I wrote a short code in .bat file - but I get an error that the if command is incorrect.
my code:
#echo off
set arg1=%1
echo %arg1%
if %arg1% == rot(
echo yes)
the output is:
rot
The syntax of the command is incorrect.
When I looked for the way to write if statement in bat file I saw lots of syntax options but non of them works for me.
I tried:
if "%arg1%" == rot
if "%arg1%" == "rot"
if %arg1% == "rot"
But no one was ok.
I Will be very happy to get help. Thanks.
Your code must look like:
#echo off
set arg1=%1
echo %arg1%
IF "%arg1%" == "rot" (
echo yes
)
both arguments must be in double quotes:
test.bat rot
rot
yes
Related
Totally new in bash programming and got a problem like this:
N empty files should be created. The file sum N and also the file name should be given from users via command line.
no loop and getopts should be used.
I've tried something like this which i found from google but it doesn't works. I got confused between linux and windows bash scripts. Hope someone can help me with the problem. Thank you!
#!/bin/bash
echo $N
:start
if [ $N > 0 ]
then
touch xyzfile_$N
$N = $N - 1
pause
goto start
else
then
echo "no data will be created"
fi
The command line and the result should be (for example if N=7) look like this:
command line:
./createfiles -n filename 7
expected result:
filename_1,filename_2,filename_3...filename_7
You can do this using a recursive shell function:
fn() {
# add some sanity checks to check parameters
touch "$1_${2}"
(($2 > 1)) && fn "$1" $(($2 - 1))
}
This part is a recursive call:
(($2 > 1)) && fn "$1" $(($2 - 1))
Where it basically calls itself by decrementing 1 from 2nd argument as long as $2 is greater than 1.
Then just call it as:
fn filename 7
It will create 7 files as:
filename_1 filename_2 filename_3 filename_4 filename_5 filename_6 filename_7
I ended up the question with a for-loop like this:
createfile() {
name="$1"
num="$2"
for((i=1;i<=num;i++))
do
touch "$1_${i}"
done
}
createfile $1 ${2}
It works great actually. And now I tried to solve the problem with getopts and it seems a function call doesn't work in the case option
When I am executing the below script, I am getting the following error :-
The script executes infintely and below line is printed everytime.
"line 9: 1=1+2: command not found". Why?
#!/bin/bash
echo "Script 1 - Linux Scripting Book"
x=1
while [ $x -le 45 ]
do
echo x : $x
$x=$x+2
done
echo "End Of Script 1"
exit 0
Also if I change the $x=$x+2 to x+$x+2 then also I am getting the below error.
line 6: [: 1+2: integer expression expected
Same script when executed like this runs fine.
#!/bin/bash
echo "Script 1 - Linux Scripting Book"
x=1
while [ $x -le 45 ]
do
echo x : $x
let x=x+2
done
echo "End Of Script 1"
exit 0
You get line 9: 1=1+2: command not found because 1=1+2 is what $x=$x+2 is expanded into.
Use expr or let or ((...)) for integer calculations and bc for floating point:
let x=x+2
((x=x+2)) #same as above
((x+=2)) #same
((x++)) #if adding just one
((++x)) #if adding just one
x=$((x+2))
x=`expr $x + 2` #space before and after +
x=$(echo $x+2|bc) #using bc
x=$(echo $x+2.1|bc) #bc also works with floating points (numbers with decimals)
Since this part of the question isn't cleared yet, and not fine to post in a comment, I add this partial answer:
x=1; for i in 1 2 3 ; do x=$x+2; echo $x; done
1+2
1+2+2
1+2+2+2
As a side note: Don't use exit 0 at the end of your script without a good reason. When the script is done, it exits by itself without your help. The exit status will be the exit status of the last command performed, in your case a simple echo, which will almost always succeed. In the rare cases it fails, you will probably without intention hide that failure.
If you source the script, the exit will throw you out of your running shell.
But you can rewrite your while loop like this:
x=0
while (($((x)) < 9))
do
echo x : $x
x=$x+2
done
echo $((x))
x : 0
x : 0+2
x : 0+2+2
x : 0+2+2+2
x : 0+2+2+2+2
10
Because that's not the Bourne shell syntax for setting a variable; it looks more like Perl or PHP. The $ is used for parameter expansion and is not part of the variable name. Variable assignment simply uses =, and let evaluates arithmetic expressions (much like $((expression))). Another syntax that should work is x=$((x+2)). Note that these arithmetic evaluations are a bash feature; standard unix shells might require use of external tools such as expr.
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
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
Can someone explain why these echo commands doesn't output [#10] and so on?
# echo [#10]
1
# echo [#11]
1
# echo [#12]
1 2
# echo [#13]
1
# echo [#14]
1
You have a file named "1" and a file named "2" in your current directory.
The shell is performing pattern matching on the glob patterns before handing the results to echo. [#10] is a character class containing a #, a 1 and a 0.
See http://www.gnu.org/software/bash/manual/bashref.html#Pattern-Matching
If you want the literal [#10], etc, you have to enclose it in quotes, single or double doesn't matter.
(to answer the question in your last comment)
You could use the printf(1) command:
printf "Error: %s went wrong. Error code [#%d]\n" "something" $[10+2]
The $[10+2] is here to show how to do arithmetic in shell. You could replace "something" with e.g. $somevariable ...