I am trying to run the below logic
if [-f $file] && [$variable -lt 1] ; then
some logic
else
print "This is wrong"
fi
It fails with the following error
MyScipt.ksh[10]: [-f: not found
Where 10th line is the if condition , I have put in .
I have also tried
if [-f $file && $variable -lt 1] ; then
which gives the same error.
I know this is a syntax mistake somehwere , but I am not sure , what is the correct syntax when I am using multiple conditions with && in a if block
[ is not an operator, it's the name of a program (or a builtin, sometimes). Use type [ to check. Regardless, you need to put a space after it so that the command line parser knows what to do:
if [ -f $file ]
The && operator might not do what you want in this case, either. You should probably read the bash(1) documentation. In this specific case, it seems like what you want is:
if [ -f $file -a $variable -lt 1 ]
Or in more modern bash syntax:
if [[ -f $file && $variable -lt 1 ]]
The [ syntax is secretly a program!
$ type [
[ is a shell builtin
$ ls -l $(which [)
-rwxr-xr-x 1 root root 35264 Nov 19 16:25 /usr/bin/[
Because of the way the shell parses (technically "lexes") your command line, it sees this:
if - keyword
[-f - the program [-f
$file] - A string argument to the [-f program, made by the value of $file and ]. If $file was "asdf", then this would be asdf]
And so forth, down your command. What you need to do is include spaces, which the shell uses to separate the different parts (tokens) of your command:
if [ -f "$file" ]; then
Now [ stands on its own, and can be recognized as a command/program. Also, ] stands on its own as an argument to [, otherwise [ will complain. A couple more notes about this:
You don't need to put a space before or after ;, because that is a special separator that the shell recognizes.
You should always "double quote" $variables because they get expanded before the shell does the lexing. This means that if an unquoted variable contains a space, the shell will see the value as separate tokens, instead of one string.
Using && in an if-test like that isn't the usual way to do it. [ (also known as test) understands -a to mean "and," so this does what you intended:
if [ -f "$file" -a "$variable" -lt 1 ]; then
Use -a in an if block to represent AND.
Note the space preceding the -f option.
if [ -f $file -a $variable -lt 1] ; then
some logic
else
print "This is wrong"
fi
Related
I'm trying to go the path and remove the files starting collectorlist*, so I write the shell script like below., however -e option always give output not found. kindly suggest me.
Script
if [ -d "/abc" ] ; then
for c in `ls -1 "/abc/"`; do
if [ -d "/abc/${c}" ] ; then
if [ -d "/abc/${c}/dynatrace/agent/linux-x86-64/agent/conf/" ] ; then
if [ -e "/bxp/${c}/dynatrace/agent/linux-x86-64/agent/conf/collectorlist* " ] && echo "Found" || echo "Not found" ; then
`ls -ltr "sat.txt"` > /users/home/abc/test.txt
fi
fi
fi
done
fi
There are some problems with your code.
There's a trailing space after the asterisk and before the closing double quote.
You are using an asterisk within double quotes; within double quotes wildcards like the asterisk are not evaluated but taken verbatim.
You are using backticks together with redirection. This will execute the stuff within the backticks, replace the backtick-expression by the output of it and execute the resulting line. So your ls command needs to give out the name of a command which then will be executed. I guess you don't want that. Just remove the backticks, you probably just want to execute ls and redirect its output into a file.
You try to find out if any file matching a pattern exists and use [ -e pattern* ] for this. The pattern* evaluates to one or more file names or (of none exists) to the pattern* itself (unless you tweak your shell to behave differently). The problem is the case of the many values. These will make a problem within your [ ... ] expression.
Result:
if [ -d "/abc" ] ; then
for c in $(ls -1 "/abc/"); do
if [ -d "/abc/${c}/dynatrace/agent/linux-x86-64/agent/conf/" ] ; then
if ls "/bxp/${c}/dynatrace/agent/linux-x86-64/agent/conf"/collectorlist* &> /dev/null; then
echo "Found"
ls -ltr "sat.txt" > /users/home/abc/test.txt
else
echo "Not Found"
fi
fi
done
fi
I am trying to create a "Lambda" style WHERE script.
I want lambdaWHERE to take piped input and pass it through if condition after given as arguments is met. Like xargs I use {} to represent what comes down the pipe.
I call command like:
ls -d EqAAL* | lambdaWHERE.sh -f {}/INFO_ACTIVETICK
I want the folder names passed through if they contain a file called INFO_ACTIVETICK
Here is the script:
#!/bin/sh
#set -x
ARGS=$*
while read i
do
CMD=`echo $ARGS | sed 's/{}/'$i'/g'`
if [[ $CMD ]]
then
echo $i
fi
done
But when I run it a mysterious "-n" appears...
