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Full Discussion: Condition problem
Top Forums Shell Programming and Scripting Condition problem Post 302972521 by znesotomayor on Friday 6th of May 2016 09:27:43 AM
Old 05-06-2016
Apologies Sir Rudic. How do i run it on if statement? i'm encountering command not found.

Code:
#!/bin/sh

for i in `ls *.txt`; do
while read line
do
DEPT=`echo $line | awk -F "," '{print $3}'`
SDEPT=`echo $line | awk -F "," '{print $4}'`
if ($SDEPT == 888 || $SDEPT == 0) && !($DEPT == 55 || $DEPT == 16 || $DEPT == 76); then
echo dept
else
class
fi
done <$i
done

Code:
output
test.sh: line 8: 888: command not found
test.sh: line 8: 888: command not found
test.sh: line 11: class: command not found
test.sh: line 8: 0: command not found
test.sh: line 8: 0: command not found
test.sh: line 11: class: command not found
test.sh: line 8: 888: command not found
test.sh: line 8: 888: command not found
test.sh: line 11: class: command not found

 

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ECHO(3) 								 1								   ECHO(3)

echo - Output one or more strings

SYNOPSIS
void echo (string $arg1, [string $...]) DESCRIPTION
Outputs all parameters. echo is not actually a function (it is a language construct), so you are not required to use parentheses with it. echo (unlike some other language constructs) does not behave like a function, so it cannot always be used in the context of a function. Additionally, if you want to pass more than one parameter to echo, the parameters must not be enclosed within parentheses. echo also has a shortcut syntax, where you can immediately follow the opening tag with an equals sign. Prior to PHP 5.4.0, this short syn- tax only works with the short_open_tag configuration setting enabled. I have <?=$foo?> foo. PARAMETERS
o $arg1 - The parameter to output. o $... - RETURN VALUES
No value is returned. EXAMPLES
Example #1 echo examples <?php echo "Hello World"; echo "This spans multiple lines. The newlines will be output as well"; echo "This spans multiple lines. The newlines will be output as well."; echo "Escaping characters is done "Like this"."; // You can use variables inside of an echo statement $foo = "foobar"; $bar = "barbaz"; echo "foo is $foo"; // foo is foobar // You can also use arrays $baz = array("value" => "foo"); echo "this is {$baz['value']} !"; // this is foo ! // Using single quotes will print the variable name, not the value echo 'foo is $foo'; // foo is $foo // If you are not using any other characters, you can just echo variables echo $foo; // foobar echo $foo,$bar; // foobarbarbaz // Some people prefer passing multiple parameters to echo over concatenation. echo 'This ', 'string ', 'was ', 'made ', 'with multiple parameters.', chr(10); echo 'This ' . 'string ' . 'was ' . 'made ' . 'with concatenation.' . " "; echo <<<END This uses the "here document" syntax to output multiple lines with $variable interpolation. Note that the here document terminator must appear on a line with just a semicolon. no extra whitespace! END; // Because echo does not behave like a function, the following code is invalid. ($some_var) ? echo 'true' : echo 'false'; // However, the following examples will work: ($some_var) ? print 'true' : print 'false'; // print is also a construct, but // it behaves like a function, so // it may be used in this context. echo $some_var ? 'true': 'false'; // changing the statement around ?> NOTES
Note Because this is a language construct and not a function, it cannot be called using variable functions. SEE ALSO
print(3), printf(3), flush(3), Heredoc syntax. PHP Documentation Group ECHO(3)
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