I was trying to execute the following commands.
de="hello world"
if [ $de -eq "hi" ]; then
....
....
because of the space between hello and the world, it out put an error. but if I define de="helloworld" it works fine. can you please tell me if there is a way I can use if statement with sentences that have spaces in it?
Quote the variable name,
de="hello world"
if [ "$de" = "hi" ]; then
-eq is for comparing numbers, so use = for text. See here for a nice overview on how to do various comparisions in bash.
Related
I'm practicing bash and honestly, it is pretty fun. However, I'm trying to write a program that compares an array's value to a variable and if they are the same then it should print the array's value with an asterisk to the left of it.
#!/bin/bash
color[0]=red
color[1]=blue
color[2]=black
color[3]=brown
color[4]=yellow
favorite="black"
for i in {0..4};do echo ${color[$i]};
if {"$favorite"=$color[i]}; then
echo"* $color[i]"
done
output should be *black
There's few incorrect statements in your code that prevent it from doing what you ask it to. The comparison in bash is done withing square brackets, leaving space around them. You correctly use the = for string comparison, but should enclose in " the string variable. Also, while you correctly address the element array in the echo statement, you don't do so inside the comparison, where it should read ${color[$i]} as well. Same error in the asterisk print. So, here a reworked code with the fixes, but read more below.
#!/bin/bash
color[0]=red
color[1]=blue
color[2]=black
color[3]=brown
color[4]=yellow
favorite=black
for i in {0..4};do
echo ${color[$i]};
if [ "$favorite" = "${color[$i]}" ]; then
echo "* ${color[$i]}"
fi
done
While that code works now, few things that probably I like and would suggest (open to more expert input of course by the SO community): always enclose strings in ", as it makes evident it is a string variable; when looping an array, no need to use index variables; enclose variables always within ${}.
So my version of the same code would be:
#!/bin/bash
color=("red" "blue" "black" "brown" "yellow")
favorite="black"
for item in ${color[#]}; do
echo ${item}
if [ "${item}" = "${favorite}" ]; then
echo "* $item"
fi
done
And a pointer to the great Advanced Bash-Scripting Guide here: http://tldp.org/LDP/abs/html/
I have a small scripts which verifies some conditions on a database server. I want to mock failures on all of those conditions to test the script, so I added the following line:
./print_results ${VAR1} ${VAR2} ... ${VARN}
If any of the variables has a value different than ZERO it because it failed.
so just for testing purpouses I added the line:
VAR1=1 ; VAR2=1 ; ... ; VARN=1
But I need to edit the file every time I want to replace the real results with the fake ones.
What's wrong with this?
[! -z $1 ] && [ "$1" == "Y"] && { echo "Debugging is ACTIVE" ; VAR1=1 ; ... ; VAR2=1 ; }
I want to have the VAR1..N = 1 after passing that line.
Thanks.
The problem is that [ is a command, but [! is not. It is probably cleaner to write your code:
test "{$1}" == Y && { echo "Debugging is ACTIVE"; VAR1=1 VAR2=1 ...; }
No need for semi-colons between the variable assignments, but they don't hurt.
This is one of the warts of sh. For some reason, it was thought to be a good idea to use the symbol [ for a command and pass it ] as an argument, trying to mimic braces in the language. Unfortunately, this leads to a great deal of confusion similar to that demonstrated in this question. It is far better to avoid [ completely and always spell it test. These two are functionally identical (except that the [ command must have ] as the final argument), and using test is much cleaner. (Would you expect test! to work?, or would you recognize that it needs to be written as ! test?)
Need a space between the "Y" and the ]. The non-zero test is pointless, but also requires a space between the [ and the !.
[ "$1" == "Y" ] && { echo "Debugging is ACTIVE" ; VAR1=1 ; ... ; VAR2=1 ; }
Also did you consider just writing this as an if...fi block?
bash provides a way to supply default values for parameters that aren't otherwise set. Presumably, your code has lines like
VAR1=$1
VAR2=$2
VAR3=$3
Replace them with
VAR1=${1-1}
VAR2=${2-1}
VAR3=${3-1}
If $1 is unset, for instance, VAR1 will be assigned the value of 1 instead of the value of $1.
