Given the following programme which reads in user input twice
function search_grep
{
if [ "$2" == "" ];then
for x in "${title[#]}"
do
value=$(echo $x | grep "$1")
if [ "$value" != "" ];then
echo "$value"
fi
done
elif [ "$1" == "" ];then
hello="123"
echo "$hello"
fi
}
echo -n "Enter title : "
read book_title
echo -n "Enter author : "
read author
title=(CatchMe HappyDay)
search_grep $book_title $author
it works as expected when i enter followed by HappyDay HOWEVER
When i enter foo followed by , I would expect console output to be
123
instead I am getting
Can someone explain to me , the programme is not executing the second elif loop though second input is
In both of your cases cases, the following:
search_grep $book_title $author
expands to a call with a single argument. Hence, the "then" clause is activated. The reason is that an unquoted argument consisting of whitespace expands to nothing and disappears. That is the way of bash.
If you want to pass search_grep two arguments, then you need to quote the variables:
search_grep "$book_title" "$author"
As shown here, you might try using = instead of ==
Or for an empty string comparison try -z
Related
This question already has answers here:
Dynamic variable names in Bash
(19 answers)
How to use a variable's value as another variable's name in bash [duplicate]
(6 answers)
Closed 5 years ago.
In my bash scripts, I often prompt users for y/n answers. Since I often use this several times in a single script, I'd like to have a function that checks if the user input is some variant of Yes / No, and then cleans this answer to "y" or "n". Something like this:
yesno(){
temp=""
if [[ "$1" =~ ^([Yy](es|ES)?|[Nn][Oo]?)$ ]] ; then
temp=$(echo "$1" | tr '[:upper:]' '[:lower:]' | sed 's/es//g' | sed 's/no//g')
break
else
echo "$1 is not a valid answer."
fi
}
I then would like to use the function as follows:
while read -p "Do you want to do this? " confirm; do # Here the user types "YES"
yesno $confirm
done
if [[ $confirm == "y" ]]; then
[do something]
fi
Basically, I want to change the value of the first argument to the value of $confirm, so that when I exit the yesno function, $confirm is either "y" or "n".
I tried using set -- "$temp" within the yesnofunction, but I can't get it to work.
You could do it by outputting the new value and overwriting the variable in the caller.
yesno() {
if [[ "$1" =~ ^([Yy](es|ES)?|[Nn][Oo]?)$ ]] ; then
local answer=${1,,}
echo "${answer::1}"
else
echo "$1 is not a valid answer." >&2
echo "$1" # output the original value
return 1 # indicate failure in case the caller cares
fi
}
confirm=$(yesno "$confirm")
However, I'd recommend a more direct approach: have the function do the prompting and looping. Move all of that repeated logic inside. Then the call site is super simple.
confirm() {
local prompt=$1
local reply
while true; do
read -p "$prompt" reply
case ${reply,,} in
y*) return 0;;
n*) return 1;;
*) echo "$reply is not a valid answer." >&2;;
esac
done
}
if confirm "Do you want to do this? "; then
# Do it.
else
# Don't do it.
fi
(${reply,,} is a bash-ism that converts $reply to lowercase.)
You could use the nameref attribute of Bash (requires Bash 4.3 or newer) as follows:
#!/bin/bash
yesno () {
# Declare arg as reference to argument provided
declare -n arg=$1
local re1='(y)(es)?'
local re2='(n)o?'
# Set to empty and return if no regex matches
[[ ${arg,,} =~ $re1 ]] || [[ ${arg,,} =~ $re2 ]] || { arg= && return; }
# Assign "y" or "n" to reference
arg=${BASH_REMATCH[1]}
}
while read -p "Prompt: " confirm; do
yesno confirm
echo "$confirm"
done
A sample test run looks like this:
Prompt: YES
y
Prompt: nOoOoOo
n
Prompt: abc
Prompt:
The expressions are anchored at the start, so yessss etc. all count as well. If this is not desired, an end anchor ($) can be added.
If neither expression matches, the string is set to empty.
I tried many different combinations with brakets, quotation marks, ||, -o, but the only way the script works without infinit loop is without the OR in while loop comparison, like this: while [ $name != Jorge ]] ; do ... This is an example of an script that I want to run:
#!/bin/bash
echo "What is my name: "
read name
while [ $name != "Jorge" ] || [ $name != "Eduardo" ] ; do
echo "Not. Try again: "
read name
done
echo "Well done!"
You should use && instead of || in the condition of while statement.
You are trying to read a name from stdin and if that name is "Jorge" or "Eduardo", you are done. when putting it in condition of while, you want to continue in the loop when the name if not "Jorge" and the name is not "Eduardo".
your current condition says that continue in the loop if the name is not "Jorge" or the name is not "Eduardo". And the name cannot be both at the same time.
