I have a script when you select a desktop file, but when I run this case function:
File=$(yad --file);
if [[ "$File" =~ *".desktop" ]]; then
echo "yes"
else
echo "no"
if
and i try this :
File=$(yad --file);
case $File in
*.desktop )
echo "yes"
;;
* )
echo "no"
;;
esac
it's always telling me that I have to try again I don't know what's the problem, can anyone help me?
I am not exactly sure what this script is supposed to do, but try this:
File="$(Yad --file)"
if [[ "$File" =~ .*[.]desktop$ ]]; then
echo "yes"
else
echo "no"
fi
Bash regular expression matching (=~) uses extended regular expressions, not glob expressions. To designate any sequence of zero or more characters, you need to use .*. The . means "any character", and the * means zero or more times. [.] designates a literal period, avoiding the "any character" meaning of . used alone. I also added an end-of-line anchor ($). This forces the pattern to match from the end of the filename, as you probably would want when matching with the extension.
There also is an error in your first line. There has to be no space between the $ sign and parentheses. And to close an if block, you need to use fi.
You can use glob-style matching with bash conditionals, just use an equal sign :
if [[ "$File" = *.desktop ]]; then
Related
I'm relatively new here and to the coding world. I'm currently taking a class in Shell Scripting and I'm a bit stuck.
I'm trying to do a little extra credit and get the script to check for command line arguments and if none or only 1 is given, prompt the user to input the missing values.
For the most part I've been able to get most of it to work except for when it comes to the numeric check part. I'm not completely sure that I am doing the nested if statements correctly because it's displaying both the "if" echo and the "else" echo.
My script so far:
q=y
# Begins loop
until [[ $q == n ]];do
# Checks command line arguments
if [[ $# -lt 2 ]];then
# Asks for second number if only 1 argument.
if [[ $# == 1 ]];then
read -r -p "Please enter your second number: " y
if [[ y =~ [1-9] ]];then
echo "You've chosen $1 as your first number and $y as your second number."
break
else
echo "This is not a valid value, please try again."
fi
# Asks for both numbers if no arguments.
else
read -r -p "Please enter your first number: " x
if [[ x =~ [1-9] ]];then
break
else
echo "This is not a valid value, please try again."
fi
read -r -p "Please enter your second number: " y
if [[ y =~ [1-9] ]];then
break
else
echo "This is not a valid value, please try again."
fi
echo "You've chosen $x as your first number and $y as your second number."
fi
# If both command line arguments are provided, echo's arguments, and sets arguments as x and y values.
else
echo "You've chosen $1 as your first number and $2 as your second number."
x=$1
y=$2
fi
read -r -p "Would you like to try again? (n to exit): " q
done
When I run it I get this for output:
Please enter your first number: 1
This is not a valid value, please try again.
Please enter your second number: 2
This is not a valid value, please try again.
You've chosen 1 as your first number and 2 as your second number.
Please enter your first number:
And will just continue to loop without breaking. Any help/guidance would be greatly appreciated, thank you.
In your expression:
if [[ x =~ [1-9] ]]; then
You are actually comparing the string literal "x" with the regex. What you want is the variable:
if [[ $x =~ [1-9] ]]; then
This will interpolate the variable first in order to compare the variable's value with the regex. I think this change also applies to some of the other comparison expressions in your code.
However, as glenn jackman and user1934428 have commented, this will also match things like foo1bar, which is probably not what you want. To fix this, you can add start/end matchers to your regex. Finally, you may want to match even if the input has leading or trailing spaces. One way to do this is to add some [[:space:]]*'s to match zero or more spaces around your [1-9]:
if [[ $x =~ ^[[:space:]]*[1-9][[:space:]]*$ ]]; then
So, to break down the regex:
^ start of input
[[:space:]]* zero or more whitespaces
[1-9] a single digit, 1-9
[[:space:]]* zero or more whitespaces
$ end of the input
I'm assuming from your question than you only want to match on a single digit, not, for example, 12, or the digit 0. To match those would require a couple more regex tweaks.
and...glob pattern
Just because glen jackman's answer led me down a bash man page adventure 🏄 and I wanted to try them out, this is a glob pattern version (note the == instead of =~):
if [[ $x == *([[:space:]])[1-9]*([[:space:]]) ]]; then
It's basically the same pattern. But notably, glob patterns seem to be implicitly anchored to the start/end of the string being matched (they are tested against the entire string) so they don't need the ^ or $, while regular expressions match against substrings by default, so they do need those additions to avoid foo1bar matching. Anyway, probably more than you cared to know.
