| count() problem [message #185705] |
Sat, 03 May 2014 18:18  |
Mr Oldies
Messages: 241
Registered: October 2013
Karma: 0
|
Senior Member |
|
|
$dir = '../audio/'.$yr.'/';
$files = scandir($dir);
$number=count(files);
echo $number;
Why does this always show the resutlt as 1?
$dir = '../audio/'.$yr.'/';
$files = scandir($dir);
echo count(files);
Shows the true count of the array.
|
|
|
|
| Re: count() problem [message #185706 is a reply to message #185705] |
Sat, 03 May 2014 18:32  |
M. Strobel
Messages: 386
Registered: December 2011
Karma: 0
|
Senior Member |
|
|
Am 03.05.2014 20:18, schrieb richard:
>
> $dir = '../audio/'.$yr.'/';
> $files = scandir($dir);
> $number=count(files);
did you mean
$number=count($files);
?
/Str.
> echo $number;
>
> Why does this always show the resutlt as 1?
>
>
> $dir = '../audio/'.$yr.'/';
> $files = scandir($dir);
> echo count(files);
>
> Shows the true count of the array.
>
|
|
|
|
|
|
| Re: count() problem [message #185708 is a reply to message #185705] |
Sat, 03 May 2014 18:36 |
M. Strobel
Messages: 386
Registered: December 2011
Karma: 0
|
Senior Member |
|
|
Am 03.05.2014 20:18, schrieb richard:
>
> $dir = '../audio/'.$yr.'/';
> $files = scandir($dir);
> $number=count(files);
> echo $number;
>
> Why does this always show the resutlt as 1?
>
>
> $dir = '../audio/'.$yr.'/';
> $files = scandir($dir);
> echo count(files);
>
> Shows the true count of the array.
>
And you still have turned error reporting off. How often have you been told to turn
it on?
php > echo count(files), "\n";
PHP Notice: Use of undefined constant files - assumed 'files' in php shell code on
line 1
Notice: Use of undefined constant files - assumed 'files' in php shell code on line 1
1
php >
/Str.
|
|
|
|
| Re: count() problem [message #185709 is a reply to message #185706] |
Sat, 03 May 2014 19:57 |
Mr Oldies
Messages: 241
Registered: October 2013
Karma: 0
|
Senior Member |
|
|
On Sat, 03 May 2014 20:32:56 +0200, M. Strobel wrote:
> Am 03.05.2014 20:18, schrieb richard:
>>
>> $dir = '../audio/'.$yr.'/';
>> $files = scandir($dir);
>> $number=count(files);
> did you mean
> $number=count($files);
> ?
> /Str.
>> echo $number;
>>
>> Why does this always show the resutlt as 1?
>>
>>
>> $dir = '../audio/'.$yr.'/';
>> $files = scandir($dir);
>> echo count(files);
>>
>> Shows the true count of the array.
>>
thanks. many times I overlook the obvious.
|
|
|
|
| Re: count() problem [message #185712 is a reply to message #185709] |
Sat, 03 May 2014 22:14 |
Denis McMahon
Messages: 634
Registered: September 2010
Karma: 0
|
Senior Member |
|
|
On Sat, 03 May 2014 15:57:26 -0400, richard wrote:
> On Sat, 03 May 2014 20:32:56 +0200, M. Strobel wrote:
>
>> Am 03.05.2014 20:18, schrieb richard:
>>>
>>> $dir = '../audio/'.$yr.'/';
>>> $files = scandir($dir); $number=count(files);
>> did you mean
>> $number=count($files);
>> ?
>> /Str.
>>> echo $number;
>>>
>>> Why does this always show the resutlt as 1?
>>>
>>>
>>> $dir = '../audio/'.$yr.'/';
>>> $files = scandir($dir);
>>> echo count(files);
>>>
>>> Shows the true count of the array.
> thanks. many times I overlook the obvious.
Did you even think to check for any warning messages? Did you have
warnings turned on? PHP will usually generate a warning if warnings are
on and you appear to have left a $ off of a variable name.
--
Denis McMahon, denismfmcmahon(at)gmail(dot)com
|
|
|
|
Calendar
Search
Help
Members
Register
Login
Home
]