What's the difference between $# and $* in UNIX? When echoed in a script, they both seem to produce the same output.
Please see the bash man page under Special Parameters.
Special Parameters
The shell treats several parameters specially. These parameters may
only be referenced; assignment to them is not allowed.
* Expands to the positional parameters, starting from one. When
the expansion occurs within double quotes, it expands to a sin‐
gle word with the value of each parameter separated by the first
character of the IFS special variable. That is, "$*" is equiva‐
lent to "$1c$2c...", where c is the first character of the value
of the IFS variable. If IFS is unset, the parameters are sepa‐
rated by spaces. If IFS is null, the parameters are joined
without intervening separators.
# Expands to the positional parameters, starting from one. When
the expansion occurs within double quotes, each parameter
expands to a separate word. That is, "$#" is equivalent to "$1"
"$2" ... If the double-quoted expansion occurs within a word,
the expansion of the first parameter is joined with the begin‐
ning part of the original word, and the expansion of the last
parameter is joined with the last part of the original word.
When there are no positional parameters, "$#" and $# expand to
nothing (i.e., they are removed).
One difference is in how they handle the IFS variable on output.
#!/bin/sh
echo "unquoted asterisk " $*
echo "quoted asterisk $*"
echo "unquoted at " $#
echo "quoted at $#"
IFS="X"
echo "IFS is now $IFS"
echo "unquoted asterisk " $*
echo "quoted asterisk $*"
echo "unquoted at " $#
echo "quoted at $#"
If you run this like this: ./demo abc def ghi, you get this output:
unquoted asterisk abc def ghi
quoted asterisk abc def ghi
unquoted at abc def ghi
quoted at abc def ghi
IFS is now X
unquoted asterisk abc def ghi
quoted asterisk abcXdefXghi
unquoted at abc def ghi
quoted at abc def ghi
Notice that (only) the "quoted asterisk" line shows an X between each "word" after IFS is changed to "X". If the value of IFS contains multiple characters, only the first character is used for this purpose.
This feature can also be used for other arrays:
$ array=(123 456 789)
$ saveIFS=$IFS; IFS="|"
$ echo "${array[*]}"
123|456|789
$ IFS=$saveIFS
for i in "$#"
do
echo $i # loop $# times
done
for i in "$*"
do
echo $i # loop 1 times
done
It's safer to use "$#" instead of $*. When you use multiword strings as arguments to a shell script, it's only "$#" that interprets each quoted argument as a separate argument.
As the output above suggests, if you use $*, the shell makes a wrong count of the arguments.
Related
I have a log file and I would like to divide the result of one grep and count by another grep and count.
$ echo $((cat log2.txt | grep timed\|error\|Error | wc -l)/(cat log2.txt | grep Duration | wc -l))
zsh: bad math expression: operator expected at `log2.txt |...'
It's ugly, doesn't work and I can probably do it in a better way but I don't know how.
Also I would like to know if it possible to id incrementaly on a log stream read by tail for example.
First of all, you should know that, both grep|wc -l will count number of matched lines instead of occurrences, I hope this is what you really want.
Regarding your requirement, indeed, your approach is ugly (7 processes), apart from the mistakes. The job can be done by a single awk line:
awk '/timed|[Ee]rror/{a++}/Duration/{b++}END{printf "%.2f\n",a/b}' log2.txt
The above line calculates the result based on matched number of lines, same as your grep|wc -l.
You have several problems:
You are trying to run shell commands directly inside an arithmetic expression.
You aren't passing the correct regular expression to grep.
You need to make sure at least one of the operands is a floating-point value to trigger zsh's floating-point division.
Each pipeline can also be reduced to a single command; use input redirection instead of cat, and use the -c option to get the number of lines that match the regular expression.
echo $(( 1.0 * $(grep -c 'timed\|error\|Error' log2.txt) / $(grep -c Duration log2.txt))
Basic regular expressions treat unescaped | as a literal character, not an alteration operator.
