Do you know why below shell script just generates the output file named 'Loading_EMP.sql' instead of 'Loading_1_EMP.sql'?
#!/bin/bash
JOBID="1"
TABLE="EMP"
echo 'test' > Loading_$JOBID_$TABLE.sql;
# Output
Loading_EMP.sql
# Expected Output
Loading_1_EMP.sql
echo 'test' > Loading_${JOBID}_${TABLE}.sql;
should do it, or better
echo 'test' > "Loading_${JOBID}_${TABLE}.sql" # to avoid word splitting
In Loading_$JOBID_$TABLE, shell trates $JOBID_ as a single variable and since it is not set, it substitutes $JOBID_ with nothing resulting in Loading_EMP.sql
This [ answer ] must read for you.
_ is not a special character in shell. So the shell sees this as you echoing a $JOBID_ variable, which is undefined, and thus empty. Instead, you should explicitly show the shell what your variables are by using {}:
echo 'test' > Loading_${JOBID}_${TABLE}.sql;
# Here -----------------^-----^--^-----^
The interpreter thinks you are referring to a variable named JOBID_. Enclose the name with {}.
echo 'test' > Loading_${JOBID}_$TABLE.sql
Related
If I am in a Linux terminal and I start setting variables such as export AGE=45.
Then I have a script to read user data from terminal variables and process it, is this possible to do?
IE:
user#linux$ export AGE=45
user#linux$ ./age.sh
#script asks for input
read -p "what is your age?" scriptAGE
#user inputs variable set in terminal
$AGE
#echo output
echo "your age is: " $scriptAGE"
#should say your age is: 45
There is no such thing as a terminal variable. read just assigns a string to your variable scriptAGE.
If this string contains some $NAME you want to expand, you could apply eval to it, but this is of course extremely dangerous because of possible code injection.
A safer way to do this is using envsubst, but this requires that the variables to be substituted must be environment variables. In your case, AGE is in the environment, so this condition is met.
In your case, you would have to do therefore a
envsubst <<<"$scriptAGE"
which would print on stdout the content of scriptAGE with all environment variables in it substituted.
Variables are not expanded in input, only in the script itself.
You could use eval to force it to process the variable value as shell syntax.
eval "echo 'your age is:' $scriptAGE"
But this will also process other shell syntax. If they enter $AGE; rm * it will say their age is 45 and then delete all their files.
you could just do
age=$1
echo "Your age is $1"
where $1, $2, $3, .., $N are the passed arguments by order
And then run your script
bash script sh Noureldin
For more Info read this:
passing names args
In shell script, How can I set 0 as default value to parameter.
param2=0
${2:-0}
and
${2:-param2}
These result is 0: not found
and
param=$5
S{1:-$param}
is active. but $0 mean shell script name. I want to use number 0.
echo "param: $param" # param is not defined
echo "param: ${param-0}" # use 0 as default
param=99 # set a value for param
echo "param: ${param-0}" # ignore the default
output:
param:
param: 0
param: 99
To understand problems like this, add set -x before the code. That will tell you that ${2:-0} is expanded to 0. The shell thinks this is a command and tries to execute it.
This will be easier to understand if you look at this code:
cmd=echo
${cmd} test
If you run this, ${cmd} is replaced with echo.
In your code, you probably want to assign the result of ${2:-0} to a new variable:
foo=${2:-0}
I am trying to create a shell script for logs and trying to append data into a text file. I have write this sample "test.sh" code for testing:
#!/bin/sh -e
touch /home/sample.txt
SPTH = '/home/sample'.txt
echo "MY LOG FILE" >> "$SPTH"
echo "DUMP started at $(date +'%d-%m-%Y %H:%M:%S')" >> /home/sample.txt
echo "DUMP finished at $(date +'%d-%m-%Y %H:%M:%S')" >> /home/sample.txt
but in above code all lines are working correct except one line of code i.e.
echo "MY LOG FILE" >> "$SPTH"
It is giving error:
test.sh: line 6: : No such file or directory
I want to replace this full path of file "/home/sample.txt" to variable "$SPATH".
I am executing my shell script using
sh test.sh
What I am doing wrong.
Variable assignments in bash shell does not allow you to have spaces within. It will be actually interpreted as command with = and the subsequent keywords as arguments to the first word, which is wrong.
Change your code to
SPTH="/home/sample.txt"
That is the reason why SPTH was not assigned to the actual path you intended it to have. And you have no reason to have single-quote here and excluding the extension part. Using it fully within double-quotes is absolutely fine.
The syntax for the command line is that the first token is a command, tokens are separated by whitespace. So:
SPTH = '/home/sample'.txt
Has the command as SPTH, the second token is =, and so on. You might think this is daft, but most shells behave like this for historical reasons.
So you need to remove the whitespace:
SPTH='/home/sample'.txt
If I am writing a bash script, and I choose to use a config file for parameters. Can I still pass in parameters for it via the command line? I guess I'm asking can I do both on the same command?
The watered down code:
#!/bin/bash
source builder.conf
function xmitBuildFile {
for IP in "{SERVER_LIST[#]}"
do
echo $1#$IP
done
}
xmitBuildFile
builder.conf:
SERVER_LIST=( 192.168.2.119 10.20.205.67 )
$bash> ./builder.sh myname
My expected output should be myname#192.168.2.119 and myname#10.20.205.67, but when I do an $ echo $#, I am getting 0, even when I passed in 'myname' on the command line.
Assuming the "config file" is just a piece of shell sourced into the main script (usually containing definitions of some variables), like this:
. /etc/script.conf
of course you can use the positional parameters anywhere (before or after ". /etc/..."):
echo "$#"
test -n "$1" && ...
you can even define them in the script or in the very same config file:
test $# = 0 && set -- a b c
Yes, you can. Furthemore, it depends on your architecture of script. You can overwrite parametrs with values from config and vice versa.
By the way shflags may be pretty useful in writing such script.
The colon command is a null command.
The : construct is also useful in the conditional setting of variables. For example,
: ${var:=value}
Without the :, the shell would try to evaluate $var as a command. <=???
I don't quite understand the last sentence in above statement. Can anyone give me some details?
Thank you
Try
var=badcommand
$var
you will get
bash: badcommand: command not found
Try
var=
${var:=badcommand}
and you will get the same.
The shell (e.g. bash) always tries to run the first word on each command line as a command, even after doing variable expansion.
The only exception to this is
var=value
which the shell treats specially.
The trick in the example you provide is that ${var:=value} works anywhere on a command line, e.g.
# set newvar to somevalue if it isn't already set
echo ${newvar:=somevalue}
# show that newvar has been set by the above command
echo $newvar
But we don't really even want to echo the value, so we want something better than
echo ${newvar:=somevalue}.
The : command lets us do the assignment without any other action.
I suppose what the man page writers meant was
: ${var:=value}
Can be used as a short cut instead of say
if [ -z "$var" ]; then
var=value
fi
${var} on its own executes the command stored in $var. Adding substitution parameters does not change this, so you use : to neutralize this.
Try this:
$ help :
:: :
Null command.
No effect; the command does nothing.
Exit Status:
Always succeeds.