.
lines that have non-whitespace characters starting at ^ are new log lines.
If a line matching #2 doesn't match the date format, we have an error, so print the error, and include the line number.
count=0
processLine() {
count=$(( count + 1 ))
line="$#"
result=$( echo $line | egrep '^[0-9]{4}-[0-9]{2}-[0-9]{2} [012][0-9]:[0-9]{2}:[0-9]{2},[0-9]{3}Z' -a -c )
if (( $result == 0 )); then
# if result = 0, then my line did not start with the proper date.
# if the line starts with whitespace, then it may be a continuation
# of a multi-line log entry (like a java stacktrace)
continues=$( echo $line | egrep "^ |^ " -a -c )
if (( $continues == 0 )); then
# if we got here, then the line did not start with a proper date,
# AND the line did not start with white space. This is a bad line.
echo "The line is not with correct date format: "
echo "$count: $line"
exit 1
fi
fi
}
Create a condition to check if the line starts with a date. If not, skip that line as it is part of a multi-line log.
processLine(){
# get all args
line="$#"
result=`echo $line | egrep "[0-9]{4}-[0-9]{2}-[0-9]{2} [012][0-9]:[0-9]{2}:[0-9]{2},[0-9]{3}Z" -a -c`
if [ "$result" == "0" ]; then
echo "Log entry is multi-lined - continuing."
fi
}