I have been running through my code for a while, and can't seem to find the reason this is failing because it is failing on line 10 apparently which is the if statement, but it is correctly finding the value of line.
#!/bin/bash
#a script that reads the largest number from a file
file="$1"
largest=""
while IFS= read -r line
do
if("$line" > "$largest")
then
"$largest"="$line"
fi
done <"$file"
echo "$largest"
This is incorrect:
if("$line" > "$largest")
then
"$largest"="$line"
fi
Change to:
if [ "$line" -gt "$largest" ]
then
largest="$line"
fi
First, as the pointed out in the comment, > is a redirection operator, and bash is trying to run the "$line" command. Parentheses are not test operators, the square brackets are.
Finally, the "$largest" is incorrect as the target of an assignment. The $ tells bash to provide the value of the variable, and we want to assign to largest, not to the VALUE of largest.
Related
This question already has answers here:
How to iterate over arguments in a Bash script
(9 answers)
Closed 5 years ago.
I have this code below:
#!/bin/bash
filename=$1
file_extension=$( echo $1 | cut -d. -f2 )
directory=${filename%.*}
if [[ -z $filename ]]; then
echo "You forgot to include the file name, like this:"
echo "./convert-pdf.sh my_document.pdf"
else
if [[ $file_extension = 'pdf' ]]; then
[[ ! -d $directory ]] && mkdir $directory
convert $filename -density 300 $directory/page_%04d.jpg
else
echo "ERROR! You must use ONLY PDF files!"
fi
fi
And it is working perfectly well!
I would like to create a script which I can do something like this: ./script.sh *.pdf
How can I do it? Using asterisk.
Thank you for your time!
Firstly realize that the shell will expand *.pdf to a list of arguments. This means that your shell script will never ever see the *. Instead it will get a list of arguments.
You can use a construction like the following:
#!/bin/bash
function convert() {
local filename=$1
# do your thing here
}
if (( $# < 1 )); then
# give your error message about missing arguments
fi
while (( $# > 0 )); do
convert "$1"
shift
done
What this does is first wrap your functionality in a function called convert. Then for the main code it first checks the number of arguments passed to the script, if this is less than 1 (i.e. none) you give the error that a filename should be passed. Then you go into a while loop which is executed as long as there are arguments remaining. The first argument you pass to the convert function which does what your script already does. Then the shift operation is performed, what this does is it throws away the first argument and then shifts all the remaining arguments "left" by one place, that is what was $2 now is $1, what was $3 now is $2, etc. By doing this in the while loop until the argument list is empty you go through all the arguments.
By the way, your initial assignments have a few issues:
you can't assume that the filename has an extension, your code could match a dot in some directory path instead.
your directory assignment seems to be splitting on . instead of /
your directory assignment will contain the filename if no absolute or relative path was given, i.e. only a bare filename
...
I think you should spend a bit more time on robustness
Wrap your code in a loop. That is, instead of:
filename=$1
: code goes here
use:
for filename in "$#"; do
: put your code here
done
I am trying to parse each line of a file and look for a particular string. The script seems to be doing its intended job, however, in parallel it tries to execute the if command on line 6:
#!/bin/bash
for line in $(cat $1)
do
echo $line | grep -e "Oct/2015"
if($?==0); then
echo "current line is: $line"
fi
done
and I get the following (my script is readlines.sh)
./readlines.sh: line 6: 0==0: command not found
First: As Mr. Llama says, you need more spaces. Right now your script tries to look for a file named something like /usr/bin/0==0 to run. Instead:
[ "$?" -eq 0 ] # POSIX-compliant numeric comparison
[ "$?" = 0 ] # POSIX-compliant string comparison
(( $? == 0 )) # bash-extended numeric comparison
Second: Don't test $? at all in this case. In fact, you don't even have good cause to use grep; the following is both more efficient (because it uses only functionality built into bash and requires no invocation of external commands) and more readable:
if [[ $line = *"Oct/2015"* ]]; then
echo "Current line is: $line"
fi
If you really do need to use grep, write it like so:
if echo "$line" | grep -q "Oct/2015"; then
echo "Current line is: $line"
fi
That way if operates directly on the pipeline's exit status, rather than running a second command testing $? and operating on that command's exit status.
#Charles Duffy has a good answer which I have up-voted as correct (and it is), but here's a detailed, line by line breakdown of your script and the correct thing to do for each part of it.
for line in $(cat $1)
As I noted in my comment elsewhere this should be done as a while read construct instead of a for cat construct.
This construct will wordsplit each line making spaces in the file separate "lines" in the output.
All empty lines will be skipped.
In addition when you cat $1 the variable should be quoted. If it is not quoted spaces and other less-usual characters appearing in the file name will cause the cat to fail and the loop will not process the file.
The complete line would read:
while IFS= read -r line
An illustrative example of the tradeoffs can be found here. The linked test script follows. I tried to include an indication of why IFS= and -r are important.
