error in shell script: unexpected end of file - shell

The following script is showing me "unexpected end of file" error. I have no clue why am I facing this error. My all the quotes are closed properly.
#!/usr/bin/sh
insertsql(){
#sqlite3 /mnt/rd/stats_flow_db.sqlite <<EOF
echo "insert into flow values($1,$2,$3,$4,$5,$6,$7,$8,$9,$10,$11,$12,$13,$14,$15,$16,$17,$18)"
#.quit
}
for i in {1..100}
do
src_ip = "10.1.2."+$i
echo $src_ip
src_ip_octets = ${src_ip//,/}
src_ip_int = $src_ip_octets[0]*1<<24+$src_ip_octets[1]*1<<16+$src_ip_octets[2]*1<<8+$src_ip_octets[3]
dst_ip = "10.1.1."+$i
dst_ip_octets = ${dst_ip//,/}
dst_ip_int = $dst_ip_octets[0]*1<<24+$dst_ip_octets[1]*1<<16+$dst_ip_octets[2]*1<<8+$dst_ip_octets[3]
insertsql(1, 10000, $dst_ip, 20000, $src_ip, "2012-08-02,12:30:25.0","2012-08-02,12:45:25.0",0,0,0,"flow_a010105_a010104_47173_5005_1_50183d19.rrd",0,12,$src_ip_int,$dst_ip_int,3,50000000,80000000)
done

That error is caused by <<. When encountering that, the script tries to read until it finds a line which has exactly (starting in the first column) what is found after the <<. As that is never found, the script searches to the end and then complains that the file ended unexpectedly.
That will not be your only problem, however. I see at least the following other problems:
You can only use $1 to $9 for positional parameters. If you want to go beyond that, the use of the shift command is required or, if your version of the shell supports it, use braces around the variable name; e.g. ${10}, ${11}...
Variable assignments must not have whitespace arount the equal sign
To call your insertsql you must not use ( and ); you'd define a new function that way.
The cass to your insertsql function must pass the parameters whitespace separated, not comma separated.

A couple of problems:
There should be no space between equal sign and two sides of an assignment: e.g.,: dst_ip="10.1.1.$i"
String concatenation is not done using plus sign e.g., dst_ip="10.1.1.$i"
There is no shift operator in bash, no <<: $dst_ip_octets[0]*1<<24 can be done with expr $dst_ip_octets[0] * 16777216 `
Functions are called just like shell scripts, arguments are separated by space and no parenthesis: insertsql 1 10000 ...

That is because you don't follow shell syntax.
To ser variable you are not allowed to use space around = and to concatenate two parts of string you shouldn't use +. So the string
src_ip = "10.1.2."+$i
become
src_ip="10.1.2.$i"
Why you're using the string
src_ip_octets = ${src_ip//,/}
I don't know. There is absolutely no commas in you variable. So even to delete all commas it should look like (the last / is not required in case you're just deleting symbols):
src_ip_octets=${src_ip//,}
The next string has a lot of symbols that shell intepreter at its own way and that's why you get the error about unexpected end of file (especially due to heredoc <<)
src_ip_int = $src_ip_octets[0]*1<<24+$src_ip_octets[1]*1<<16+$src_ip_octets[2]*1<<8+$src_ip_octets[3]
So I don't know what exactly did you mean, though it seems to me it should be something like
src_ip_int=$(( ${src_ip_octets%%*.}+$(echo $src_ip_octets|sed 's/[0-9]\+\.\(\[0-9]\+\)\..*/\1/')+$(echo $src_ip_octets|sed 's/\([0-9]\+\.\)\{2\}\(\[0-9]\+\)\..*/\1/')+${src_ip_octets##*.} ))
The same stuff is with the next strings.

You can't do this:
dst_ip_int = $dst_ip_octets[0]*1<<24+$dst_ip_octets[1]*1<<16+$dst_ip_octets[2]*1<<8+$dst_ip_octets[3]
The shell doesn't do math. This isn't C. If you want to do this sort of calculation, you'll need to use something like bc, dc or some other tool that can do the sort of math you're attempting here.
Most of those operators are actually shell metacharacters that mean something entirely different. For example, << is input redirection, and [ and ] are used for filename globbing.

