Why am I getting this parse error?

jason1074 · 10-02-2006, 01:29 PM

Hello,

I am terrible with codes, I am in no way a programmer or developer.
I keep getting this parse error. I posted a message before but thought I would include a link to the complete code, since I didn't understand what or how I was being told to fix the problem.

Error I Receive:
Parse error: syntax error, unexpected T_IF in /home/shoutout/public_html/thumb/class.php on line 1043

Link to a copy of the source code:
http://www.jdelbert.com/class.txt

Is anyone willing to take a look at the code and fix the problem for me? I would really appreciate any help.

Jay

acid_kewpie · 10-02-2006, 01:36 PM

you're not even going to tell us where line 1043 is??

jason1074 · 10-02-2006, 01:39 PM

Sorry Chris, Here are lines 1042-1049

{
$GIFtemp = if ($this->getimagesizeinfo[2] == 1);
$ImageCreateFunction = ($GIFtemp ? 'ImageCreateFromGIF' : 'ImageCreateFromPNG');
if (ini_get ('safe_mode'))
{
$this->DebugMessage ('ImageMagickThumbnailToGD() aborting because safe_mode is enabled', __FILE__, 852);
return false;
}

graemef · 10-02-2006, 10:18 PM

The if syntax is as follows:
if (condition) statement;
Which means that you have the if keyword followed by the condition in parenthesis. If (and only if) the condition is true will the following statement be executed. There is an extension to the if syntax being:
if (condition) statement1; else statement2;
Where statement1 will be performed if the condition is true and statement2 will be performed if the condition is false.
So the code that you want will be as follows:

PHP Code:



if ($this->getimagesizeinfo[2] == 1) 
   $GIFtemp = true; 
else 
   $GIFtemp = false;

An alternative approach (and I suspect the one that you were trying to achieve) is as follows. Because the comparison $this->getimagesizeinfo[2] == 1 will return either true or false this can be assigned directly into the variable $GIFtemp, hence you will have:

PHP Code:



$GIFtemp = $this->getimagesizeinfo[2] == 1;

(notice in this case there is no need for the if keyword, it is a direct assignment)

graemef · 10-02-2006, 10:25 PM

I should also add that I don't think that the variable name $GIFtemp is the best thought out name. The variable appears to be used to determine if the image is a gif, maybe $GIFtype would be a better variable name?

paulsm4 · 10-02-2006, 10:25 PM

Hi, Jason1074 -

This is the same question you asked about the other day:
http://www.linuxquestions.org/questi...d.php?t=488040

The answer's the same, too: use "if/then/else" conditionals instead of terniary (?

expressions.

Terniary expressions certainly have their place ... but this isn't it! Just Say No ;-)

graemef · 10-02-2006, 10:52 PM

I tend not to use the ternary expression but I think that the following would work:

PHP Code:



$GIFtemp  
? $ImageCreateFunction = 'ImageCreateFromGIF'  
: $ImageCreateFunction = 'ImageCreateFromPNG';

Again the problem is because if or ?: doesn't return a value, so the assignment needs to be in the statement.