Hi everybody,
I'm sorry If I ask a silly question. I have a simple code like this
I have the following error:
Can anyone explain for me why I have this error, and how can I correct it?
Thanks in advance.
First thing, indent your code properly.
#!/bin/bash
pattern()
{
if [ $1 -eq 100 ]
then
if [ $2 -eq 100 ]
then
echo debug 1
else
echo debug 2
fi
elif [ $1 -eq 50 ]
then
echo debug 3
else
echo debug 4
fi
}
pattern 2 3
And if you want to use "then" in the same line as "if" then I think you should put a semi-colon after the condition, something like this: "if [ condition ]; then"
I'm sorry. I tried to indent, but it seems that the space is automatically deleted. I will tried with your advice.
Thanks a lot.
---------- Post updated at 11:11 PM ---------- Previous update was at 11:03 PM ----------
I have another problem, If my code is like below:
verify()
{// do something
return $(grep -q 'VERIFICATION SUCCESSFUL' result.txt)
}
if [ verify 'a Condition' ]; thenecho debug
fi
Then I have the following error:
line 37: [: verify: unary operator expected
Would you please explain for me?
Try something like this:
verify()
{
// do something
grep -q 'VERIFICATION SUCCESSFUL' result.txt
if [ $? -eq 0 ]; then
echo debug
fi
}
Sorry, but what is $?.
I simplified the code, actually I need to use the function verify at many condition, I cannot do anything with the if statement inside the function.
Return is exit from function. Not for return values. Stdout is method to return values.
$? = last command exit status, 0=ok, <>0 not so ok.
pattern()
{
val1=$1
val2=$2
(( val1 == 100 && val2 == 100)) && return 1
(( val1 == 100 )) && return 2
(( val1 == 50 )) && return 3
return 4
}
some()
{
input="$1"
infile="$2"
data=$(grep "$input" $infile 2>/dev/null )
stat=$?
# if stat = 0, grep found it
# data is empty, nothing founded or if founded, data include it
[ "$data" = "" ] && echo "0" && return 1
# or
((stat !=0 )) && echo "0" && return 1
#
#return passwd line
echo "$data"
}
##############
result=$(some "xyz" /etc/passwd)
stat=$?
echo "stat:$stat" # 1 if xyz not in passwd
echo "result:$result" # 0 or passwd line
result=$(some "root" /etc/passwd)
stat=$?
echo "stat:$stat" # 1 if root not in passwd
echo "result:$result" # 0 or passwd line
pattern 100 100
echo $?
pattern 50 1
echo $?
1 Like
$? is the exit status of the last command executed. Check the below post:
You may try something like this:
#! /bin/bash
verify()
{
# do something
grep -q 'VERIFICATION SUCCESSFUL' result.txt
verify_exit_status=$?
}
verify
if [ $verify_exit_status -eq 0 ]; then
echo debug
fi
By the way, what did you mean by this:
1 Like
Thank you very much for your replies.
What I want to do is: I will run a program and write the result of its execution to the file "result.txt". Then I grep in the output file, if it contains the phrase "VERIFICATION SUCCESSFUL", it means the verification process is successful.
Then I need to run a complicated multi-level if statement whose conditions are the verification result. Something like this:
pattern()
{ if [ verify 'condition1' ] then
if [ verify 'condition2' ] then
echo debug 1
else
echo debug 2
fi
elif [ verify 'condition3' ] then
echo debug 3
else
echo debug 4
fi
}
I can test the exist of the phrase by this way:
if grep -q 'VERIFICATION SUCCESSFUL' result.txt
But how can I make a wrapper of it, to use the verify function as above?
You're rather ambiguous about this verify() function. It takes an argument? I don't see where...
'if' takes a command as an argument, [ is a command, grep is also a command. A function by default returns the exit code of it's last command...
verify() {
local condition=$1
# do something with $condition
grep -q 'SUCCESS' results.txt
}
if verify 'condition1'; then
whatever
elif verify 'condition2'; then
something else...
else
i don't know ...
fi
1 Like
Yes, it worked for me. Don't know how to thank you enough.
I didn't know that is also a command, i thought that it just like "if (condition)" in C.
Just one last question, if [ is a command, what is its meaning in this statement if [ $i -gt $j ]
[ is a command, but in bash it's a shell builtin.
$ ls -l /usr/bin/[
-rwxr-xr-x 1 root root 28300 Apr 28 2010 /usr/bin/[
test $i greater than $j. So that exit is 0 (success) if $i > $j, and non-zero otherwise.
Since you're using bash, it's worth noting that [[ is a better and preferred test. Quoting isn't needed, and it offers more (regex matching)..
# using "[":
if [ "$i" -gt "$j" ]; then
# using bash's "[["
if [[ $i -gt $j ]]; then
# using arithmetic
if ((i > j)); then
1 Like
I understand. Thanks again.
After if it's always command in shells: [, [[, (( are only some commands.
if commandline
then # exit status 0
:
else # exit status != 0
:
fi
[ and test are same command, usually builtin command today. If you use command [, then the last argument must be ] = line looks like as in some programming languages.
grep "some" file >/dev/null 2>&1
stat=$?
if [ $stat = 0 ] ; then
echo founded
else
echo not
fi
Same result [ and test are same command
grep "some" file >/dev/null 2>&1
stat=$?
if test $stat = 0 ; then
echo founded
else
echo not
fi
Same result, grep output is not needed => output to the /dev/null
if grep "some" file >/dev/null 2>&1 ; then
echo founded
else
echo not
fi
Same result
grep "some" file >/dev/null 2>&1 && echo founded || echo not
1 Like