Error when trying to compare strings in shell script - linux

I have the following in my script:
fw=$2
if [ "$fw" == "zf" ]
then
public_dir=$dir/current/public
else
public_dir=$dir/current
fi
But I get the following error when I run it:
./generatevhost.sh: 15: [: zf: unexpected operator
I've lookup pages explaining how to compare strings and they describe like so. What am I doing wrong?

I believe you're not using BASH or ksh. Try this if condition by replacing == by =:
fw="$2"
if [ "$fw" = "zf" ]
then
public_dir=$dir/current/public
else
public_dir=$dir/current
fi

Related

linux shell script variable "not found"

Just trying to exit a loop once no input is entered at the prompt, but I'm having trouble testing for the value in an if statement?
CODE:
SQL="?";
while true
do
if [ "$SQL" == "" ]
then
break
else
read -p "SQL: " SQL
clear
php -f sql.php "$SQL"
fi
done
OUTPUT:
sql.sh: 5: [: ?: unexpected operator
SQL:
Although it looks like part of a scripting language, [ is actually the name of a command, also known as test. The if statement runs that command, and acts based on its result. (The same is true, incidentally, of the true in your while loop - it's a command that "does nothing, successfully".)
As such, you need a space between the command and its parameters, as well as between the if and the command. You also need to use the correct arguments; the standard spelling for comparing two strings is = rather than ==
if [ "$SQL" = "" ]
Which is equivalent to:
if test "$SQL" = ""

Syntax error near unexpected token `done' whis done written

I have a problem with the done.
It says I have some typo error but I can't figure what's wrong at all.
Here is the code:
#./bin/bash
until [$err == 0];
do
java -Xms512m -Xmx512m -cp lib/*:lib/uMad/*:mysql-connector-java-5.1.15-bin.jar:l2jfrozen-core.jar com.l2jfrozen.gameserver.GameServer
err=$?
sleep 5
done
Your shebang line is wrong. #./bin/bash will not execute bash.
It should read #!/bin/bash. You are probably using a shell other than bash to invoke this script.
Also, beware that the [$err == 0] line expands the value of $err, which is probably an empty string, unless it has been exported. If it's empty, this will result in an error, because Bash will be interpreting [ == 0].
The safest approach is this:
unset err
until [ "$err" == "0" ];
do
# etc...
done
From my experience when working with brackets and if loops, you need proper spacing and double, not single brackets. There needs to be space on each side of the double brackets with the exception of the semi-colon. Here is an example block:
#!/bin/bash
err=5
until [[ $err == 0 ]]; do
((err-=1));
echo -e "$err\n";
sleep 3
done
I do not see why the same would not apply to a do until loop.
You're probably aware but your heading has a period in it instead of a shebang.
#./bin/bash

Having trouble with simple Bash if/elif/else statement

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
}

how to declare variable name with "-" char (dash ) in linux bash script

I wrote simple script as follow
#!/bin/bash
auth_type=""
SM_Read-only="Yes"
SM_write-only="No"
echo -e ${SM_Read-only}
echo -e ${SM_Write-only}
if [ "${SM_Read-only}" == "Yes" ] && [ "${SM_Write-only}" == "Yes" ]
then
auth_type="Read Write"
else
auth_type="Read"
fi
echo -e $auth_type
And when i execute it i got following output with errors.
./script.bash: line 5: SM_Read-only=Yes: command not found
./script.bash: line 6: SM_write-only=No: command not found
only
only
Read
Any one know correct way to declare the variable with "-" (dash)?
EDIT:
have getting response from c code and evaluate the variables for example
RESP=`getValue SM_ Read-only ,Write-only 2>${ERR_DEV}`
RC=$?
eval "$RESP"
from above scripts code my c binary getValue know that script want Read-only and Write-only and return value to script.So during eval $RESP in cause error and in my script i access variable by
echo -e ${SM_Read-only}
echo -e ${SM_Write-only}
which also cause error.
Rename the variable name as follows:
SM_Read_only="Yes"
SM_write_only="No"
Please, don't use - minus sign in variable names in bash, please refer to the answer, on how to set the proper variable name in bash.
However if you generate the code, based on others output, you can simply process their output with sed:
RESP=$(getValue SM_ Read-rule,Write-rule 2>${ERR_DEV}|sed "s/-/_/g")
RC=$?
eval "$RESP"
- is not allowed in shell variable names. Only letters, numbers, and underscore, and the first character must be a letter or underscore.
I think you cant have a dash in your variables names, only letters, digits and "_"
Try:
SM_Read_only
Or
SM_ReadOnly

Shell script showing errors while executing in Linux

SCRIPT :
IMAGE=$imgvalue;
if [ $imgvalue :=1 ]
then
echo DO=ABC;
elif [ $imgvalue :=2 ]
then
echo DO=ETC;
elif [ $imgvalue :=3 ]
then
echo DO=XYZ;
else
echo "$imgvalue is unsupported";
exit 1;
fi
In the script above, IMAGE=1, IMAGE=2, IMAGE=3 whatever may be the value I have assigned. It's showing only DO=ABC. Other conditions not working. Can anyone explain what's wrong with my script?
If $imgvalue is not an empty string, your first test is a syntax error, so I am assuming it is empty in the tests you are doing. In that case, your first test is equivalent to:
if [ :=1 ]
which is always true because :=1 is not an empty string. You probably meant to write:
if [ "$imgvalue" = 1 ]

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