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ok im building facebook app and im just back to php after months not touched it.

 

this is the code:

$image = $facebook->api_client->photos_get(null,null, echo $uid);
print_r ($image);
echo "<BR>";
echo $image[0]["src"];

 

the problem is that echo $uid

i tried every variation and it does not work!

 

i maybe forget something that i should have done?

 

the $uid containing user id numbers so what i did wrong and if im not wrong i wanted the numbers($uid) Wrapped in quotation marks

 

so i tried: echo ." $uid ".

and alot of other

 

plz help tnx :P

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https://forums.phpfreaks.com/topic/226993-echo-syntax-problem-please-help/
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You can't echo in a function argument list.  Echo literally means "echo this to the screen", so you can see how that won't make sense.  If you need quotes around the variable, simply do:

 

$uid = '"' . $uid . '"';

(that's a single quote, followed by a double quote, followed by another single quote)

 

That said, are you sure you need the value of $uid to be wrapped in quotes?

You can't echo in a function argument list.  Echo literally means "echo this to the screen", so you can see how that won't make sense.  If you need quotes around the variable, simply do:

 

$uid = '"' . $uid . '"';

(that's a single quote, followed by a double quote, followed by another single quote)

 

That said, are you sure you need the value of $uid to be wrapped in quotes?

tnx

silly me!

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