05-11-2011
STDIN and STDOUT
Hallo,
i have a script like:
PHP Code:
if [$1] ;then
echo "OK"
else
echo "ERROR $2 is missing"
fi;
if [$2];then
touch $2
fi;
if [$2];then
cat $1 | grep xy > $2
fi;
but i must do that with STDIN and STDOUT, when i want to run the script it must not give parameters. It must run like
PHP Code:
cat zz | ./rrr.sh > zz_solution.txt
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LEARN ABOUT PHP
php_logo_guid
PHP_LOGO_GUID(3) 1 PHP_LOGO_GUID(3)
php_logo_guid - Gets the logo guid
SYNOPSIS
string php_logo_guid (void )
DESCRIPTION
This function returns the ID which can be used to display the PHP logo using the built-in image. Logo is displayed only if expose_php is
On.
Warning
This function has been DEPRECATED and REMOVED as of PHP 5.5.0.
RETURN VALUES
Returns PHPE9568F34-D428-11d2-A769-00AA001ACF42.
CHANGELOG
+--------+---------------------------------------------+
|Version | |
| | |
| | Description |
| | |
+--------+---------------------------------------------+
| 5.5.0 | |
| | |
| | php_logo_guid(3) has been removed from PHP. |
| | |
+--------+---------------------------------------------+
EXAMPLES
Example #1
php_logo_guid(3) example
<?php
echo '<img src="' . $_SERVER['PHP_SELF'] .
'?=' . php_logo_guid() . '" alt="PHP Logo !" />';
?>
SEE ALSO
phpinfo(3), phpversion(3), phpcredits(3), zend_logo_guid(3).
PHP Documentation Group PHP_LOGO_GUID(3)