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Top Forums Shell Programming and Scripting problem with single quotes in a string and findbug Post 302102446 by bob122480 on Wednesday 10th of January 2007 01:23:14 PM
Old 01-10-2007
problem with single quotes in a string and findbug in ksh

I'm having trouble manipulating a string that contains single quotes (') in it. I'm writing a ksh script to parse in a few queries from a config file, such as this:

findbug \(\(Project 'in' "Deployment,HDRCI,LHS,LSS,WUCI" '&&' Status 'in' "N" '&&' New_on 'lessthan' "070107" \)\) '&&' \(Class 'isequal' "Development" \)

if i physically call this command it works fine and I get my database results, but I am parsing this string from the config file and storing it in a string variable and then trying to use it and it doesn't work.

For instance I have a variable NewQuery which contains the string.
I do echo $NewQuery and this prints out:
findbug \(\(Project 'in' "Deployment,HDRCI,LHS,LSS,WUCI" '&&' Status 'in' "N" '&&' New_on 'lessthan' "070107" \)\) '&&' \(Class 'isequal' "Development" \)

but if i just do $NewQuery to execute that command it complains:
findbug: Syntax Error: Bad operator 'in'.

Usage: findbug [options] [expr \&\& expr \|\| expr ...]

any idea what i'm doing wrong?

Last edited by bob122480; 01-10-2007 at 03:08 PM..
 

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EXPR(1) 						      General Commands Manual							   EXPR(1)

NAME
expr - evaluate arguments as an expression SYNOPSIS
expr arg ... DESCRIPTION
The arguments are taken as an expression. After evaluation, the result is written on the standard output. Each token of the expression is a separate argument. The operators and keywords are listed below. The list is in order of increasing precedence, with equal precedence operators grouped. expr | expr yields the first expr if it is neither null nor `0', otherwise yields the second expr. expr & expr yields the first expr if neither expr is null or `0', otherwise yields `0'. expr relop expr where relop is one of < <= = != >= >, yields `1' if the indicated comparison is true, `0' if false. The comparison is numeric if both expr are integers, otherwise lexicographic. expr + expr expr - expr addition or subtraction of the arguments. expr * expr expr / expr expr % expr multiplication, division, or remainder of the arguments. expr : expr The matching operator compares the string first argument with the regular expression second argument; regular expression syntax is the same as that of ed(1). The (...) pattern symbols can be used to select a portion of the first argument. Otherwise, the matching operator yields the number of characters matched (`0' on failure). ( expr ) parentheses for grouping. Examples: To add 1 to the Shell variable a: a=`expr $a + 1` To find the filename part (least significant part) of the pathname stored in variable a, which may or may not contain `/': expr $a : '.*/(.*)' '|' $a Note the quoted Shell metacharacters. SEE ALSO
ed(1), sh(1), test(1) DIAGNOSTICS
Expr returns the following exit codes: 0 if the expression is neither null nor `0', 1 if the expression is null or `0', 2 for invalid expressions. EXPR(1)
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