I'm writing bash scripts that need to work both on Linux and on Mac.
I'm writing a function that will return a directory path depending on which environment I'm in.
Here is the pseudo code:
If I'm on a Mac OS X machine, I need my function to return the path:
/usr/local/share/
Else if I'm on a Linux machine, I need my function to return the path:
/home/share/
Else, you are neither on a Linux or a Mac...sorry.
I'm very new to Bash, so I apologize in advance for the really simple question.
Below is the function I have written. Whether I'm on a Mac or Linux, it always returns
/usr/local/share/
Please take a look and enlighten me with the subtleties of Bash.
function get_path(){
os_type=`uname`
if [ $os_type=="Darwin" ]; then
path="/usr/local/share/"
elif [ $os_type=="Linux" ]; then
path="/home/share/"
else
echo "${os_type} is not supported"
exit 1
fi
echo $path
}
You need spaces around the operator in a test command: [ $os_type == "Darwin" ] instead of [ $os_type=="Darwin" ]. Actually, you should also use = instead of == (the double-equal is a bashism, and will not work in all shells). Also, the function keyword is also nonstandard, you should leave it off. Also, you should double-quote variable references (like "$os_type") just in case they contain spaces or any other funny characters. Finally, echoing an error message ("...not supported") to standard output may confuse whatever's calling the function, because it'll appear where it expected to find a path; redirect it to standard error (>&2) instead. Here's what I get with these cleaned up:
get_path(){
os_type=`uname`
if [ "$os_type" = "Darwin" ]; then
path="/usr/local/share/"
elif [ "$os_type" = "Linux" ]; then
path="/home/share/"
else
echo "${os_type} is not supported" >&2
exit 1
fi
echo "$path"
}
EDIT: My explanation of the difference between assignments and comparisons got too long for a comment, so I'm adding it here. In many languages, there's a standard expression syntax that'll be the same when it's used independently vs. in test. For example, in C a = b does the same thing whether it's alone on a line, or in a context like if ( a = b ). The shell isn't like that -- its syntax and semantics vary wildly depending on the exact context, and it's the context (not the number of equal signs) that determines the meaning. Here are some examples:
a=b by itself is an assignment
a = b by itself will run a as a command, and pass it the arguments "=" and "b".
[ a = b ] runs the [ command (which is a synonym for the test command) with the arguments "a", "=", "b", and "]" -- it ignores the "]", and parses the others as a comparison expression.
[ a=b ] also runs the [ (test) command, but this time after removing the "]" it only sees a single argument, "a=b" -- and when test is given a single argument it returns true if the argument isn't blank, which this one isn't.
bash's builtin version of [ (test) accepts == as a synonym for =, but not all other versions do.
BTW, just to make things more complicated bash also has [[ ]] expressions (like test, but cleaner and more powerful) and (( )) expressions (which are totally different from everything else), and even ( ) (which runs its contents as a command, but in a subshell).
You need to understand what [ means. Originally, this was a synonym for the /bin/test command. These are identical:
if test -z "$foo"
then
echo "String '$foo' is null."
fi
if [ -z "$foo" ]
then
echo "String '$foo' is null."
fi
Now, you can see why spaces are needed for all of the parameters. These are parameters and not merely boolean expressions. In fact, the test manpage is a great place to learn about the various tests. (Note: The test and [ are built in commands to the BASH shell.)
if [ $os_type=="Darwin" ]
then
This should be three parameters:
"$os_type"
= and not ==
"Darwin"
if [ "$os_type" = "Darwin" ] # Three parameters to the [ command
then
If you use single square brackets, you should be in the habit to surround your parameters with quotation marks. Otherwise, you will run into trouble:
foo="The value of FOO"
bar="The value of BAR"
if [ $foo != $bar ] #This won't work
then
...
