I am trying to do a script where i need to check if the user input is valid. I just can't figure it out. Have been trying different ways but can't find a solution. So if there are a busybox ash guru out there i am happy for all help.
if ! [[ $ANS =~ ^[0-9][.0-9]*$ ]]; then
echo "abort"
else
echo "go on"
fi
I want to see if the user inputs a number. A number with decimal is also allowed. If not then it should abort.
Same goes with..
if ! [[ $ANS =~ ^[0-9A-Fa-f]{6}$ ]] ; then
echo "abort"
else
echo "go on"
fi
Where i need it to see if hexadecimal is used. All i get is "unknown operand".
It feels a bit of a hack, but you can use egrep for this:
$ ANS=10.2
$ echo -n $ANS | egrep -q '^[0-9]*[.]?[0-9]*$' && echo success || echo failure
success
$ ANS=10.2.2
$ echo -n $ANS | egrep -q '^[0-9]*[.]?[0-9]*$' && echo success || echo failure
failure
Related
I have a file a.txt contains a string like:
Axxx-Bxxxx
Rules for checking if it is valid or not include:
length is 10 characters.
x here is digits only.
Then, I try to check with:
#!/bin/bash
exp_len=10;
file=a.txt;
msg="checking string";
tmp="File not exist";
echo $msg;
if[ -f $file];then
tmp=$(cat $file);
if[[${#tmp} != $exp_len ]];then
msg="invalid length";
elif [[ $tmp =~ ^[A[0-9]{3}-B[0-9]{4}]$]];then
msg="valid";
else
msg="invalid";
fi
else
msg="file not exist";
fi
echo $msg;
But in valid case it doesn't work...
Is there someone help to correct me?
Thanks :)
Other than the regex fix, your code can be refactored as well, moreover there are syntax issues as well. Consider this code:
file="a.txt"
msg="checking string"
tmp="File not exist"
echo "$msg"
if [[ -f $file ]]; then
s="$(<$file)"
if [[ $s =~ ^A[0-9]{3}-B[0-9]{4}$ ]]; then
msg="valid"
else
msg="invalid"
fi
else
msg="file not exist"
fi
echo "$msg"
Changes are:
Remove unnecessary cat
Use [[ ... ]] when using bash
Spaces inside [[ ... ]] are required (your code was missing them)
There is no need to check length of 10 as regex will make sure that part as well
As mentioned in comments earlier correct regex should be ^A[0-9]{3}-B[0-9]{4}$ or ^A[[:digit:]]{3}-B[[:digit:]]{4}$
Note that a regex like ^[A[0-9]{3}-B[0-9]{4}]$ matches
^ - start of string
[A[0-9]{3} - three occurrences of A, [ or a digit
-B - a -B string
[0-9]{4} - four digits
] - a ] char
$ - end of string.
So, it matches strings like [A[-B1234], [[[-B1939], etc.
Your regex checking line must look like
if [[ $tmp =~ ^A[0-9]{3}-B[0-9]{4}$ ]];then
See the online demo:
#!/bin/bash
tmp="A123-B1234";
if [[ $tmp =~ ^A[0-9]{3}-B[0-9]{4}$ ]];then
msg="valid";
else
msg="invalid";
fi
echo $msg;
Output:
valid
Using just grep might be easier:
$ echo A123-B1234 > valid.txt
$ echo 123 > invalid.txt
$ grep -Pq 'A\d{3}-B\d{4}' valid.txt && echo valid || echo invalid
valid
$ grep -Pq 'A\d{3}-B\d{4}' invalid.txt && echo valid || echo invalid
invalid
With your shown samples and attempts, please try following code also.
#!/bin/bash
exp_len=10;
file=a.txt;
msg="checking string";
tmp="File not exist";
if [[ -f "$file" ]]
then
echo "File named $file is existing.."
awk '/^A[0-9]{3}-B[0-9]{4}$/{print "valid";next} {print "invalid"}' "$file"
else
echo "Please do check File named $file is not existing, exiting from script now..."
exit 1;
fi
OR In case you want to check if line in your Input_file should be 10 characters long(by seeing OP's attempted code's exp_len shell variable) then try following code, where an additional condition is also added in awk code.
#!/bin/bash
exp_len=10;
file=a.txt;
msg="checking string";
tmp="File not exist";
if [[ -f "$file" ]]
then
echo "File named $file is existing.."
awk -v len="$exp_len" 'length($0) == len && /^A[0-9]{3}-B[0-9]{4}$/{print "valid";next} {print "invalid"}' "$file"
else
echo "Please do check File named $file is not existing, exiting from script now..."
exit 1;
fi
NOTE: I am using here -f flag to test if file is existing or not, you can change it to -s eg: -s "$file" in case you want to check file is present and is of NOT NULL size.
