Currently, I am studying crond on Centos7 and want to take a close look at run-parts.sh. But I find some strange scripts such as:
"${i%,v}"
What does it mean? I know ${i%%.\*}, ${i##.*}, ${i,}, and ${i,,}.
Here is the part script, thanks every one in advance.
for i in $(LC_ALL=C;echo ${1%/}/*[^~,]) ; do
[ -d $i ] && continue
# Don't run *.{rpmsave,rpmorig,rpmnew,swp,cfsaved} scripts
[ "${i%.cfsaved}" != "${i}" ] && continue
[ "${i%.rpmsave}" != "${i}" ] && continue
[ "${i%.rpmorig}" != "${i}" ] && continue
[ "${i%.rpmnew}" != "${i}" ] && continue
[ "${i%.swp}" != "${i}" ] && continue
[ "${i%,v}" != "${i}" ] && continue
The syntax ${var%pattern} just expands to the value of $var with the glob pattern matcing 'pattern' removed from the end of the string. So ${i%,v} is just $i with the trailing ,v removed. The only difference between ${i%%,v} and ${i%,v} is that the former will match the longest possible match, but that's irrelevant here since ,v can only expand to the literal string ,v. It's stated best in the documentation (http://pubs.opengroup.org/onlinepubs/9699919799/):
${parameter%[word]}
Remove Smallest Suffix Pattern. The word shall be expanded to produce apattern. The parameter expansion shall then result in parameter, with the smallest portion of the suffix matched by the pattern deleted. If present, word shall not begin with an unquoted '%'.
${parameter%%[word]}
Remove Largest Suffix Pattern. The word shall be expanded to produce a pattern. The parameter expansion shall then result in parameter, with the largest portion of the suffix matched by the pattern deleted
Related
I need help to find the appropriate way to write my conditional statement because it is not working.
cumonth= date +%m
cudinamic=
if [ $cumonth =10 ];
then
cudinamic=a
elif [ $cumonth =11];
then
cudinamic=b
elif [ $cumonth =12];
then
cudinamic=c
else
cudinamic=$cumonth
fi
#Echo display message
$echo $ytday
$echo $cmonth
echo "$cudinamic"
To save the output of a command, use command substitution $(...):
cumonth=$(date +%m)
The arguments to [ must be separated by spaces, spaces aren't optional:
elif [ "$cumonth" = 11 ];
Note that there are 4 parameters: "$cumonth", =, 11, and ]. It's a good habit to quote the variable, in case it's empty or contains spaces, it will still be considered a single word.
my#comp:~/wtfdir$ cat wtf.sh
str1=$(echo "")
str2=$(echo "")
if [ $str1 != $str2 ]; then
echo "WTF?!?!"
fi
my#comp:~/wtfdir$ ./wtf.sh
WTF?!?!
my#comp:~/wtfdir$
WTF is going on here?!
How I wrote the above code: Googling "bash compare strings" brought me to this website which says:
You can check the equality and inequality of two strings in bash by using if statement. “==” is used to check equality and “!=” is used to check inequality of the strings.
Yet I'm getting the above?
What am I not understanding? What am I doing wrong?
You aren't running a comparison at all, because you aren't using quotes where they're mandatory. See the warning from http://shellcheck.net/ about unquoted expansions at SC2086.
If both string are empty, then:
[ $str1 != $str2 ]
...evaluates to...
[ != ]
...which is a test for whether the string != is nonempty, which is true. Change your code to:
[ "$str1" != "$str2" ]
...and the exact values of those strings will actually be passed through to the [ command.
Another alternative is using [[; as described in BashFAQ #31 and the conditional expression page on the bash-hackers' wiki, this is extended shell syntax (in ksh, bash, and other common shells extending the POSIX sh standard) which suppresses the string-splitting behavior that's tripping you up:
[[ $str1 != "$str2" ]]
...requires quotes only on the right-hand side, and even those aren't needed for the empty-string case, but to prevent that right-hand side from being treated as a glob (causing the comparison to always reflect a match if str2='*').