Using regex:
while [[ ! "$name" =~ ^(Jorge|Eduardo)$ ]]; do
or Bash extented globbing:
shopt -s extglob
while [[ "$name" != #(Jorge|Eduardo) ]]; do
I have an issue in finding a part of string variable in another string variable, I tried many methods but none worked out..
for example:
echo -e " > Required_keyword: $required_keyword"
send_func GUI WhereAmI
echo -e " > FUNCVALUE: $FUNCVALUE"
flag=`echo $FUNCVALUE|awk '{print match($0,"$required_keyword")}'`;
if [ $flag -gt 0 ];then
echo "Success";
else
echo "fail";
fi
But it always gives fail though there are certain words in variable which matches like
0_Menu/BAA_Record ($required_keyword output string)
Trying to connect to 169.254.98.226 ... OK! Executing sendFunc GUI
WhereAmI Sent Function WhereAmI [OK PageName:
"_0_Menu__47__BAA_Record" ($FUNCVALUE output string)
As we can see here the BAA_Record is common in both of the output still, it always give FAIL
The output echo is
> Required_keyword: 0_Menu/BAA_Record
> FUNCVALUE:
Trying to connect to 169.254.98.226 ... OK!
Executing sendFunc GUI WhereAmI
Sent Function WhereAmI [OK]
PageName: "_0_Menu__47__BAA_Record"
Bash can do wildcard and regex matches inside double square brackets.
if [[ foobar == *oba* ]] # wildcard
if [[ foobar =~ fo*b.r ]] # regex
In your example:
if [[ $FUNCVALUE = *$required_keyword* ]]
if [[ $FUNCVALUE =~ .*$required_keyword.* ]]
Not sure if I understand what you want, but if you need to find out if there's part of string "a" present in variable "b" you can use simply just grep.
grep -q "a" <<< "$b"
[[ "$?" -eq 0 ]] && echo "Found" || echo "Not found"
EDIT: To clarify, grep searches for string a in variable b and returns exit status (see man grep, hence the -q switch). After that you can check for exit status and do whatever you want (either with my example or with regular if statement).
I've spent 2 hours with an if statement, that never works like I want:
#should return true
if [ "$1" == "355258054414904" ]; then
Here is the whole script:
#!/bin/bash
param=$1
INPUT=simu_900_imei_user_pass.csv
OLDIFS=$IFS
IFS=,
[ ! -f $INPUT ] && { echo "$INPUT ime not found"; exit 99; }
while read imei email pass
do
echo "First Parameter-IMEI: $1"
if [ "$1" == "355258054414904" ]; then
echo "GOOD"
fi
done < $INPUT
IFS=$OLDIFS
This is the output of the script:
First Parameter-IMEI: 355258054414904
First Parameter-IMEI: 355258054414904
First Parameter-IMEI: 355258054414904
I have seen a lot of pages about the subject, but I can't make it work :(
EDIT: I Join the content of csv for better understanding ! Tx for your help !
4790057be1803096,user1,pass1
355258054414904,juju,capp
4790057be1803096,user2,pass2
358854053154579,user3,pass3
The reason $1 does not match is because $1 means the first parameter given to the script on the command line, while you want it to match the first field read from the file. That value is in $imei.
You probably meant:
if [ "$imei" == "355258054414904" ]; then
echo "GOOD"
fi
Since it is inside the loop where you read input file line by line.
To check content of $1 use:
cat -vet <<< "$1"
UPDATE: To strip \r from $1 have this at top:
param=$(tr -d '\r' <<< "$1")
And then use "$param" in rest of your script.
To test string equality with [ you want to use a single '=' sign.
I couldn't find any one simple straightforward resource spelling out the meaning of and fix for the following BASH shell error, so I'm posting what I found after researching it.
The error:
-bash: [: too many arguments
Google-friendly version: bash open square bracket colon too many arguments.
Context: an if condition in single square brackets with a simple comparison operator like equals, greater than etc, for example:
VARIABLE=$(/some/command);
if [ $VARIABLE == 0 ]; then
# some action
fi
If your $VARIABLE is a string containing spaces or other special characters, and single square brackets are used (which is a shortcut for the test command), then the string may be split out into multiple words. Each of these is treated as a separate argument.
So that one variable is split out into many arguments:
VARIABLE=$(/some/command);
# returns "hello world"
if [ $VARIABLE == 0 ]; then
# fails as if you wrote:
# if [ hello world == 0 ]
fi
The same will be true for any function call that puts down a string containing spaces or other special characters.
Easy fix
Wrap the variable output in double quotes, forcing it to stay as one string (therefore one argument). For example,
VARIABLE=$(/some/command);
if [ "$VARIABLE" == 0 ]; then
# some action
fi
Simple as that. But skip to "Also beware..." below if you also can't guarantee your variable won't be an empty string, or a string that contains nothing but whitespace.
Or, an alternate fix is to use double square brackets (which is a shortcut for the new test command).
This exists only in bash (and apparently korn and zsh) however, and so may not be compatible with default shells called by /bin/sh etc.
This means on some systems, it might work from the console but not when called elsewhere, like from cron, depending on how everything is configured.
It would look like this:
VARIABLE=$(/some/command);
if [[ $VARIABLE == 0 ]]; then
# some action
fi
If your command contains double square brackets like this and you get errors in logs but it works from the console, try swapping out the [[ for an alternative suggested here, or, ensure that whatever runs your script uses a shell that supports [[ aka new test.
