I have a string
$VAR="I-UAT";
in my shell script code. I need a conditional statement to check if "UAT" is present in that string.
What command should I use to get either true or false boolean as output?
Or is there any other way of checking it?
What shell? Using bash:
if [[ "$VAR" =~ "UAT" ]]; then
echo "matched"
else
echo "didn't match"
fi
You can do it this way:
case "$VAR" in
*UAT*)
# code when var has UAT
;;
esac
The classic way, if you know ahead of time what string you're looking for, is a case statement:
case "$VAR" in
*UAT*) : OK;;
*) : Oops;;
esac
You can use an appropriate command in place of the : command. This will work with Bourne and Korn shells too, not just with Bash.
found=`echo $VAR | grep -c UAT`
Then test for $found non-zero.
In bash script you could use
if [ "$VAR" != "${VAR/UAT/}" ]; then
# UAT present in $VAR
fi
try with grep:
$ echo I\-UAT | grep UAT
$ echo $?
0
$ echo I\-UAT | grep UAX
$ echo $?
1
so testing
if [ $? -ne 0 ]; then
# not found
else
# found
fi
I like this a little better than using case/esac (and it'll work with non-bash shells):
#!/bin/sh
full_string="I-UAT"
substring="UAT"
if [ -z "${full_string##*$substring*}" ]; then
echo "Found substring!"
else
echo "Substring is MIA!"
fi
If the string returned is zero-length (-z), then the substring was found.
Related
I want to ensure that the given arguments are flags, by which I mean -x where x is some character or sequence of.
I have tried to do this with the following:
if [[ "$(echo '$1' | sed 's/[^-]//g')" -ne "-" ]];
then
echo "$usage"
exit
fi
Where my reasoning is that if - is not present when other characters are stripped it's not a flag.
This doesn't work though, and is obviously flimsy, but I don't know how to do this correctly.
# valid
script.sh -asdf
# invalid
script.sh sdf
You can do it this way to make sure $1 is starting with -:
if [[ "${1?}" != -* ]]; then
echo "$usage"
exit 1
fi
${1?} will fail the script if $1 is not available.
I am taking baby steps at learning bash and I am developing a piece of code which takes an input and checks if it contains any spaces. The idea is that the variable should NOT contain any spaces and the code should consequently echo a suitable message.
Try this:
#!/bin/bash
if [[ $1 = *[[:space:]]* ]]
then
echo "space exist"
fi
You can use grep, like this:
echo " foo" | grep '\s' -c
# 1
echo "foo" | grep '\s' -c
# 0
Or you may use something like this:
s=' foo'
if [[ $s =~ " " ]]; then
echo 'contains space'
else
echo 'ok'
fi
You can test simple glob patterns in portable shell by using case, without needing any external programs or Bash extensions (that's a good thing, because then your scripts are useful to more people).
#!/bin/sh
case "$1" in
*' '*)
printf 'Invalid argument %s (contains space)\n' "$1" >&2
exit 1
;;
esac
You might want to include other whitespace characters in your check - in which case, use *[[:space:]]* as the pattern instead of *' '*.
You can use wc -w command to check if there are any words. If the result of this output is a number greater than 1, then it means that there are more than 1 words in the input. Here's an example:
#!/bin/bash
read var1
var2=`echo $var1 | wc -w`
if [ $var2 -gt 1 ]
then
echo "Spaces"
else
echo "No spaces"
fi
Note: there is a | (pipe symbol) which means that the result of echo $var1 will be given as input to wc -w via the pipe.
Here is the link where I tested the above code: https://ideone.com/aKJdyN
You could use parameter expansion to remove everything that isn't a space and see if what's left is the empty string or not:
var1='has space'
var2='nospace'
for var in "$var1" "$var2"; do
if [[ ${var//[^[:space:]]} ]]; then
echo "'$var' contains a space"
fi
done
The key is [[ ${var//[^[:space:]]} ]]:
With ${var//[^[:space:]]}, everything that isn't a space is removed from the expansion of $var.
[[ string ]] has a non-zero exit status if string is empty. It's a shorthand for the equivalent [[ -n string ]].
We could also quote the expansion of ${var//[^[:space:]]}, but [[ ... ]] takes care of the quoting for us.
I have next string comparison in bash:
if [[ $1 == "/"* ]]; then echo YES; else echo "$1"; fi
How to do same in sh?
You could do it with a case statement.
case $1 in
/*) echo YES;;
*) echo $1
esac
The direct translation would be if [ "$1" = "/*" ], but it wouldn't work because sh doesn't support glob matches there. You'd need to invoke an external command like grep.
if printf '%s\n' "$1" | grep -e '^/' >/dev/null 2>&1; then
...
fi
This would be shorter with grep -q, but if you don't have bash then you may not have grep -q either.
The case statement handles this elegantly.
case $1 in
'/'* ) echo YES;;
*) echo "$1";;
esac
The lack of quoting before in and the general syntax with the double semicolons and unpaired right parentheses is jarring to the newcomer, but you quickly get used to it. It's quite versatile and much under-appreciated.
If you insist on using [ you could perhaps do something like
if temp=${1#?}; [ "${1%$temp}" -eq '*' ]; then
...
which uses a couple of parameter expansions to extract the first character of the variable; but case has glob pattern matching built in, so it's considerably more readable.