$ echo foo | grep foo\|bar
$ echo foo | grep foo\\\|bar # Pass a literal backslash as part of the regex
foo
$ echo foo | grep 'foo\|bar' # Use '...' instead of explicitly escaping \ and |
foo
$ echo foo | grep -E 'foo|bar' # Use extended regular expressions instead
I am attempting to use the sqrt function from awk command in my script, but all it returns is 0. Is there anything wrong with my script below?
echo "enter number"
read root
awk 'BEGIN{ print sqrt($root) }'
This is my first time using the awk command, are there any mistakes that I am not understanding here?
Maybe you can try this.
echo "enter number"
read root
echo "$root" | awk '{print sqrt($0)}'
You have to give a data input to awk. So, you can pipe 'echo'.
The BEGIN statement is to do things, like print a header...etc before
awk starts reading the input.
$ echo "enter number"
enter number
$ read root
3
$ awk -v root="$root" 'BEGIN{ print sqrt(root) }'
1.73205
See the comp.unix.shell FAQ for the 2 correct ways to pass the value of a shell variable to an awk script.
UPDATE : My proposed solution turns out to be potentially dangerous. See Ed Morton's answer for a better solution. I'll leave this answer here as a warning.
Because of the single quotes, $root is interpreted by awk, not by the shell. awk treats root as an uninitialized variable, whose value is the empty string, treated as 0 in a numeric context. $root is the root'th field of the current line -- in this case, as $0, which is the entire line. Since it's in a BEGIN block, there is no current line, so $root is the empty string -- which again is treated as 0 when passed to sqrt().
You can see this by changing your command line a bit:
$ awk 'BEGIN { print sqrt("") }'
0
$ echo 2 | awk '{ print sqrt($root) }'
1.41421
NOTE: The above is merely to show what's wrong with the original command, and how it's interpreted by the shell and by awk.
One solution is to use double quotes rather than single quotes. The shell expands variable references within double quotes:
$ echo "enter number"
enter number
$ read x
2
$ awk "BEGIN { print sqrt($x) }" # DANGEROUS
1.41421
You'll need to be careful when doing this kind of thing. The interaction between quoting and variable expansion in the shell vs. awk can be complicated.
UPDATE: In fact, you need to be extremely careful. As Ed Morton points out in a comment, this method can result in arbitrary code execution given a malicious value for $x, which is always a risk for a value read from user input. His answer avoids that problem.
(Note that I've changed the name of your shell variable from $root to $x, since it's the number whose square root you want, not the root itself.)
Not a long question, what does this mean?
LogMsg "File:${#}"
LogMsg() is a method that logs a message with a timestamp.
But what the heck does
${#}
mean? I should also mention the script also has $1 and $2 as well. Google produces no results.
Literally:
f() { printf '%s\n' "File: $#"; }
f "First Argument" "Second Argument" "Third Argument"
will expand to and run the command:
printf '%s\n' "File: First Argument" "Second Argument" "Third Argument"
That is to say: It expands your argument list ($1, $2, $3, etc) while maintaining separation between subsequent arguments (not throwing away any information provided by the user by way of quoting).
This is different from:
printf '%s\n' File: $#
or
printf '%s\n' File: $*
which are both the same as:
printf '%s\n' "File:" "First" "Argument" "Second" "Argument" "Third" "Argument"
...these both string-split and glob-expand the argument list, so if the user had passed, say, "*" (inside quotes intended to make it literal), the unquoted use here would replace that character with the results of expanding it as a glob, ie. the list of files in the current directory. Also, string-splitting has other side effects such as changing newlines or tabs to spaces.
It is also different from:
printf '%s\n' "File: $*"
which is the same as:
printf '%s\n' "File: First Argument Second Argument Third Argument"
...which, as you can see above, combines all arguments by putting the first character in IFS (which is by default a space) between them.
in KSH there is two positional paremeters * and #
"$*" is a single string that consists of all of the positional parameters, separated by the first character in the variable IFS (internal field separator), which is a space, TAB, and newline by default.
On the other hand, "$#" is equal to "$1" "$2" … "$N ", where N is the number of positional parameters.
