Getting unknown operator error while using a function

Shell Programming and Scripting
1

Hi,

I wrote a function for the first time and not able to get the desired result.

I have requirement to execute 10 queries. For this i wrote a function like below.

function Command {
  typeset var SOL;
  if [ ${1:-0} -ne 0 ]; then CONNECTION="${CONNECTION} -e -time";
  else SOL="`nzsql ${CONNECTION} -c "${1}"`";
  fi;
  echo "${SOL}";
}

CHECK1="$(Command "SELECT COUNT(*) FROM ${DATABASE}..EMPLOYEE WHERE NVL(EMPLOYEE_ID, 0) <= 0;")"
if [ ${?} -ne 0 ] ; then
echo "CHECK1 : ${CHECK1}";
fi;

Above code is throwing an error.

.[16]: Command[10]: [: COUNT(*): unknown operator

I need to put the check 1 result in a file and need to attach the file to email and send once all the 10 checks performed.

Could any one please tell me what went wrong in this function!?

2

Your double quotes are not doing what's intended.

3

Thanks for the response Aia.

You mean to say, there is no problem with defined function and only the double quotes are causing issue?

Even i tried without quotes, it is not working.

4

The $() allows quite normal shell code.

But this one is problematic:

if [ ${1:-0} -ne 0 ]

$1 must be in quotes, and comparing it with a number looks odd.
This one looks more appropriate:

if [ -z "$1" ]
5

This is not correct, they are in fact doing what is intended.

Shell Command Language: Double-Quotes

In fact, the command is equivalent to:

CHECK1=$(Command "SELECT COUNT(*) FROM ${DATABASE}..EMPLOYEE WHERE NVL(EMPLOYEE_ID, 0) <= 0;")
6

Many Thanks MadeInGermany,

if [ -z "$1" ]

Above one is working.

Thanks Scrutinizer, for a detailed Double- Quotes explanation.

:b: Thanks again both of you!

7

Dear Samah,

You are trying to program with many levels of language at a time.

I suggest:

A) use set -x to debug

B) Try one level at a time

C) Try this command line:

eval $(Command "SELECT SYSDATE FROM DUAL")

D) If it works, try:

CHECK1=$(eval $(Command "SELECT SYSDATE FROM DUAL") )

E) The error parameter is $?, or ${?}; $(?) does not work.

Regards.

8

@AbellLuis, what is the sense of introducing "eval" here. I do not understand your statement that $(?) does not work, when no-one was suggesting that. Besides , the OP has already suggested that he found an answer in the thread..

9

@Scrutinizer
<<what is the sense of introducing "eval" here.>>
The function Command build the name of a call to SQL, but don't execute it; so it need to be parsed and executed. "eval" does it.
<< I do not understand your statement that $(?) does not work, when no-one was suggesting that.>>
The code quoted by the initiator, had this, may be a typegraphical mistake, or a visual one of mine.
<< Besides , the OP has already suggested that he found an answer in the thread.>>
It is my alternative answer, that's all.

Regards.

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