Re: [PHP] What am I not understanding about $GLOBALS['myvar'] vs global $myvar?

2003-03-27 Thread Ernest E Vogelsinger
At 09:24 27.03.2003, Marek Kilimajer said: [snip] echo GLOBALS['$key'] = $valuebr\n; is right because $key is in double quotes (the outermost qoutes) [snip] ...almost... the array deref should be in curly quotes

Re: [PHP] What am I not understanding about $GLOBALS['myvar'] vs global $myvar?

2003-03-27 Thread Ernest E Vogelsinger
At 09:46 27.03.2003, Marek Kilimajer said: [snip] there is no $ in front of GLOBALS ;) [snip] right you are - I noticed it as soon as the message was out... sometimes one's too fast hitting the send button ;-) --

RE: [PHP] What am I not understanding about $GLOBALS['myvar'] vs global $myvar?

2003-03-26 Thread Daevid Vincent
. -Original Message- From: Marek Kilimajer [mailto:[EMAIL PROTECTED] Sent: Wednesday, March 26, 2003 1:16 PM To: Daevid Vincent Cc: [EMAIL PROTECTED] Subject: Re: [PHP] What am I not understanding about $GLOBALS['myvar'] vs global $myvar? Daevid Vincent wrote: This one

RE: [PHP] What am I not understanding about $GLOBALS['myvar'] vs global $myvar?

2003-03-26 Thread Johnson, Kirk
I think you either want to use no quotes or double quotes, but not single quotes. Double quotes will interpolate the variable, single quotes will not, i.e. $key becomes a string literal rather than a variable. No quotes will work, although the docs seem to indicate it is deprecated syntax. Not

RE: [PHP] What am I not understanding about $GLOBALS['myvar'] vs global $myvar?

2003-03-26 Thread Jennifer Goodie
Actually I didn't. ;-) $GLOBALS[$key] is incorrect and depricated AFAIK. $GLOBALS['$key'] (with the single quotes) is the proper way to write these Wouldn't it need to be $GLOBALS[$key] because single quotes tell the parser not to expand variables? I think you are misinterpreting the part of

Re: [PHP] What am I not understanding about $GLOBALS['myvar'] vs global $myvar?

2003-03-26 Thread Leif K-Brooks
No, you misunderstand. The key is simply a string. Just as print foo; is wrong, $array[foo] is wrong (unless foo is a constant, of course). If you do print '$variable';, it will print the literal string $variable. It will not print the value of $variable. You should either use double