$ ls -d EqAAL* | /q/lambdaWHERE.sh -f {}/INFO_ACTIVETICK
+ ARGS='-f {}/INFO_ACTIVETICK'
+ read i
++ echo -f '{}/INFO_ACTIVETICK'
++ sed 's/{}/EqAAL-1m/g'
+ CMD='-f EqAAL-1m/INFO_ACTIVETICK'
+ [[ -n -f EqAAL-1m/INFO_ACTIVETICK ]]
+ echo EqAAL-1m
EqAAL-1m
+ read i
How can I get the bit in the [[ ]] correct?
You were quite close. you only need to switch to the standard POSIX [ $CMD ] and it will work.
The main difference between using [[ $CMD ]] and [ $CMD ] is that the first has fewer surprises and you need not quote variables. That also means that a variable is though of as one token and cannot have a whole expression in it like you are trying. [ $CMD ] however works the same way as the original shell where [ was just a command an thus need explicit quotations in order to interpret something with spaces as one argument.
There is a relevant question about the differences between [[ ...]] and [ ..]
I'm trying to read user input and compare it against a stored value as follows:
read confirm
if [ echo $confirm | grep -q y ]; then
...
fi
However, this results in a pair of errors:
[: missing `]'
grep: ]: No such file or directory
Why does this happen, and what's the appropriate alternative?
Short Answer
For your immediate use case, you simply want:
if echo "$confirm" | grep -q y; then
...or its much more efficient equivalent (if your shell is bash):
if [[ $confirm = *y* ]]; then
...or its much more efficient equivalent (for any POSIX shell):
case $confirm in *y*) echo "Put your code for the yes branch here" ;; esac
Why was the original wrong?
[ is not part of if syntax: if simply takes a (potentially compound) command as its argument before the then. [ is different name for the test command, which runs checks on its arguments; however, if what you want to test is the exit status of grep -q, then the test command doesn't need to be invoked for this purpose at all.
If you put a | inside a [ command, that makes your compound command a pipeline, and starts a new simple command. Arguments after the | are thus no longer passed to [.
With your original code:
if [ echo $confirm | grep -q y ]; then
...this was running two commands, with a pipeline between them:
[ echo $confirm # first command
grep -q y ] # second command
Since [ requires that its last argument be ], it reported that that mandatory argument was missing; and since grep treats extra arguments as filenames to read, it complained that no file named ] could be found.
Also, [ "$foo" ] checks whether the contents of foo is nonempty. Since the output of grep -q is always empty, [ "$(echo "$confirm" | grep -q y)" ], while syntactically correct, would always evaluate to false, even while exit status of grep -q changes to indicate whether a match was found. ([ "$(echo "$confirm" | grep y)" ], by contrast, is an alternative that emits a correct result - using [ ] to test whether the output from grep is or is not empty -- but is much less efficient than the best-practice approaches).
Formal if syntax
From help if:
if: if COMMANDS; then COMMANDS; [ elif COMMANDS; then COMMANDS; ]... [ else COMMANDS; ] fi
The if COMMANDS list is executed. If its exit status is zero, then the
then COMMANDS list is executed. Otherwise, each elif COMMANDS list is
executed in turn, and if its exit status is zero, the corresponding
then COMMANDS list is executed and the if command completes. Otherwise,
the else COMMANDS list is executed, if present. The exit status of the
entire construct is the exit status of the last command executed, or zero
if no condition tested true.
Notably, if takes a list of COMMANDS, and no [ is included in the syntax specification.
I'd like to add an argument to a command in bash only if a variable evaluates to a certain value. For example this works:
test=1
if [ "${test}" == 1 ]; then
ls -la -R
else
ls -R
fi
The problem with this approach is that I have to duplicate ls -R both when test is 1 or if it's something else. I'd prefer if I could write this in one line instead such as this (pseudo code that doesn't work):
ls (if ${test} == 1 then -la) -R
I've tried the following but it doesn't work:
test=1
ls `if [ $test -eq 1 ]; then -la; fi` -R
This gives me the following error:
./test.sh: line 3: -la: command not found
A more idiomatic version of svlasov's answer:
ls $( (( test == 1 )) && printf %s '-la' ) -R
Since echo understands a few options itself, it's safer to use printf %s to make sure that the text to print is not mistaken for an option.
Note that the command substitution must not be quoted here - which is fine in the case at hand, but calls for a more robust approach in general - see below.
However, in general, the more robust approach is to build up arguments in an array and pass it as a whole:
# Build up array of arguments...
args=()
(( test == 1 )) && args+=( '-la' )
args+=( '-R' )
# ... and pass it to `ls`.
ls "${args[#]}"
Update: The OP asks how to conditionally add an additional, variable-based argument to yield ls -R -la "$PWD".