If you would like to match a pattern, or regex in if statements if sh or fish you can use
if echo "$1" | grep -q ^/; then echo yes; else echo "$1"; fi
I am trying to see if I found something using grep or not with this
found=`grep -F "something" somefile.txt`
if ((${#found} == 0)); then
echo "Not Found"
else
echo "Found"
fi
I succeeded using above logic that if grep found something it stores the output in found variable but the issue I am facing is with if condition. Whenever found=0 it gives me some error like that
final.sh: 13: final.sh: 0: not found
FYI: final.sh is the script name
The problem is that you're writing bash specific code, but running it with sh. In bash, (( .. )) is an arithmetic context, while in POSIX sh, it's merely two nested subshells, causing it to try to execute the number as a command.
You can run it with bash instead of sh by specifying #!/bin/bash in the shebang, and/or using bash yourfile instead of sh yourfile if you invoke it that way.
The correct way for your example, however, is to use grep's exit status directly:
if grep -q something somefile
then
echo "found"
else
echo "not found"
fi
To check whether some string is in your file, you can use the return status from grep
grep -q something somefile.txt
if [ $? -eq 0 ]
then
echo "found"
else
echo "not found"
fi
a shorter form will be
grep -q something somefile.txt && echo found || echo not found
found=$(grep -F "something" somefile.txt)
if [ $? = 0 ]; then # $? is the return status of a previous command. Grep will return 0 if it found something, and 1 if nothing was found.
echo "Something was found. Found=$found"
else
echo 'Nothing was found'
fi
I find this code more elegant than other answers.
But anyway, why are you writing in sh? Why don't you use bash? Are you sure that you need that portability? Check out this link to see if you really need sh
Here's how I do that sort of thing:
found=$(grep -F "something" somefile.txt)
if [[ -z $found ]]; then
echo "Not found"
else
echo "Found"
fi
I want to check if a file contains a specific string or not in bash. I used this script, but it doesn't work:
if [[ 'grep 'SomeString' $File' ]];then
# Some Actions
fi
What's wrong in my code?
if grep -q SomeString "$File"; then
Some Actions # SomeString was found
fi
You don't need [[ ]] here. Just run the command directly. Add -q option when you don't need the string displayed when it was found.
The grep command returns 0 or 1 in the exit code depending on
the result of search. 0 if something was found; 1 otherwise.
$ echo hello | grep hi ; echo $?
1
$ echo hello | grep he ; echo $?
hello
0
$ echo hello | grep -q he ; echo $?
0
You can specify commands as an condition of if. If the command returns 0 in its exitcode that means that the condition is true; otherwise false.
$ if /bin/true; then echo that is true; fi
that is true
$ if /bin/false; then echo that is true; fi
$
As you can see you run here the programs directly. No additional [] or [[]].
In case if you want to check whether file does not contain a specific string, you can do it as follows.
if ! grep -q SomeString "$File"; then
Some Actions # SomeString was not found
fi
In addition to other answers, which told you how to do what you wanted, I try to explain what was wrong (which is what you wanted.
In Bash, if is to be followed with a command. If the exit code of this command is equal to 0, then the then part is executed, else the else part if any is executed.
You can do that with any command as explained in other answers: if /bin/true; then ...; fi
[[ is an internal bash command dedicated to some tests, like file existence, variable comparisons. Similarly [ is an external command (it is located typically in /usr/bin/[) that performs roughly the same tests but needs ] as a final argument, which is why ] must be padded with a space on the left, which is not the case with ]].
Here you needn't [[ nor [.
Another thing is the way you quote things. In bash, there is only one case where pairs of quotes do nest, it is "$(command "argument")". But in 'grep 'SomeString' $File' you have only one word, because 'grep ' is a quoted unit, which is concatenated with SomeString and then again concatenated with ' $File'. The variable $File is not even replaced with its value because of the use of single quotes. The proper way to do that is grep 'SomeString' "$File".
Shortest (correct) version:
grep -q "something" file; [ $? -eq 0 ] && echo "yes" || echo "no"
can be also written as
grep -q "something" file; test $? -eq 0 && echo "yes" || echo "no"
but you dont need to explicitly test it in this case, so the same with:
grep -q "something" file && echo "yes" || echo "no"
##To check for a particular string in a file
cd PATH_TO_YOUR_DIRECTORY #Changing directory to your working directory
File=YOUR_FILENAME
if grep -q STRING_YOU_ARE_CHECKING_FOR "$File"; ##note the space after the string you are searching for
then
echo "Hooray!!It's available"
else
echo "Oops!!Not available"
fi
grep -q [PATTERN] [FILE] && echo $?
The exit status is 0 (true) if the pattern was found; otherwise blankstring.
if grep -q [string] [filename]
then
[whatever action]
fi
Example
if grep -q 'my cat is in a tree' /tmp/cat.txt
then
mkdir cat
fi
In case you want to checkif the string matches the whole line and if it is a fixed string, You can do it this way
grep -Fxq [String] [filePath]
example
searchString="Hello World"
file="./test.log"
if grep -Fxq "$searchString" $file
then
echo "String found in $file"
else
echo "String not found in $file"
fi
From the man file:
-F, --fixed-strings
Interpret PATTERN as a list of fixed strings, separated by newlines, any of
which is to be matched.
(-F is specified by POSIX.)
-x, --line-regexp
Select only those matches that exactly match the whole line. (-x is specified by
POSIX.)
-q, --quiet, --silent
Quiet; do not write anything to standard output. Exit immediately with zero
status if any match is
found, even if an error was detected. Also see the -s or --no-messages
option. (-q is specified by
POSIX.)
Try this:
if [[ $(grep "SomeString" $File) ]] ; then
echo "Found"
else
echo "Not Found"
fi
I done this, seems to work fine
if grep $SearchTerm $FileToSearch; then
echo "$SearchTerm found OK"
else
echo "$SearchTerm not found"
fi
grep -q "something" file
[[ !? -eq 0 ]] && echo "yes" || echo "no"