| echo other way to output a constant? [message #169476] |
Tue, 14 September 2010 22:12  |
MikeB
Messages: 65
Registered: September 2010
Karma:
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Member |
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Say I have the following code:
<?php
define('MAX_FILE_SIZE', 300);
echo <<<_END
<html><head><title>PHP Form Upload</title></head><body>
<form method='post' action='UploadFile2.php' enctype='multipart/form-data'>
<!-- MAX_FILE_SIZE must precede the file input field -->
<input type="hidden" name="MAX_FILE_SIZE" value="" />
Select a JPG, GIF, PNG or TIF File:
<input type='file' name='filename' size='60' />
<br/><input type='submit' value='Upload' /></form>
_END;
echo "</body></html>";
?>
How can I get PHP to place the value of the constant MAX_FILE_SIZE in
the value attribute for the hidden field MAX_FILE_SIZE?
Since it does not start with a $-sign it is not interpreted as a
variable. I tried putting it in curly braces {}, which sometimes work,
but that didn't do the trick either.
I was hoping that print_r() would do it, but I could not get PHP to
interpret that either.
So any way or do I have to a) put the constant in a variable or b) echo
each line individually so that I can do concatenation of the output lines?
Thanks.
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