Possible Duplicate:
Reference - What does this symbol mean in PHP?
I have this assignment:
$buffer=@data_value[$i];
what does the @ mean?
Asked
Active
Viewed 4,173 times
4
-
15It's to prevent you from figuring out what's wrong with your code. – Ignacio Vazquez-Abrams Dec 08 '10 at 23:01
-
1Actually that statement does not even compile – Matteo Riva Dec 08 '10 at 23:02
-
@Ignacio I second that. Our phpcs code standards are configured to pick up any developers trying to do this and yell at them. – El Yobo Dec 08 '10 at 23:03
-
@nico: `PHP Parse error: syntax error, unexpected '[' in...` – Matteo Riva Dec 08 '10 at 23:05
-
Top tip - Only use the error suppressor when you absolutely have to - a reasonable example being a file operation you expect to fail, want to fail silently and intend to check yourself post-operation. – John Parker Dec 08 '10 at 23:05
-
@nico - no, it doesn't, there's no $ on data_value. – El Yobo Dec 08 '10 at 23:06
-
should be: `$buffer=@$data_value[$i];` – Stoic Dec 08 '10 at 23:19
-
1@kemp: None of my PHP code compiles! – webbiedave Dec 08 '10 at 23:23
5 Answers
5
The @ will suppress errors about variable not being initialized (will evaluate as null instead).
Also, your code is probably missing a $ after @:
$buffer=@$data_value[$i];
rustyx
- 80,671
- 25
- 200
- 267
1
It is called the "error-control operator". Since it's an assignment, I believe you should do the rest yourself.
Jon
- 428,835
- 81
- 738
- 806
0
the @ in front of a statement means that no warnings/errors should be reported from the result of that statement. To put simply, Error Reporting is suppressed for this statement.
This is particularly useful, when e.g. @fclose(fopen("file.txt",w")) which can throw several warnings/errors depending on the situation, but with an @ in front of it, all these warnings or errors will be suppressed.
Stoic
- 10,536
- 6
- 41
- 60
-
-
Agreed. What I meant here was, this is useful when you are in a production environment and do not want to show error reports to end-user. – Stoic Dec 08 '10 at 23:18
0
As above, it suppresses the error if the array key doesn't exist. A version which will do the same without resorting to dodgy error suppression is
$buffer = array_key_exists($i, $data_value) ? $data_value[$i] : null;
El Yobo
- 14,823
- 5
- 60
- 78
-
That's just error suppression via syntactic salt, not a significantly different outcome. – mario Dec 08 '10 at 23:50
-
No, it's correctly checking that a value exists before attempting to make use of it. – El Yobo Dec 09 '10 at 00:34