Here's an alternate implementation, for your consideration: hit me up with any questions
#!/usr/bin/env bash
get_number() {
local n
while true; do
read -rp "Enter a number between 1 and 9: " n
if [[ $n == [1-9] ]]; then
echo "$n"
return
fi
done
}
case $# in
0) first=$(get_number)
second=$(get_number)
;;
1) first=$1
second=$(get_number)
;;
*) first=$1
second=$2
;;
esac
# or, more compact but harder to grok
[[ -z ${first:=$1} ]] && first=$(get_number)
[[ -z ${second:=$2} ]] && second=$(get_number)
echo "You've chosen $first as your first number and $second as your second number."
This uses:
a function to get a a number from the user, so you don't have so much duplicated code,
a case statement to switch over the $# variable
input validation with the == operator within [[...]] -- this operator is a pattern matching operator, not string equality (unless the right-hand operand is quoted)
Note that [[ $x =~ [1-9] ]] means: "$x contains a character in the range 1 to 9" -- it does not mean that the variable is a single digit. If x=foo1bar, then the regex test passes.
I want to write a bash script to read an input from user and check whether the input matches a single character. Below is the code.
read letter
if [[ $letter =~ ([a-zA-Z]) ]]
then
echo correct
fi
When I input two characters it still says "correct", see below output:
$ sh tmp.sh
aa
correct
how to write regular expression to matches exactly one character?
Your regex is not using anchors hence it will match any input that has [a-zA-Z] but it will match anything that is greater than 1 in length.
Having said that you don't even need regex and can use glob pattern here which will match exactly one alphabet:
if [[ $letter == [a-zA-Z] ]]
then
echo 'correct'
fi
If you must use regex then use:
if [[ $letter =~ ^[a-zA-Z]$ ]]
then
echo 'correct'
fi
Given that you probably want to take different actions depending on the value, you may want to consider a simple case statement:
case $letter in [a-zA-Z]) echo correct;; *) : do nothing;; esac
This will make it easy change to things like
case $letter in
a) : do something for a;;
[b-m]) : do something else;;
[n-zA-Z) : ... ;;
* ) ... ;;
esac
I'm looking at a simple for loop with the following logic:
variable=`some piped string`
array_value=(1.1 2.9)
for i in ${array_value[#]}; do
if [[ "$variable" == *some_text*"$array_value" ]]; then
echo -e "Info: Found a matching string"
fi
The problem is that I cannot get this to show me when it finds either the string ending in 1.1 or 2.9 as sample data.
If I do an echo $array_value in the for loop I can see that the array values are being taken so its values are being parsed, though the if loop doesn't return that echo message although the string is present.
LE:
Based on the comments received I've abstracted the code to something like this, which still doesn't work if I want to use wildcards inside the comparison quote
versions=(1.1 2.9)
string="system is running version:2.9"
for i in ${versions[#]}; do
if [[ "$string" == "system*${i}" ]]; then
echo "match found"
fi
done
Any construction similar to "system* ${i}" or "* ${i}" will not work, though if I specify the full string pattern it will work.
The problem with the test construct has to you with your if statement. To construct the if statement in a form that will evaluate, use:
if [[ "$variable" == "*some_text*${i}" ]]; then
Note: *some_text* will need to be replaced with actual text without * wildcards. If the * is needed in the text, then you will need to turn globbing off to prevent expansion by the shell. If expansion is your goal, then protect the variable i by braces.
There is nothing wrong with putting *some_text* up against the variable i, but it is cleaner, depending on the length of some_text, to assign it to a variable itself. The easiest way to accommodate this would be to define a variable to hold the some_text you are needing. E.g.:
prefix="some_text"
if [[ "$variable" == "${prefix}${i}" ]]; then
If you have additional questions, just ask.
Change "system*${i}" to system*$i.
Wrapping with quotes inside [[ ... ]] nullifies the wildcard * by treating it as a literal character.