I have a bash script that has a function like so:
sanitize(){
rb_reg="^([0-9]{1,3})\.([0-9]{1,3})\.([0-9]{1,3})$"
if grep -Ex "${rb_reg}" "${1}/.ruby-version" > /dev/null 2>&1; then
sanitize_tmp="$(<"${1}"/.ruby-version)" &&
ruby_version="${sanitize_tmp//[^0-9\.]/}" &&
echo "Setting Ruby Version: ${ruby_version}"
else
echo "There was an error trying to sanitize a .ruby-version file"
echo "The file was: ${1}/.ruby-version"
exit 7
fi
}
I'm using it to check a .ruby-version file and then set the version in there as a variable.
Mostly these files will contain something sensible like: 2.0.0 which works OK. I want to be defensive and not trust the input file, so check/sanitize it as much as possible.
Two questions:
If for some reason there were multiple version numbers in the file on multiple lines, say:
'2.0.0
1.0.0'
That's going to smash them together currently removing white space and end up with a variable like: '2.0.01.0.0'
What's a good way to only pick up the first version number that matches the regex?
Is there a better way to do this, maybe entirely in bash without grep? Appreciate any examples people have of checking for a version like this but not trusting the input file.
I'm still playing around with this a little, but here is what I ended up doing.
I'm passing in the file name as an argument to the function elsewhere in the script. Really liked the concept of BASH_REMATCH, so tried to avoid using grep, sed, awk etc and do it this way.
You can view the latest version of the code here: https://github.com/octopusnz/scripts
sanitize(){
if [[ "${#}" -ne 1 ]]; then
echo "[ERROR 7]: We expected 1 argument to the sanitize() function."
echo "But we got ${#} instead."
exit 7
fi
rbv_reg="^([0-9]{1,2})\.([0-9]{1,2})\.([0-9]{1,2})(-([a-z]{1,10}))?$"
reg_matches=0
while read -r rbv_line || [[ -n "$rbv_line" ]]; do
if [[ "${rbv_line}" =~ ${rbv_reg} ]]; then
ruby_version="${BASH_REMATCH[0]//[^0-9a-z\.\-]/}" &&
((reg_matches="${reg_matches}"+1)) &&
echo "" &&
echo "Setting Ruby version: ${ruby_version}" &&
break
fi
done < "${1}"
if [[ "${reg_matches}" -lt 1 ]]; then
if [[ -v ruby_version ]]; then
echo "We couldn't parse ${1} and set a valid Ruby version."
echo "Using default: ${ruby_version}"
else
echo "We couldn't parse ${1} and set a default Ruby version."
echo "[ERROR 4]: No valid .ruby-version file found."
exit 4
fi
fi
}
In Unix, I need an input validation which use grep:
echo $INPUT| grep -E -q '^foo1foo2foo3' || echo "no"
What I really need is if input doesn't match at least one of these values: foo1 foo2 or foo3, exit the program.
Source: syntax is taken from Validating parameters to a Bash script
You need to use alternation:
echo "$INPUT" | grep -Eq 'foo1|foo2|foo3' || echo "no"
Do you really need grep? If you're scripting in bash:
[[ $INPUT == #(foo1|foo2|foo3) ]] || echo "no"
or
[[ $INPUT == foo[123] ]] || echo "no"
If you want "$INPUT contains one of those patterns
[[ $INPUT == *#(foo1|foo2|foo3)* ]] || echo "no"
Does this solve your problem:
echo $INPUT | grep -E 'foo1|foo2|foo3' || echo "no"
?
I'm trying to write a bash script that helps solving crosswords. For example, the question is "Alcoholic Drink in German". I already have a 'B' at the first place, an 'R' at the last place and two gaps in between. So a regex would be $B..R^
Since I live in Switzerland, I'd like to use the ngerman dictionary (DICT=/usr/share/dict/ngerman).
Here's how I'd do it directly on the shell:
grep -i '^B...$' /usr/share/dict/ngerman
That works perfectly, and the word 'Bier' appears among three others. Since this syntax is cumbersome, I'd like to write a little batch script, that allows me to enter it like this:
crosswords 'B..R'
Here's my approach:
#!/bin/bash
DICT=/usr/share/dict/ngerman
usage () {
progname=$(basename $0)
echo "usage: $progname regex"
}
if [ $# -le 0 ]; then
usage
exit 1
fi
regex="'^$1$'"
cmd="grep -i $regex $DICT"
echo $regex
echo $cmd
$($cmd) | while read word; do
echo "$word"
done
But nothing appears, it doesn't work. I also output the $regex and the $cmd variable for debugging reasons. Here's what comes out:
'^B..R$'
grep -i '^B..R$' /usr/share/dict/ngerman
That's exactly what I need. If I copy/paste the command above, it works perfectly. But if i call it with $($cmd), it fails.