#!/bin/bash
mkdir -p /tmp/testcase
pushd /tmp/testcase >/dev/null
printf '%s\n' '' two 'three three' '' ' five with leading spaces' 'c:\some\dos\path' '' > testfile
printf '\nwc -l testfile:\n'
wc -l testfile
printf '\n\nfor line in $(cat) ... \n\n'
let n=1
for line in $(cat testfile) ; do
echo line $n: "$line"
let n++
done
printf '\n\nfor line in "$(cat)" ... \n\n'
let n=1
for line in "$(cat testfile)" ; do
echo line $n: "$line"
let n++
done
let n=1
printf '\n\nwhile read ... \n\n'
while read line ; do
echo line $n: "$line"
let n++
done < testfile
printf '\n\nwhile IFS= read ... \n\n'
let n=1
while IFS= read line ; do
echo line $n: "$line"
let n++
done < testfile
printf '\n\nwhile IFS= read -r ... \n\n'
let n=1
while IFS= read -r line ; do
echo line $n: "$line"
let n++
done < testfile
rm -- testfile
popd >/dev/null
rmdir /tmp/testcase
Note that this is a bash-heavy example. Other shells do not tend to support -r for read, for example, nor is let portable. On to the next line of your script.
do
As a matter of style I prefer do on the same line as the for or while declaration, but there's no convention on this.
echo $line | grep -e "Oct/2015"
The variable $line should be quoted here. In general, meaning always unless you specifically know better, you should double-quote all expansion--and that means subshells as well as variables. This insulates you from most unexpected shell weirdness.
You decclared your shell as bash which means you will have there "Here string" operator <<< available to you. When available it can be used to avoid the pipe; each element of a pipeline executes in a subshell, which incurs extra overhead and can lead to unexpected behavior if you try to modify variables. This would be written as
grep -e "Oct/2015" <<<"$line"
Note that I have quoted the line expansion.
You have called grep with -e, which is not incorrect but is needless since your pattern does not begin with -. In addition you have full-quoted a string in shell but you don't attempt to expand a variable or use other shell interpolation inside of it. When you don't expect and don't want the contents of a quoted string to be treated as special by the shell you should single quote them. Furthermore, your use of grep is inefficient: because your pattern is a fixed string and not a regular expression you could have used fgrep or grep -F, which does string contains rather than regular expression matching (and is far faster because of this). So this could be
grep -F 'Oct/2015' <<<"$line"
Without altering the behavior.
if($?==0); then
This is the source of your original problem. In shell scripts commands are separated by whitespace; when you say if($?==0) the $? expands, probably to 0, and bash will try to execute a command called if(0==0) which is a legal command name. What you wanted to do was invoke the if command and give it some parameters, which requires more whitespace. I believe others have covered this sufficiently.
You should never need to test the value of $? in a shell script. The if command exists for branching behavior based on the return code of whatever command you pass to it, so you can inline your grep call and have if check its return code directly, thus:
if grep -F 'Oct/2015` <<<"$line" ; then
Note the generous whitespace around the ; delimiter. I do this because in shell whitespace is usually required and can only sometiems be omitted. Rather than try to remember when you can do which I recommend an extra one space padding around everything. It's never wrong and can make other mistakes easier to notice.
As others have noted this grep will print matched lines to stdout, which is probably not something you want. If you are using GNU grep, which is standard on Linux, you will have the -q switch available to you. This will suppress the output from grep
if grep -q -F 'Oct/2015' <<<"$line" ; then
If you are trying to be strictly standards compliant or are in any environment with a grep that doesn't know -q the standard way to achieve this effect is to redirect stdout to /dev/null/
if printf "$line" | grep -F 'Oct/2015' >/dev/null ; then
In this example I also removed the here string bashism just to show a portable version of this line.
echo "current line is: $line"
There is nothing wrong with this line of your script, except that although echo is standard implementations vary to such an extent that it's not possible to absolutely rely on its behavior. You can use printf anywhere you would use echo and you can be fairly confident of what it will print. Even printf has some caveats: Some uncommon escape sequences are not evenly supported. See mascheck for details.
printf 'current line is: %s\n' "$line"
Note the explicit newline at the end; printf doesn't add one automatically.
fi
No comment on this line.
done
In the case where you did as I recommended and replaced the for line with a while read construct this line would change to:
done < "$1"
This directs the contents of the file in the $1 variable to the stdin of the while loop, which in turn passes the data to read.
In the interests of clarity I recommend copying the value from $1 into another variable first. That way when you read this line the purpose is more clear.
I hope no one takes great offense at the stylistic choices made above, which I have attempted to note; there are many ways to do this (but not a great many correct) ways.
Be sure to always run interesting snippets through the excellent shellcheck and explain shell when you run into difficulties like this in the future.
And finally, here's everything put together:
#!/bin/bash
input_file="$1"
while IFS= read -r line ; do
if grep -q -F 'Oct/2015' <<<"$line" ; then
printf 'current line is %s\n' "$line"
fi
done < "$input_file"
If you like one-liners, you may use AND operator (&&), for example:
echo "$line" | grep -e "Oct/2015" && echo "current line is: $line"
or:
grep -qe "Oct/2015" <<<"$line" && echo "current line is: $line"
Spacing is important in shell scripting.
