I've been happily using the error suppression operator on my PHP dev setup. But recently have gotten hit with Notices like so:
Notice: Uninitialized string offset: 0 in C:\websites\xxx\htdocs\includes\myscript.php on line 35
Line 35:
$file_name = @$File['file_name'];
I have display_errors on, and error_reporting set to 6143 (E_ALL).
Am I missing something? Shouldn't the error be suppressed?
Edit:
Tested in virgin script:
$a = array(); $b = @$a['f5'];
Suppressed the error. So I'm thinking we're changing the error_reporting value somehow. (Film at 11)
Thx for yr help.
Not a big fan of error suppression here except for throw-away scripts or instances where there really is no good way to catch an error.
Let me explain the behavior of the Uninitialized string offset
error. It's not a bug:
Example #1
$a = 0;
$b = $a['f5'];
$a
is a numeric scalar value. In the second line PHP is implicitly casting this numeric value to a string. The string '0'
has a length of 1.
In PHP you can lookup a character in a string using an array index, as PHP stores strings as arrays internally. For instance:
$s= 'abcd';
print_r($s[1]);
The output of this code will be b
as it is the second character in the string. In example #1 the lookup 'f5'
is being converted to a number as strings can only be indexed by character position. echo intval('f5');
shows us what PHP interprets the 'f5'
string as 0
in a numeric context.
With me so far? Here's what happens when we apply this to example #2
Example #2
$a = '';
$b = $a['f5'];
$a
is zero-length string. The second line is the same as $b= $a[0];
- i.e., the second line is asking for the first character of a zero-length string, but the string contains no characters. So PHP throws the following error, letting you know the index simply does not exist:
Notice: Uninitialized string offset: 0 in C:\websites\tcv3\wc2009\htdocs\aatest_array.php on line 3
These are the hard knocks of programming in a loosely typed language.
Since the error involves a string offset of 0, it must not apply to the line of code provided. There is no integer offset; you're using the string 'file_name' there.
Also, using '@' is a really terrible idea. Functions can fail, causing the script to exit with absolutely no indication as to where or why. A better way to handle the line of code you've pasted is this:
if (isset($File['file_name'])) {
$file_name = $File['file_name'];
} else {
// throw exception, return FALSE, print an error and exit, whatever
}
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