Php Parse error: syntax error, unexpected T_CONST, expecting T_OLD_FUNCTION or T_FUNC

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February 23rd, 2013, 06:05 PM

we5inelgr

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Php Parse error: syntax error, unexpected T_CONST, expecting T_OLD_FUNCTION or T_FUNC

Hi all,

Getting this error while trying to create a constant inside an object:

Quote:

php Parse error: syntax error, unexpected T_CONST, expecting T_OLD_FUNCTION or T_FUNCTION or T_VAR or '}'

The line it's complaining about is this, which is inside a class:

PHP Code:


		
			
const SI_IMAGE_JPEG = 1;

b.t.w. I'm running PHP version 5.2.17.

Why would I be getting this error?

February 23rd, 2013, 06:37 PM

requinix

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What's the rest of the code?

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February 23rd, 2013, 06:59 PM

we5inelgr

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the object is being called from another file:

PHP Code:


		
			
include_once ('securimage.php');

$secureimage = new Securimage();

The file is there, named exactly that.

The Securimage() object, in the securimage.php is:

PHP Code:


		
			
class Securimage

{

    // All of the public variables below are securimage options

    // They can be passed as an array to the Securimage constructor, set below,

    // or set from securimage_show.php and securimage_play.php



    /**

     * Renders captcha as a JPEG image

     * @var int

     */

    const SI_IMAGE_JPEG = 1; <----

    /**

     * Renders captcha as a PNG image (default)

     * @var int

     */

    const SI_IMAGE_PNG  = 2;

and continues.

The line with the arrow is the line the error is occuring on according to php.

February 23rd, 2013, 08:19 PM

ManiacDan

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That code works for me. Are you sure you're in the right file?

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February 24th, 2013, 12:34 AM

we5inelgr

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Quote:

Originally Posted by ManiacDan

That code works for me. Are you sure you're in the right file?

Yeah, because if I comment out that line, then the error occurs on the next constant:

PHP Code:


		
			
const SI_IMAGE_PNG  = 2;

February 24th, 2013, 12:45 AM

E-Oreo

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You would only receive that error message in this context if you were not actually running PHP 5.

Add:

PHP Code:


		
			
die(phpversion());

immediately before your require statement and determining what value it outputs. I would be very surprised if it's actually 5.2.17.

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February 24th, 2013, 01:07 AM

we5inelgr

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Quote:

Originally Posted by E-Oreo

You would only receive that error message in this context if you were not actually running PHP 5.

Add:

PHP Code:


		
			
die(phpversion());

immediately before your require statement and determining what value it outputs. I would be very surprised if it's actually 5.2.17.

Well isn't that interesting.

When using (at the top of the page):

PHP Code:


		
			
echo phpinfo();

I get PHP version 5.2.17.

And using:

PHP Code:


		
			
die(phpversion());

I get 4.4.9

Why the difference? Perhaps because I'm on a shared server? How does that work? Some on the server have 5.2.17, and my site doesn't?

It makes sense now why the OO PHP doesn't work if I'm really running 4.4.9 and not 5.2.17. But still, two different answers on the same page?

February 24th, 2013, 06:26 AM

requinix

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There's some difference. Maybe the way the run is running, maybe the file extension, maybe where the file is located... Something is different.

February 25th, 2013, 08:27 AM

ManiacDan

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Yeah...it might be a good idea to uninstall/reinstall PHP.

Are you really saying that if you put phpinfo() at the top of the page, you get 5+, and if you put phpversion() at the bottom of the page, you get 4? You're accessing the page in the same way?

Last edited by ManiacDan : February 25th, 2013 at 08:39 AM.

#10

February 25th, 2013, 06:40 PM

we5inelgr

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Quote:

Originally Posted by ManiacDan

Yeah.

#11

February 25th, 2013, 06:44 PM

we5inelgr

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I finally figured out what was going on.

If the page calling the OO PHP was a .html page, phpversion() shows 4.4.9.
If the page calling the OO PHP was a .php page, phpversion() shows 5.2.17.

And, either page extension type would show version 5.2.17 with phpinfo();

Even though the page (either the .html or the .php version) were accessed in the same way and they had identical code...only the .php extension page would get phpversion() to show the 5.2.17 version whereas the .html page would give the 4.4.9 version.

Why that's the case, I don't know yet (probably server configuration as to how .html pages are rendered with the php engine), but I got it working now by simply changing the page extension from .html to .php.

The server host company is making changes to allow .html pages to behave the same with this regards.

Last edited by we5inelgr : February 25th, 2013 at 06:49 PM.

Php Parse error: syntax error, unexpected T_CONST, expecting T_OLD_FUNCTION or T_FUNC

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