In that case, the array approach is a must: each argument must become its own array element, which is crucial for supporting arguments that may have embedded whitespace:
(( test == 1 )) && args+= ( '-la' "$PWD" ) # Add each argument as its own array element.
As for why your command,
ls `if [ $test -eq 1 ]; then -la; fi` -R
didn't work:
A command between backticks (or its modern, nestable equivalent, $(...)) - a so-called command substitution - is executed just like any other shell command (albeit in a sub-shell) and the whole construct is replaced with the command's stdout output.
Thus, your command tries to execute the string -la, which fails. To send it to stdout, as is needed here, you must use a command such as echo or printf.
Print the argument with echo:
test=1
ls `if [ $test -eq 1 ]; then echo "-la"; fi` -R
I can't say how acceptable this is, but:
test=1
ls ${test:+'-la'} -R
See https://stackoverflow.com/revisions/16753536/1 for a conditional truth table.
Another answer without using eval and using BASH arrays:
myls() { local arr=(ls); [[ $1 -eq 1 ]] && arr+=(-la); arr+=(-R); "${arr[#]}"; }
Use it as:
myls
myls "$test"
This script builds whole command in an array arr and preserves the original order of command options.
I'm trying to get the first character of a variable, but I'm getting a Bad substitution error. Can anyone help me fix it?
code is:
while IFS=$'\n' read line
do
if [ ! ${line:0:1} == "#"] # Error on this line
then
eval echo "$line"
eval createSymlink $line
fi
done < /some/file.txt
Am I doing something wrong or is there a better way of doing this?
-- EDIT --
As requested - here's some sample input which is stored in /some/file.txt
$MOZ_HOME/mobile/android/chrome/content/browser.js
$MOZ_HOME/mobile/android/locales/en-US/chrome/browser.properties
$MOZ_HOME/mobile/android/components/ContentPermissionPrompt.js
To get the first character of a variable you need to say:
v="hello"
$ echo "${v:0:1}"
h
However, your code has a syntax error:
[ ! ${line:0:1} == "#"]
# ^-- missing space
So this can do the trick:
$ a="123456"
$ [ ! "${a:0:1}" == "#" ] && echo "doesnt start with #"
doesnt start with #
$ a="#123456"
$ [ ! "${a:0:1}" == "#" ] && echo "doesnt start with #"
$
Also it can be done like this:
$ a="#123456"
$ [ "$(expr substr $a 1 1)" != "#" ] && echo "does not start with #"
$
$ a="123456"
$ [ "$(expr substr $a 1 1)" != "#" ] && echo "does not start with #"
does not start with #
Update
Based on your update, this works to me:
while IFS=$'\n' read line
do
echo $line
if [ ! "${line:0:1}" == "#" ] # Error on this line
then
eval echo "$line"
eval createSymlink $line
fi
done < file
Adding the missing space (as suggested in fedorqui's answer ;) ) works for me.
An alternative method/syntax
Here's what I would do in Bash if I want to check the first character of a string
if [[ $line != "#"* ]]
On the right hand side of ==, the quoted part is treated literally whereas * is a wildcard for any sequence of character.
For more information, see the last part of Conditional Constructs of Bash reference manual:
When the ‘==’ and ‘!=’ operators are used, the string to the right of the operator is considered a pattern and matched according to the rules described below in Pattern Matching
Checking that you're using the right shell
If you are getting errors such as "Bad substitution error" and "[[: not found" (see comment) even though your syntax is fine (and works fine for others), it might indicate that you are using the wrong shell (i.e. not Bash).
So to make sure you are using Bash to run the script, either
make the script executable and use an appropriate shebang e.g. #!/bin/bash
or execute it via bash my_script
Also note that sh is not necessarily bash, sometimes it can be dash (e.g. in Ubuntu) or just plain ol' Bourne shell.
Try this:
while IFS=$'\n' read line
do
if ! [ "${line:0:1}" = "#" ]; then
eval echo "$line"
eval createSymlink $line
fi
done < /some/file.txt
or you can use the following for your if syntax:
if [[ ! ${line:0:1} == "#" ]]; then
TIMTOWTDI ^^
while IFS='' read -r line
do
case "${line}" in
"#"*) echo "${line}"
;;
*) createSymlink ${line}
;;
esac
done < /some/file.txt
Note: I dropped the eval, which could be needed in some (rare!) cases (and are dangerous usually).
Note2: I added a "safer" IFS & read (-r, raw) but you can revert to your own if it is better suited. Note that it still reads line by line.
Note3: I took the habit of using always ${var} instead of $var ... works for me (easy to find out vars in complex text, and easy to see where they begin and end at all times) but not necessary here.
Note4: you can also change the test to : *"#"*) if some of the (comments?) lines can have spaces or tabs before the '#' (and none of the symlink lines does contain a '#')