Related

Using a bash script to insert into an SQL table with $s

I'm using a bash script to make changes to an SQL database. One of the values i'm updating uses dollar signs. The current value being something like "$$$$$" and i need to change it to "$$$$$$$$$$". However, a a $ in a bash script is used for variables.
How can i allow this small section of my bash script to used a $ as a normal character?
function prep() {
DATE_STAMP=$(date +%m%d%Y)
log "Changing mask to 10 characters"
log "$(/opt/CPU/bin/connx-query -q "update TYPE set TYPE.MASK = '$$$$$$$$$$'")"
}
As it stands right now, its just replacing each dollar sign with some random number found earlier in my script.
Bash provides different types of quoting, each with different rules about substitution (single quote ', double quote ", here document/string <<<"string" and and $'.
The double quote (used in the log ... update) will enable variable substitution, replacing each pair of $$ with the current shell PID (looks like random number).
Few options:
Consider quoting each '$' to prevent expansion
log "$(/opt/CPU/bin/connx-query -q "update TYPE set TYPE.MASK = '\$\$\$\$\$\$\$\$\$\$'")"
Over thought my own question. I can just escape the $. '\$\$\$\$\$\$\$\$\$\$'

Why expanding on left hand side arg is leading to error

I accidentally wrote next assignment in one of my scripts:
$X=$(echo 'astring')
which fails with =astring: command not found.
The correct and intended assignment was X=$(echo 'astring') which works and sets X='astring'.
The question is what happens with the first one? Is $ trying to execute the result of the right hand side? And if that is so then why is it also incorporating = in it? I'm confused.
The behaviour of $X=$(echo 'astring') depends on the contents of $X. When it's empty (which it probably was), it expands to an empty string, and the remaining string is interpreted as a command
$X=$(echo 'astring')
=astring
If $X contains something, e.g. "astring", the string is expanded to
astring=astring
But it doesn't set the $astring variable as one might think, because of the order of expansions. Assignments are identified before any expansion happens. So, it's interpreted as a command again
astring=astring: command not found

Zsh array of strings in for loop

Am trying to print a bunch of strings in a script (in zsh) and it doesn't seem to work. The code would work if I place the array in a variable and use it instead. Any ideas why this doesn't work otherwise?
for string in (some random strings to print) ; echo $string
The default form of the for command in zsh does not use parentheses (if there are any they are not interpreted as part of the for statement):
for string in some random strings to show
do
echo _$string
done
This results in the following output:
_some
_random
_strings
_to
_show
So, echo _$string was run for each word after in. The list ends with the newline.
It is possible to write the whole statement in a single line:
for string in some random strings to show; do echo _$string; done
As usual when putting multiple shell commands in the same line, newlines just need to be replaced by ;. The exception here is the newline after do; while zsh allows a ; to be placed after do, it is usually not done, and in bash it would be a syntax error.
There are also several short forms available for for, all of which are equivalent to the default form above and produce the same output:
for single commands (to be exact: single pipelines or multiple pipelines linked with && or ||, where a pipeline can also be just a single command), there are two options:
the default form, just without do or done:
for string in some random strings to show ; echo _$string
without in but with parentheses, also without do or done
for string (some random strings to show) ; echo _$string
for a list of commands (like in the default form), foreach instead of for, no in, with parentheses and terminated by end:
foreach string (some random strings to show) echo _$string ; end
In your case, you mixed the two short forms for single commands. Due to the presence of in, zsh did not take the parentheses as a syntactic element of the for command. Instead they are interpreted as a glob qualifier. Aside from the fact that you did not intend any filename expansions, this fails for two reasons:
there is no pattern (with or without actual globs) before the glob qualifier. So any matching filename would have to exactly match an empty string, which is just not possible
but mainly "some random strings to print" is not a valid glob qualifier. You probably get an error like "zsh: unknown file attribute: i" (at least with zsh 5.0.5, it may depend on the zsh version).
Check the zsh forloop documentation:
for x (1 2 3); do echo $x; done
for x in 1 2 3; do echo $x; done
You are probably trying to do this:
for string in some random strings to print ;do
echo $string
done

What does this variable assignment do?