I'm writing bash scripts that need to work both on Linux and on Mac.
I'm writing a function that will return a directory path depending on which environment I'm in.
Here is the pseudo code:
If I'm on a Mac OS X machine, I need my function to return the path:
/usr/local/share/
Else if I'm on a Linux machine, I need my function to return the path:
/home/share/
Else, you are neither on a Linux or a Mac...sorry.
I'm very new to Bash, so I apologize in advance for the really simple question.
Below is the function I have written. Whether I'm on a Mac or Linux, it always returns
/usr/local/share/
Please take a look and enlighten me with the subtleties of Bash.
function get_path(){
os_type=`uname`
if [ $os_type=="Darwin" ]; then
path="/usr/local/share/"
elif [ $os_type=="Linux" ]; then
path="/home/share/"
else
echo "${os_type} is not supported"
exit 1
fi
echo $path
}
You need spaces around the operator in a test command: [ $os_type == "Darwin" ] instead of [ $os_type=="Darwin" ]. Actually, you should also use = instead of == (the double-equal is a bashism, and will not work in all shells). Also, the function keyword is also nonstandard, you should leave it off. Also, you should double-quote variable references (like "$os_type") just in case they contain spaces or any other funny characters. Finally, echoing an error message ("...not supported") to standard output may confuse whatever's calling the function, because it'll appear where it expected to find a path; redirect it to standard error (>&2) instead. Here's what I get with these cleaned up:
get_path(){
os_type=`uname`
if [ "$os_type" = "Darwin" ]; then
path="/usr/local/share/"
elif [ "$os_type" = "Linux" ]; then
path="/home/share/"
else
echo "${os_type} is not supported" >&2
exit 1
fi
echo "$path"
}
EDIT: My explanation of the difference between assignments and comparisons got too long for a comment, so I'm adding it here. In many languages, there's a standard expression syntax that'll be the same when it's used independently vs. in test. For example, in C a = b does the same thing whether it's alone on a line, or in a context like if ( a = b ). The shell isn't like that -- its syntax and semantics vary wildly depending on the exact context, and it's the context (not the number of equal signs) that determines the meaning. Here are some examples:
a=b by itself is an assignment
a = b by itself will run a as a command, and pass it the arguments "=" and "b".
[ a = b ] runs the [ command (which is a synonym for the test command) with the arguments "a", "=", "b", and "]" -- it ignores the "]", and parses the others as a comparison expression.
[ a=b ] also runs the [ (test) command, but this time after removing the "]" it only sees a single argument, "a=b" -- and when test is given a single argument it returns true if the argument isn't blank, which this one isn't.
bash's builtin version of [ (test) accepts == as a synonym for =, but not all other versions do.
BTW, just to make things more complicated bash also has [[ ]] expressions (like test, but cleaner and more powerful) and (( )) expressions (which are totally different from everything else), and even ( ) (which runs its contents as a command, but in a subshell).
You need to understand what [ means. Originally, this was a synonym for the /bin/test command. These are identical:
if test -z "$foo"
then
echo "String '$foo' is null."
fi
if [ -z "$foo" ]
then
echo "String '$foo' is null."
fi
Now, you can see why spaces are needed for all of the parameters. These are parameters and not merely boolean expressions. In fact, the test manpage is a great place to learn about the various tests. (Note: The test and [ are built in commands to the BASH shell.)
if [ $os_type=="Darwin" ]
then
This should be three parameters:
"$os_type"
= and not ==
"Darwin"
if [ "$os_type" = "Darwin" ] # Three parameters to the [ command
then
If you use single square brackets, you should be in the habit to surround your parameters with quotation marks. Otherwise, you will run into trouble:
foo="The value of FOO"
bar="The value of BAR"
if [ $foo != $bar ] #This won't work
then
...