Or if you want the match to be assigned to a variable:
match="system*"
you can then do:
[[ $string == $match$i ]]
You actually don't need quotes around $string either as word splitting is not performed inside [[ ... ]].
From man bash:
[[ expression ]]
...
Word splitting and pathname expansion are not
performed on the words between the [[ and ]]
...
Any part of the pattern may be quoted to force
the quoted portion to be matched as a string.
I'm trying to match any strings that start with /John/ but does not contain / after /John/
if
[ $string == /John/[!/]+ ]; then ....
fi
This is what I got and it doesn't seem to be working.
So I tried
if
[[ $string =~ ^/John/[!/]+$ ]]; then ....
fi
It still didn't work, and so I changed it to
if
[[ $string =~ /John/[^/] ]]; then ....
fi
It worked but will match with all the strings that has / behind /John/ too.
For bash you want [[ $string =~ /John/[^/]*$ ]] -- the end-of-line anchor ensures there are no slashes after the last acceptable slash.
How about "the string starts with '/John/' and doesn't contain any slashes after '/John/'"?
[[ $string = /John/* && $string != /John/*/* ]]
Or you could compare against a parameter expansion that only expands if the conditions are met. This says "after stripping off everything including and after the last slash, the string is /John":
[[ ${string%/*} = /John ]]
In fact, this last solution is the only entirely POSIXLY_STRICT one I can come up with without multiple test expressions.
[ "${string%/*}" = /John ]
By the way, your problem is probably simply be using double-equals inside a single-bracket test expression. bash actually does accept them inside double-bracket test expressions, but a single equals is a better idea.
You can also use plain old grep:
string='/John Lennon/Yoko Ono'
if echo "$string" | grep -q "/John[^/]" ; then
echo "matched"
else
echo "no match found"
fi
This only fails if /John is at the very end of the string... if that's a possibility then you can tweak to handle that case, for instance:
string='/John Lennon/Yoko Ono'
if echo "$string" | grep -qP "(/John[^/])|(/John$)" ; then
echo "matched"
else
echo "no match found"
fi
Not sure what language you're using, but normal negative character classes are prefixed with a ^
e.g.
[^/]
You can also put in start/end qualifiers (clojure example, so Java's regex engine). Usually ^ at beginning and $ at end.
user => (re-matches #"^/[a-zA-Z]+[^/]$" "/John/")
nil
I coded the below code and if no -l or -L option is passes to the script I need to assume (detect) whether a filename was passed as a param. The below third condition only matches if filename is one lowercase character. How can I make it flexible to match upper and lower case letters or variable length of the string?
while [ $# -gt 0 ]
do
case $1 in
-l) echo ls;;
-L) echo ls -l;;
[a-z]) echo filename;;
*) echo usage
exit 1;;
esac
shift
done
Also, how can I include a condition in that case statement that would react to empty $1?
If for example the script is called without any options or filenames.
You can match an empty string with a '') or "") case.
A file name can contain any character--even weird ones likes symbols, spaces, newlines, and control characters--so trying to figure out if you have a file name by looking for letters and numbers isn't the right way to do it. Instead you can use the [ -e filename ] test to check if a string is a valid file name.
You should, by the way, put "$1" in double quotes so your script will work if the file name does contain spaces.
case "$1" in
'') echo empty;;
-l) echo ls;;
-L) echo ls -l;;
*) if [ -e "$1" ]; then
echo filename
else
echo usage >&2 # echo to stderr
exit 1
fi;;
esac
Use getopts to parse options, then treat remaining non-option arguments however you like (such as by testing if they're a file).
you find the information in the bash man page if you search for "Pattern Matching" without the quotes.this does the trick: [a-zA-Z0-9]*)
you should probably read on about pattern matching, regular expressions and so on.
furthermore you should honour john kugelmans hint about the double quotes.
in the following code snippet you can see how to check if no parameter got passed.
#!/bin/sh
case "$1" in
[a-zA-Z0-9]*)
echo "filename"
;;
"")
echo "no data"
;;
esac
#OP, generally if you are using bash, to match case insensitivity you can use shopt and set nocasematch
$ shopt -s nocasematch
to check null in your case statement
case "$var" in
"") echo "empty value";;
esac
You are better off falling into the default case *) and there you can at least check if the file exists with [ -e "$1" ] ... if it doesn't then echo usage and exit 1.