For more detailed information and example : http://oreilly.com/catalog/korn2/chapter/ch04.html
This is the set of the arguments of the command line.
If you launch a script via a command like cmd a b c d, there is 5 arguments, $0 will be the command cmd, $1the first argument a, $2 the second b, etc. ${#} will be all the arguments except the command.
The one piece that was not explained by the other posts is the use of {. $# is the same as ${#} but allows you to add letters, etc if needed and those letters will not have a space added in. e.g. you could say ${foo}dog and if $foo was set to little the result would be littledog with no spaces. In the case of ${#}dogdog and $# is set to a b c d the result is "a" "b" "c" "ddogdog".
Goal
In ZSH script, for a given args, I want to obtain the first string and the rest.
For instance, when the script is named test
sh test hello
supposed to extract h and ello.
ZSH manual
http://zsh.sourceforge.net/Doc/zsh_a4.pdf
says:
Subscripting may also be performed on non-array values, in which case the subscripts specify a
substring to be extracted. For example, if FOO is set to ‘foobar’, then ‘echo $FOO[2,5]’ prints
‘ooba’.
Q1
So, I wrote a shell script in a file named test
echo $1
echo $1[1,1]
terminal:
$ sh test hello
hello
hello[1,1]
the result fails. What's wrong with the code?
Q2
Also I don't know how to extract subString from n to the last. Perhaps do I have to use Array split by regex?
EDIT: Q3
This may be another question, so if it's proper to start new Thread, I will do so.
Thanks to #skishore Here is the further code
#! /bin/zsh
echo $1
ARG_FIRST=`echo $1 | cut -c1`
ARG_REST=`echo $1 | cut -c2-`
echo ARG_FIRST=$ARG_FIRST
echo ARG_REST=$ARG_REST
if $ARG_FIRST = ""; then
echo nullArgs
else
if $ARG_FIRST = "#"; then
echo #Args
else
echo regularArgs
fi
fi
I'm not sure how to compare string valuables to string, but for a given args hello
result:
command not found: h
What's wrong with the code?
EDIT2:
What I've found right
#! /bin/zsh
echo $1
ARG_FIRST=`echo $1 | cut -c1`
ARG_REST=`echo $1 | cut -c2-`
echo ARG_FIRST=$ARG_FIRST
echo ARG_REST=$ARG_REST
if [ $ARG_FIRST ]; then
if [ $ARG_FIRST = "#" ]; then
echo #Args
else
echo regularArgs
fi
else
echo nullArgs
fi
EDIT3:
As the result of whole, this is what I've done with this question.
https://github.com/kenokabe/GitSnapShot
GitSnapShot is a ZSH thin wrapper for Git commands for easier and simpler usage
A1
As others have said, you need to wrap it in curly braces. Also, use a command interpreter (#!...), mark the file as executable, and call it directly.
#!/bin/zsh
echo $1
echo ${1[1,1]}
A2
The easiest way to extract a substring from a parameter (zsh calls variables parameters) is to use parameter expansion. Using the square brackets tells zsh to treat the scalar (i.e. string) parameter as an array. For a single character, this makes sense. For the rest of the string, you can use the simpler ${parameter:start:length} notation instead. If you omit the :length part (as we will here), then it will give you the rest of the scalar.
File test:
#!/bin/zsh
echo ${1[1]}
echo ${1:1}
Terminal:
$ ./test Hello
H
ello
A3
As others have said, you need (preferably double) square brackets to test. Also, to test if a string is NULL use -z, and to test if it is not NULL use -n. You can just put a string in double brackets ([[ ... ]]), but it is preferable to make your intentions clear with -n.
if [[ -z "${ARG_FIRST}" ]]; then
...
fi
Also remove the space between #! and /bin/zsh.
And if you are checking for equality, use ==; if you are assigning a value, use =.
RE:EDIT2:
Declare all parameters to set the scope. If you do not, you may clobber or use a parameter inherited from the shell, which may cause unexpected behavior. Google's shell style guide is a good resource for stuff like this.
Use builtins over external commands.