Also beware of the [: unary operator expected error
If you're seeing the "too many arguments" error, chances are you're getting a string from a function with unpredictable output. If it's also possible to get an empty string (or all whitespace string), this would be treated as zero arguments even with the above "quick fix", and would fail with [: unary operator expected
It's the same 'gotcha' if you're used to other languages - you don't expect the contents of a variable to be effectively printed into the code like this before it is evaluated.
Here's an example that prevents both the [: too many arguments and the [: unary operator expected errors: replacing the output with a default value if it is empty (in this example, 0), with double quotes wrapped around the whole thing:
VARIABLE=$(/some/command);
if [ "${VARIABLE:-0}" == 0 ]; then
# some action
fi
(here, the action will happen if $VARIABLE is 0, or empty. Naturally, you should change the 0 (the default value) to a different default value if different behaviour is wanted)
Final note: Since [ is a shortcut for test, all the above is also true for the error test: too many arguments (and also test: unary operator expected)
Just bumped into this post, by getting the same error, trying to test if two variables are both empty (or non-empty). That turns out to be a compound comparison - 7.3. Other Comparison Operators - Advanced Bash-Scripting Guide; and I thought I should note the following:
I used -e thinking it means "empty" at first; but that means "file exists" - use -z for testing empty variable (string)
String variables need to be quoted
For compound logical AND comparison, either:
use two tests and && them: [ ... ] && [ ... ]
or use the -a operator in a single test: [ ... -a ... ]
Here is a working command (searching through all txt files in a directory, and dumping those that grep finds contain both of two words):
find /usr/share/doc -name '*.txt' | while read file; do \
a1=$(grep -H "description" $file); \
a2=$(grep -H "changes" $file); \
[ ! -z "$a1" -a ! -z "$a2" ] && echo -e "$a1 \n $a2" ; \
done
Edit 12 Aug 2013: related problem note:
Note that when checking string equality with classic test (single square bracket [), you MUST have a space between the "is equal" operator, which in this case is a single "equals" = sign (although two equals' signs == seem to be accepted as equality operator too). Thus, this fails (silently):
$ if [ "1"=="" ] ; then echo A; else echo B; fi
A
$ if [ "1"="" ] ; then echo A; else echo B; fi
A
$ if [ "1"="" ] && [ "1"="1" ] ; then echo A; else echo B; fi
A
$ if [ "1"=="" ] && [ "1"=="1" ] ; then echo A; else echo B; fi
A
... but add the space - and all looks good:
$ if [ "1" = "" ] ; then echo A; else echo B; fi
B
$ if [ "1" == "" ] ; then echo A; else echo B; fi
B
$ if [ "1" = "" -a "1" = "1" ] ; then echo A; else echo B; fi
B
$ if [ "1" == "" -a "1" == "1" ] ; then echo A; else echo B; fi
B
Another scenario that you can get the [: too many arguments or [: a: binary operator expected errors is if you try to test for all arguments "$#"
if [ -z "$#" ]
then
echo "Argument required."
fi
It works correctly if you call foo.sh or foo.sh arg1. But if you pass multiple args like foo.sh arg1 arg2, you will get errors. This is because it's being expanded to [ -z arg1 arg2 ], which is not a valid syntax.
The correct way to check for existence of arguments is [ "$#" -eq 0 ]. ($# is the number of arguments).
I also faced same problem. #sdaau answer helped me in logical way. Here what I was doing which seems syntactically correct to me but getting too many arguments error.
Wrong Syntax:
if [ $Name != '' ] && [ $age != '' ] && [ $sex != '' ] && [ $birthyear != '' ] && [ $gender != '' ]
then
echo "$Name"
echo "$age"
echo "$sex"
echo "$birthyear"
echo "$gender"
else
echo "Enter all the values"
fi
in above if statement, if I pass the values of variable as mentioned below then also I was getting syntax error
export "Name"="John"
export "age"="31"
export "birthyear"="1990"
export "gender"="M"
With below syntax I am getting expected output.
Correct syntax:
if [ "$Name" != "" -a "$age" != "" -a "$sex" != "" -a "$birthyear" != "" -a "$gender" != "" ]
then
echo "$Name"
echo "$age"
echo "$sex"
echo "$birthyear"
echo "$gender"
else
echo "it failed"
fi
There are few points which we need to keep in mind
use "" instead of ''
use -a instead of &&
put space before and after operator sign like [ a = b], don't use as [ a=b ] in if condition
Hence above solution worked for me !!!
Some times If you touch the keyboard accidentally and removed a space.
if [ "$myvar" = "something"]; then
do something
fi
Will trigger this error message. Note the space before ']' is required.
I have had same problem with my scripts. But when I did some modifications it worked for me. I did like this :-
export k=$(date "+%k");
if [ $k -ge 16 ]
then exit 0;
else
echo "good job for nothing";
fi;
that way I resolved my problem. Hope that will help for you too.