I'm having to code a subversion hook script, and I found a few examples online, mostly python and perl. I found one or two shell scripts (bash) as well. I am confused by a line and am sorry this is so basic a question.
FILTER=".(sh|SH|exe|EXE|bat|BAT)$"
The script later uses this to perform a test, such as (assume EXT=ex):
if [[ "$FILTER" == *"$EXT"* ]]; then blah
My problem is the above test is true. However, I'm not asking you to assist in writing the script, just explaining the initial assignment of FILTER. I don't understand that line.
Editing in a closer example FILTER line. Of course the script, as written does not work, because 'ex' returns true, and not just 'exe'. My problem here is only, however, that I don't understant the layout of the variable assignment itself.
Why is there a period at the beginning? ".(sh..."
Why is there a dollar sign at the end? "...BAT)$"
Why are there pipes between each pattern? "sh|SH|exe"
You probably looking for something as next:
FILTER="\.(sh|SH|exe|EXE|bat|BAT)$"
for EXT
do
if [[ "$EXT" =~ $FILTER ]];
then
echo $EXT extension disallowed
else
echo $EXT is allowed
fi
done
save it to myscript.sh and run it as
myscript.sh bash ba.sh
and will get
bash is allowed
ba.sh extension disallowed
If you don't escape the "dot", e.g. with the FILTER=".(sh|SH|exe|EXE|bat|BAT)$" you will get
bash extension disallowed
ba.sh extension disallowed
What is (of course) wrong.
For the questions:
Why is there a period at the beginning? ".(sh..."
Because you want match .sh (as extension) and not for example bash (without the dot). And therefore the . must be escaped, like \. because the . in regex mean "any character.
Why is there a dollar sign at the end? "...BAT)$"
The $ mean = end of string. You want match file.sh and not file.sh.jpg. The .sh should be at the end of string.
Why are there pipes between each pattern? "sh|SH|exe"
In the rexex, the (...|...|...) construction delimites the "alternatives". As you sure quessed.
You really need read some "regex tutorial" - it is more complicated - and can't be explained in one answer.
Ps: NEVER use UPPERCASE variable names, they can collide with environment variables.
This just assigns a string to FILTER; the contents of that string have no special meaning. When you try to match it against the pattern *ex*, the result is true assuming that the value of $FILTER consists the string ex surrounded by anything on either side. This is true; ex is a substring of exe.
FILTER=".(sh|SH|exe|EXE|bat|BAT)$"
^^
|
+---- here is the "ex" from the pattern.
As I can this is similar to regular expression pattern:
In regular expressions the string start with can be show with ^, similarly in this case . represent seems doing that.
In the bracket you have exact string, which represents what the exact file extensions would be matched, they are 'Or' by using the '|'.
And at the end the expression should only pick the string will '$' or end point and not more than.
I would say that way original author might have looked at it and implemented it.

Shell script input containing asterisk

How do I write a shell script (bash on HPUX) that receives a string as an argument containing an asterisk?
e.g. myscript my_db_name "SELECT * FROM table;"
The asterisk gets expanded to all the file names in the current directory, also if I assign a variable like this.
DB_QUERY="$2"
echo $DB_QUERY
The asterisk "*" is not the only character you have to watch out for, there's lots of other shell meta-charaters that can cause problems, like < > $ | ; &
The simple answer is always to put your arguments in quotes (that's the double-quote, " ) when you don't know what they might contain.
For your example, you should write:
DB_QUERY="$2"
echo "$DB_QUERY"
It starts getting awkward when you want your argument to be used as multiple parameters or you start using eval, but you can ask about that separately.
You always need to put double quotes around a variable reference if you want to prevent it from triggering filename expansion. So, in your example, use:
DB_QUERY="$2"
echo "$DB_QUERY"
In the first example, use single quotes:
myscript my_db_name 'SELECT * FROM table;'
In the second example, use double quotes:
echo "$DB_QUERY"

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