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Top Forums Shell Programming and Scripting Need help on If statement for comparison Post 302475017 by jmathew99 on Friday 26th of November 2010 07:44:39 AM
Old 11-26-2010
Need help on If statement for comparison

Hi,
We are using AIX and while using KSH Unix script, we are reading data in a file and then using the if statement for comparison. The code with if statement is as follows;

Code:
 
cat $FILETMP | while read a
do
        jobstr1=`echo $a | awk '{print $1}'`
        var1=`echo $jobstr1 | cut -c 9`
        var2=`echo $jobstr1 | cut -c 17`
        if [ $var1 = _ -a $var2 = _ ]
        then
                echo "$a" >> $outfile
        fi
done

The code is working fine and writing the data to the file "outfile", but the code gives errors. The errors are when the characters are "empty" or "blank" then while comparing the system is giving errors as follows;

ksh: _: unknown test operator
ksh: test: argument expected

Please help to have the code without having error messages. I would also like to know if there is a better way of doing this.
Thanks for the help.
 

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DEBUG_ZVAL_DUMP(3)							 1							DEBUG_ZVAL_DUMP(3)

debug_zval_dump - Dumps a string representation of an internal zend value to output

SYNOPSIS
void debug_zval_dump (mixed $variable, [mixed $...]) DESCRIPTION
Dumps a string representation of an internal zend value to output. PARAMETERS
o $variable - The variable being evaluated. RETURN VALUES
No value is returned. EXAMPLES
Example #1 debug_zval_dump(3) example <?php $var1 = 'Hello World'; $var2 = ''; $var2 =& $var1; debug_zval_dump(&$var1); ?> The above example will output: &string(11) "Hello World" refcount(3) Note Beware the refcount The refcount value returned by this function is non-obvious in certain circumstances. For example, a developer might expect the above example to indicate a refcount of 2. The third reference is created when actually calling debug_zval_dump(3). This behavior is further compounded when a variable is not passed to debug_zval_dump(3) by reference. To illustrate, consider a slightly modified version of the above example: Example #2 <?php $var1 = 'Hello World'; $var2 = ''; $var2 =& $var1; debug_zval_dump($var1); // not passed by reference, this time ?> The above example will output: string(11) "Hello World" refcount(1) Why refcount(1)? Because a copy of $var1 is being made, when the function is called. This function becomes even more confusing when a variable with a refcount of 1 is passed (by copy/value): Example #3 <?php $var1 = 'Hello World'; debug_zval_dump($var1); ?> The above example will output: string(11) "Hello World" refcount(2) A refcount of 2, here, is extremely non-obvious. Especially considering the above examples. So what's happening? When a variable has a single reference (as did $var1 before it was used as an argument to debug_zval_dump(3)), PHP's engine opti- mizes the manner in which it is passed to a function. Internally, PHP treats $var1 like a reference (in that the refcount is increased for the scope of this function), with the caveat that if the passed reference happens to be written to, a copy is made, but only at the moment of writing. This is known as "copy on write." So, if debug_zval_dump(3) happened to write to its sole parameter (and it doesn't), then a copy would be made. Until then, the parameter remains a reference, causing the refcount to be incremented to 2 for the scope of the function call. SEE ALSO
var_dump(3), debug_backtrace(3), References Explained, References Explained (by Derick Rethans). PHP Documentation Group DEBUG_ZVAL_DUMP(3)
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