In the above, the shell will interpolate $foo and $bar with their values before evaluating the expressions. You'll get:
if [ The value of FOO != The value of BAR ]
The [ will look at this and realize that neither The or value are correct parameters, and will complain. Using quotes will prevent this:
if [ "$foo" != "$bar" ] #This will work
then
This becomes:
if [ "The value of FOO" != "The value of BAR" ]
This is why it's highly recommended that you use double square brackets for your tests: [[ ... ]]. The test looks at the parameters before the shell interpolates them:
if [[ $foo = $bar ]] #This will work even without quotation marks
Also, the [[ ... ]] allows for pattern matching:
if [[ $os_type = D* ]] # Single equals is supported
then
path="/usr/local/share/"
elif [[ $os_type == L* ]] # Double equals is also supported
then
path="/home/share/"
else
echo "${os_type} is not supported"
exit 1
fi
This way, if the string is Darwin32 or Darwin64, the if statement still functions. Again, notice that there has to be white spaces around everything because these are parameters to a command (actually, not anymore, but that's the way the shell parses them).
Adding spaces between the arguments for the conditionals fixed the problem.
This works
function get_path(){
os_type=`uname`
if [ $os_type == "Darwin" ]; then
path="/usr/local/share/"
elif [ $os_type == "Linux" ]; then
path="/home/share/"
else
echo "${os_type} is not supported"
exit 1
fi
echo $path
}
It is my understanding that when writing a Unix shell program you can iterate through a string like a list with a for loop. Does this mean you can access elements of the string by their index as well?
For example:
foo="fruit vegetable bread"
How could I access the first word of this sentence? I've tried using brackets like the C-based languages to no avail, and solutions I've read online require regular expressions, which I would like to avoid for now.
Pass $foo as argument to a function. Than you can use $1, $2 and so on to access the corresponding word in the function.
function try {
echo $1
}
a="one two three"
try $a
EDIT: another better version is:
a="one two three"
b=( $a )
echo ${b[0]}
EDIT(2): have a look at this thread.
Using arrays is the best solution.
Here's a tricky way using indirect variables
get() { local idx=${!#}; echo "${!idx}"; }
foo="one two three"
get $foo 1 # one
get $foo 2 # two
get $foo 3 # three
Notes:
$# is the number of parameters given to the function (4 in all these cases)
${!#} is the value of the last parameter
${!idx} is the value of the idx'th parameter
You must not quote $foo so the shell can split the string into words.
With a bit of error checking:
get() {
local idx=${!#}
if (( $idx < 1 || $idx >= $# )); then
echo "index out of bounds" >&2
return 1
fi
echo "${!idx}"
}
Please don't actually use this function. Use an array.
I'm attempting to write a KornShell (ksh) function that uses printf to pad a string to a certain width.
Examples:
Call
padSpaces Hello 10
Output
'Hello '
I currently have:
padSpaces(){
WIDTH=$2
FORMAT="%-${WIDTH}.${WIDTH}s"
printf $FORMAT $1
}
Edit: This seems to be working, in and of itself, but when I assign this in the script it seems to lose all but the first space.
TEXT=`padSpaces "TEST" 10`
TEXT="${TEXT}A"
echo ${TEXT}
Output:
TEST A
I'm also open to suggestions that don't use printf. What I'm really trying to get at is a way to make a fixed width file from ksh.
Your function works fine for me. Your assignment won't work with spaces around the equal sign. It should be:
SOME_STRING=$(padSpaces TEST 10)
I took the liberty of replacing the backticks, too.
You don't show how you are using the variable or how you obtain the output you showed. However, your problem may be that you need to quote your variables. Here's a demonstration:
$ SOME_STRING=$(padSpaces TEST 10)
$ sq=\'
$ echo $sq$SOME_STRING$sq
'TEST '
$ echo "$sq$SOME_STRING$sq"
'TEST '
Are you aware that you define a function called padSpaces, yet call one named padString? Anyway, try this:
padString() {
WIDTH=$2
FORMAT="%-${WIDTH}s"
printf $FORMAT $1
}
Or, the more compact:
padString() {
printf "%-${2}s" $1
}
The minus sign tells printf to left align (instead of the default right alignment). As the manpage states about the command printf format [ arg ... ],
The arguments arg are printed on standard output in accordance with the
ANSI-C formatting rules associated with the format string format.
(I just installed ksh to test this code; it works on my machineTM.)