reading variables in a filename

[phpnewbie26]

I believe this is a simple problem but I'm not able to figure it out. I have a file that has a filename on each line. I read through this file setting the string to a variable. Then I am trying to use this variable in opening the file but it won't read it. This is my current code.

while (! feof($ts_proj)){
        $filename= fgets($ts_proj);
                $codes=fopen('/projects/$filename/ts_codes',"r");
}
 

The error says it cant open /project/$filename/ts_codes instead of /project/project1/tscodes which means the variable is passing through. Any help is appreciated.

[phpnewbie26]

oh..okay..i figured it out. I forgot to add the "." in before and after my variable.=)

[Markus]

@phpnewbie26
Single quotations will not parse the string for variables. Double quotes, however, will.

$name = "Mark";
echo "My name is $name.";
// Output: My name is Mark.
 

For readability's sake, if you use this technique, wrap the variable in curly braces.

$name = "Mark";
echo "My name is {$name}.";
 

[dlite922]

besides better readability I think the curly braces were used in cases where your variable is not surrounded by space.

e.g. : this won't work:
echo "6584564$someNum2345";

solution:
echo "6584564${someNum}2345";

:)

Cheers,


Dan

[Atli]

@dlite922
Interesting. I didn't know putting the $ outside the braces worked to.

I always do

echo "6584564{$someNum}2345";

But they both seem to work :D

I wonder if there are any differences...
* Goes for the Google bookmark *

[Dormilich]

@Atli
there are some edge cases, mostly concerning dynamic variables and dynamic function calls.

$$array[1] vs. ${$array[1]}
$obj->{$method}()

[phpnewbie26]

thanks guy! This is all very helpful!=)