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Full Discussion: Syntax error on script
Top Forums UNIX for Beginners Questions & Answers Syntax error on script Post 302988331 by mutley2202 on Friday 23rd of December 2016 05:20:27 PM
Old 12-23-2016
Syntax error on script

Evening All (or morning for some),
Could anyone have a look at the below and advise where i've going wrong with the syntax as i keep getting the below error while trying to run.

Any help would be really apprecaited.

Code:
./testout: line 13: syntax error near unexpected token `else'

Code:
#!/bin/sh
if ($stat == 'a')
                {
                if (($Type=='p')|| ($Type=='b') || ($Type=='c') || ($Type=='d'))
                        {
                        if      ($Con == 'c' || $Con== 'p')
                                {
                                if      ($Con== 'p')
                                        {
                                        $Da = '1'
                                        echo "Migrated"
                                        }
                                else
                                        {
                                        $Da = '1'
                                        }
                                if ($Pl != "null")
                                        {
                                        $DT = '1'
                                        }
                                }
                        if ($Mig == ' not migrated 1')
                                {
                                echo "not migrated 1"
                                }
                        }
                else
                        {
                        $dar = '1';
                        echo "$DAR Decode 3"
                        }
                }
        else
                {
                $DAR = '1'
                echo "$DAR Decode 4"
                }

 

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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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