What is wrong?
you do not need to put quotes around regex variable string. and $($cmd) should change to $cmd
so the correct code is :
#!/bin/bash
DICT=/usr/share/dict/ngerman
usage () {
progname=$(basename $0)
echo "usage: $progname regex"
}
if [ $# -le 0 ]; then
usage
exit 1
fi
regex="^$1$"
cmd="grep -i $regex $DICT"
echo $regex
echo $cmd
$cmd | while read word; do
echo "$word"
done
Change regex="^'$1$'" to regex="^$1$" and $($cmd) to $cmd
Here is a fixed version:
#!/bin/bash
DICT=/usr/share/dict/ngerman
usage () {
progname=$(basename "$0")
echo "usage: $progname regex"
}
if [ $# -le 0 ]; then
usage
exit 1
fi
regex="^$1$"
cmd="grep -i $regex $DICT"
echo "$regex"
echo "$cmd"
$cmd | while read -r word; do
echo "$word"
done
But this script has potential problems. For example try running it as ./script 'asdads * '. This will expand to all files in a directory and all of them are going to be passed to grep.
Here is a bit improved version of your code with correct quoting and also with bonus input validation:
#!/bin/bash
DICT=/usr/share/dict/ngerman
usage () {
progname=$(basename "$0")
echo "usage: $progname regex"
}
if [ $# -le 0 ]; then
usage
exit 1
fi
if ! [[ $1 =~ ^[a-zA-Z\.]+$ ]]; then
echo 'Wrong word. Please use only a-zA-Z characters and dots for unknown letters'
exit 1
fi
grep -i "^$1$" "$DICT" | while read -r word; do
echo "$word"
done
Oh, now I got it. When I do it manually, '' are expanded! Here's my test program in C (param-test.c):
#include <stdio.h>
int main(int argc, char *argv[]) {
puts(argv[1]);
return 0;
}
Then I call:
param-test 'foo'
And I see:
foo
That's the problem! grep doesn't really get 'B..R', but just B..R.
#!bin/bash
echo enter your password :
read password
passlength=$(echo ${#password})
if [ $passlength -le 8 ];
then
echo you entered correct password
else
echo entered password is incorrect
fi
if [[$password == [a-z]*[0-9][a-z]*]];
then
echo match found
else
echo match not found
fi
I am not getting what's wrong with this code. If I enter any string as a password, let's say hello123, it gives me an error:
hello123 : command not found
What is wrong with my script?
You can do the following to make it work cross-platforms with any the bourne shell (/bin/sh) based shell, no bash specific primitives -
echo "$password" | grep -q "[a-z]*[0-9][a-z]*"
if [ $? -eq 0 ] ;then
echo "match found"
else
echo "match not found"
fi
Also feel free to use quotes around the variable names. It will save you hours and hours worth of useless debugging. :)
Technically it should give you an error like [[hello123 : command not found.
The issue is that [[$password is not expanded how you think it is. Bash will first resolve the $password variable to what you entered (i.e. hello123). This will yield the string [[hello123 which bash will then try to invoke (and fail, as there is nothing with that name).
Simply add a space () after [[ and bash will recognise [[ as the command to run (although it is a builtin).
if [[ "$password" == [a-z]*[0-9][a-z]* ]]
then
...
The corrected script is below. The errors were:
#!/bin/bash, not #!bin/bash
To read password length, just do passlength=${#password}, not
passlength=$(echo ${#password})
Always put a space after [ or [[
#!/bin/bash
echo "enter your password :"
read password
passlength=${#password}
if [[ $passlength -le 8 ]]
then
echo "you entered correct password"
else
echo "entered password is incorrect"
fi
if [[ $password == [a-z]*[0-9][a-z]* ]]
then
echo "match found"
else
echo "match not found"
fi
In the bash [[ construct, the == operator will match glob-style patterns, and =~ will match regular expressions. See the documentation.
#!/bin/bash
read -s -p "Enter Password: " password
password_length=${#password}
if [ $password_length -lt 8 -o $password_length -gt 20 ] ;then
echo -e "Invalid password - should be between 8 and 20 characters in length.";
echo ;
else
# Check for invalid characters
case $password in
*[^a-zA-Z0-9]* )
echo -e "Password contains invalid characters.";
echo ;
;;
* )
echo "Password accepted.";
echo ;
break;
;;
esac
fi
More tuned example..
Try to replace line
if [[$password == [a-z]*[0-9][a-z]*]];
with following
if echo "$password" | grep -qs '[a-z]*[0-9][a-z]*'
HTH