Also, double-parens is for numerical comparison, not single-parens.
if (( $? == 0 )); then
I have this assignment to solve:
"Write a shell script that continuously reads words from the keyboard and
deletes them from all the files given in the command line."
I've tried to solve it, here's my attempt:
#!/bin/bash
echo "Enter words"
while (true)
do
read wrd
if [ "$wrd" != "exit" ]
then
for i in $#
do
sed -i -e 's/$wrd//g' $i
done
else
break
fi
done
This is the error that I receive after introducing the command: ./h84a.sh fisier1.txt
Enter words
suc
sed: can't read 1: No such file or directory
Sorry if I'm not very specific, it's my first time posting in here. I'm working in a terminal on Linux Mint which is installed on another partition of my PC. Please help me with my problem. Thanks!
I think you can simplify your script quite a lot:
#!/bin/bash
echo "Enter words"
while read -r wrd
do
[ "$wrd" = exit ] && break
sed -i "s/$wrd//g" "$#"
done
Some key changes:
The double quotes around the sed command are essential, as shell variables are not expanded within single quotes
Instead of using a loop, it is possible to pass all of the file names to sed at once, using "$#"
read -r is almost always what you want to use
I would suggest that you take care with in-place editing using the -i switch. In some versions of sed, you can specify the suffix of a backup file like -i.bak, so the original file is not lost.
In case you're not familiar with the syntax [ "$wrd" = exit ] && break, it is functionally equivalent to:
if [ "$wrd" = exit ]
then break
fi
$# expands to the number of arguments (so 1 in this case)
You probably meant to use $* or "$#"
I'm having trouble getting a script to do what I want.
I have a script that will search a file for a pattern and print the line numbers and instances of that pattern.
I want to know how to make it print the file name first before it prints the lines found
I also want to know how to write a new script that will call this one and pass two arguments to it.
The first argument being the pattern for grep and the second the location.
If the location is a directory, it will loop and search the pattern on all files in the directory using the script.
#!/bin/bash
if [[ $# -ne 2 ]]
then
echo "error: must provide 2 arguments."
exit -1
fi
if [[ ! -e $2 ]];
then
echo "error: second argument must be a file."
exit -2
fi
echo "------ File =" $2 "------"
grep -ne "$1" "$2"
This is the script i'm using that I need the new one to call. I just got a lot of help from asking a similar question but i'm still kind of lost. I know that I can use the -d command to test for the directory and then use 'for' to loop the command, but exactly how isn't panning out for me.
I think you just want to add the -H option to grep:
-H, --with-filename
Print the file name for each match. This is the default when there is more than one file to search.
grep has an option -r which can help you avoid testing for second argument being a directory and using for loop to iterate all files of that directory.
From the man page:
-R, -r, --recursive
Recursively search subdirectories listed.
It will also print the filename.
Test:
On one file:
[JS웃:~/Temp]$ grep -r '5' t
t:5 10 15
t:10 15 20
On a directory:
[JS웃:~/Temp]$ grep -r '5' perl/
perl//hello.pl:my $age=65;
perl//practice.pl:use v5.10;
perl//practice.pl:#array = (1,2,3,4,5);
perl//temp/person5.pm:#person5.pm
perl//temp/person9.pm: my #date = (localtime)[3,4,5];
perl//text.file:This is line 5
I have written a simple shell script to do some automation work. Basically the script searches for all the files in the current path and if the file is a specified one, it does some action.
Below are the relevant lines ---
#!/bin/bash
for i in `ls *`
do
if [$i =="ls.sh"]
then .... //do something
fi
done
However, the string comparision in line 3 is not working and I am getting this when I run the script --
./ls.sh: line 3: [scripth.sh: command not found
./ls.sh: line 3: [scripth.sh~: command not found
./ls.sh: line 3: [test.sh: command not found
What is the correction to be done ?
first of all, don't use ls like that. It will go bonkers if your files have spaces!.
Use shell expansion. Then, you can use case/esac to make string comparison. (or if/else)
for file in *
do
case "$file" in
"ls.sh" ) echo "do something"
;;
esac
done
There are several problems.
In line 1, you are not doing what you think you are. You should put a backquote around ls *:
for i in `ls *`
That will go through all files that list in the current directory. Your line will not run any command, but instead it will use * to get all files and your list will include a word "ls" at the front.
try this from a command line:
echo ls *
echo `ls *`
You might just want to do:
for i in *
Second problem. Put spaces inside your square brackets:
[ $i == "ls.sh" ]
The spaces are necessary.
Third problem. Use one = for string comparison
[ $i = "ls.sh" ]
Use: if [ "$i" = "ls.sh" ] - notice the spaces.
If you only want to check the existing of a file, you could do it directly in shell.
if [ -e ls.sh ]
then
# ... do something
fi
You have not included a space after ==, so your code should actually be:
#!/bin/bash
for i in `ls *`
do
if [ $i == "ls.sh" ]
then
//do something
fi
done