In the above, the shell will interpolate $foo and $bar with their values before evaluating the expressions. You'll get:
if [ The value of FOO != The value of BAR ]
The [ will look at this and realize that neither The or value are correct parameters, and will complain. Using quotes will prevent this:
if [ "$foo" != "$bar" ] #This will work
then
This becomes:
if [ "The value of FOO" != "The value of BAR" ]
This is why it's highly recommended that you use double square brackets for your tests: [[ ... ]]. The test looks at the parameters before the shell interpolates them:
if [[ $foo = $bar ]] #This will work even without quotation marks
Also, the [[ ... ]] allows for pattern matching:
if [[ $os_type = D* ]] # Single equals is supported
then
path="/usr/local/share/"
elif [[ $os_type == L* ]] # Double equals is also supported
then
path="/home/share/"
else
echo "${os_type} is not supported"
exit 1
fi
This way, if the string is Darwin32 or Darwin64, the if statement still functions. Again, notice that there has to be white spaces around everything because these are parameters to a command (actually, not anymore, but that's the way the shell parses them).
Adding spaces between the arguments for the conditionals fixed the problem.
This works
function get_path(){
os_type=`uname`
if [ $os_type == "Darwin" ]; then
path="/usr/local/share/"
elif [ $os_type == "Linux" ]; then
path="/home/share/"
else
echo "${os_type} is not supported"
exit 1
fi
echo $path
}
I'm having trouble getting grep to work properly in an if statement. In the following code segment, the if-check always comes up true (i.e. the word is not found), and the program prints NOT FOUND, even though the words are already in ~/.memory.
for (( i=0; i<${#aspellwords[*]}; i++)); do
if [ !$(grep -q "${aspellwords[$i]}" ~/.memory) ]; then
words[$i]="${aspellwords[$i]}"
printf "\nNOT FOUND\n"
fi
done
However, when I test the following code in place of the previous segment:
for (( i=0; i<${#aspellwords[*]}; i++)); do
if grep -q "${aspellwords[$i]}" ~/.memory; then echo FOUND IT; fi
done
It works perfectly fine and finds the word without any issues.
So what's wrong with the first segment of code?
A number of things are wrong with that first snippet.
You don't want [ ... ] if you want to test the return code. Drop those.
[] is not part of the if syntax (as you can see from your second snippet).
[ is a shell built-in and binary on your system. It just exits with a return code. if ...; then tests the return code of ....
$() is command substitution. It replaces itself with the output from the command that was run.
So [ !$(grep ...) ] is actually evaluating [ !output_from_grep ] and [ word ] is interpreted as [ -n word ] which will be true whenever word is non-empty. Given that ! is never non-empty that will always be true.
Simply, as indicated by #thom in his comment (a bit obliquely), add the ! negation to your second snippet with a space between it and grep.
So I have some code where I'm trying to only do the while loop if the first character of a string isn't or is a certain character.
while IFS=$'\t' read -r -a myArray
do
like=${myArray[0]}
position=${myArray[1]}
while [ ${like:0:1}=="E" ]
file=$like."Rput"
echo "$file"
So when I echo the file, the file name is ##file.output which is a file that I do not want at all. In the sense, I want it to completely skip it.
Could someone tell me what's going on?
Thanks!
It works if while is replaced with an if-then statement:
while IFS=$'\t' read -r -a myArray
do
like=${myArray[0]}
position=${myArray[1]}
if [ "${like:0:1}" == "E" ]
then
file=$like."Rput"
echo "$file"
fi
done
Note also that spaces are important. The following tests first character of like for equality with E:
[ "${like:0:1}" == "E" ]
But, the following does something unrelated:
[ "${like:0:1}"=="E" ]
Since the equal sign here is not separated by spaces, "${like:0:1}"=="E" is interpreted simply as a single string. Tests of a single string return true if the string is nonempty and false if it is empty. Since ${like:0:1}"=="E" is always non-empty, it will always return true.