Avoid backticks. Use $(...) instead.
Use single quotes when quoting a literal string. This prevents pattern matching.
Make use of elif or case to avoid nested ifs. case will be easier to read in your example here, but elif will probably be better for your actual code.
Using case:
#!/bin/zsh
typeset ARG_FIRST="${1[1]}"
typeset ARG_REST="${1:1}"
echo $1
echo 'ARG_FIRST='"${ARG_FIRST}"
echo 'ARG_REST='"${ARG_REST}"
case "${ARG_FIRST}" in
('') echo 'nullArgs' ;;
('#') echo '#Args' ;;
(*)
# Recommended formatting example with more than 1 sloc
echo 'regularArgs'
;;
esac
using elif:
#!/bin/zsh
typeset ARG_FIRST="${1[1]}"
typeset ARG_REST="${1:1}"
echo $1
echo 'ARG_FIRST='"${ARG_FIRST}"
echo 'ARG_REST='"${ARG_REST}"
if [[ -z "${ARG_FIRST}" ]]; then
echo nullArgs
elif [[ '#' == "${ARG_FIRST}" ]]; then
echo #Args
else
echo regularArgs
fi
RE:EDIT3
Use "$#" unless you really know what you are doing. Explanation.
You can use the cut command:
echo $1 | cut -c1
echo $1 | cut -c2-
Use $() to assign these values to variables:
ARG_FIRST=$(echo $1 | cut -c1)
ARG_REST=$(echo $1 | cut -c2-)
echo ARG_FIRST=$ARG_FIRST
echo ARG_REST=$ARG_REST
You can also replace $() with backticks, but the former is recommended and the latter is somewhat deprecated due to nesting issues.
So, I wrote a shell script in a file named test
$ sh test hello
This isn't a zsh script: you're calling it with sh, which is (almost certainly) bash. If you've got the shebang (#!/bin/zsh), you can make it executable (chmod +x <script>) and run it: ./script. Alternatively, you can run it with zsh <script>.
the result fails. What's wrong with the code?
You can wrap in braces:
echo ${1} # This'll work with or without the braces.
echo ${1[3,5]} # This works in the braces.
echo $1[3,5] # This doesn't work.
Running this: ./test-script hello gives:
./test-script.zsh hello
hello
llo
./test-script.zsh:5: no matches found: hello[3,5]
Also I don't know how to extract subString from n to the last. Perhaps do I have to use Array split by regex?
Use the [n,last] notation, but wrap in braces. We can determine how long our variable is with, then use the length:
# Store the length of $1 in LENGTH.
LENGTH=${#1}
echo ${1[2,${LENGTH}]} # Display from `2` to `LENGTH`.
This'll produce ello (prints from the 2nd to the last character of hello).
Script to play with:
#!/usr/local/bin/zsh
echo ${1} # Print the input
echo ${1[3,5]} # Print from 3rd->5th characters of input
LENGTH=${#1}
echo ${1[2,${LENGTH}]} # Print from 2nd -> last characters of input.
You can use the cut command:
But that would be using extra baggage - zsh is quite capable of doing all this on it's own without spawning multiple sub-shells for simplistic operations.
I have a batch file like this:
java temptable %1 %2
I need the equivalent shell script for the above. I will pass the arguments to the shell script and that should be passed to temptable.
For bash (which is one shell, but probably the most common in the Linux world), the equivalent is:
java temptable $1 $2
assuming there's no spaces in the arguments. If there are spaces, you should quote your arguments:
java temptable "$1" "$2"
You can also do:
java temptable $*
or:
java temptable "$#"
if you want all parameters passed through (again, that second one is equivalent to quoting each of the parameters: "$1" "$2" "$3" ...).
#!/bin/bash
# Call this script with at least 3 parameters, for example
# sh scriptname 1 2 3
echo "first parameter is $1"
echo "Second parameter is $2"
echo "Third parameter is $3"
exit 0
Output:
[root#localhost ~]# sh parameters.sh 47 9 34
first parameter is 47
Second parameter is 9
Third parameter is 34