How can I get the value of $fbAppPath into the PHP statement below?
<? print json_encode(array(array('text' => 'Become A Fan', 'href' => '$fbAppPath'))); ?>
views:
272answers:
5
+3
A:
<? print json_encode(array(array('text' => 'Become A Fan', 'href' => $fbAppPath))); ?>
konforce
2010-02-04 01:58:04
+3
A:
Or
<? print json_encode(array(array('text' => 'Become A Fan', 'href' => "more text ${fbAppPath} more text"))); ?>
if you wanted to embed the variable value in a string. The double quotes are important in that case.
jodonnell
2010-02-04 01:58:59
A:
Ok, I figured it out... I didn't realize I was in a function and referencing a variable outside the function... sorry to waste anyone's time
<? print json_encode(array(array('text' => 'Become A Fan', 'href' => $GLOBALS['fbAppPath']))); ?>
variable being referenced was in the global scope..
Mark
2010-02-04 02:05:27
if this is the correct solution to the problem, you can accept your own answer
Jonathan Fingland
2010-02-04 02:06:23
@Mark, nevertheless you should give proper credit to those who helped you and, in my opinion, you should accept one of the perfect 3 valid answers given. That will give points both ways.
Frankie
2010-02-04 02:08:25
In two days. Although really it isn't the answer to the problem, as it doesn't relate to the problem.
Chacha102
2010-02-04 02:08:38
@Jonathan, actually I believe he can accept his own answer but only a couple of days after it's been posted.
Frankie
2010-02-04 02:09:08
@Frankie thanks. I've never tried to accept my own so that's never come up. good to know, though.
Jonathan Fingland
2010-02-04 02:11:45
Yeah, it was a stupid mistake... It's late... The most accurate and succinct answer was the first one by konforce... had the variable been in the same scope as the statement his(her) answer would have resolved the issue. As such, that's the answer I will accept...
Mark
2010-02-04 02:27:28
+6
A:
You cannot get a variable in a single quote string. PHP interprets all single-quoted strings EXACTLY as they appear. (aside from escaping a single quote)
The ONLY way to get a variable while using single quotes is to break out of them:
$foo = 'variable';
echo 'single-quoted-string-'.$foo.'-more-single-quoted-string';
Chacha102
2010-02-04 02:05:33
A:
You do not need quotes around a variable that is already a string.
'I am a string, because I am surrounded by quotes';
$string = 'I am a string, because I am surrounded by quotes';
if (is_string($string)) {
echo 'Yes, the variable $string is a string, because it contains a string';
}
$anotherString = $string;
if (is_string($anotherString)) {
echo 'The variable $anotherString is a string as well, because it contains a string as well';
}
$notWhatYouExpect = '$string';
echo $notWhatYouExpect; // outputs the word '$string'
deceze
2010-02-04 02:10:26