if(in_array) mime check always returns FALSE

Question

Variables:

$csvFile = $_FILES['uploadfile']['tmp_name'];
$csvFile = $con->real_escape_string($csvFile);
$mimes = array('application/vnd.ms-excel','text/plain','text/csv','text/tsv','csv');

if statement:

if(in_array($csvFile, $mimes))
{
$file = fopen($csvFile, 'r');
$i = 0;
$row = 1; 
while (($line = fgetcsv($file)) !== FALSE) 
{
if($row == 1){ $row++;  continue; }
$data_array[$i] = $line[0];
$data_array[$i] = mysqli_real_escape_string($con, trim($data_array[$i]));
$encrypted_numbers[$i] = encryption::encrypt($data_array[$i]);
$encrypted_numbers[$i] .= mysqli_real_escape_string($con, trim($line[1]));
$i++;
}
fclose($file);
} 
else 
{
die("Sorry, file type not allowed");
};

I've been trying stuff for 30 min, anyone know why I always get false even if the file is named csv.csv?

what does var_dump($csvFile) before the in_array call give you ? — Maximus2012, Aug 12 '13 at 01:27
most likely $csvFile will give you the actual file name and not the mime type for that file which is what you are trying to do with in_array. — Maximus2012, Aug 12 '13 at 01:29
@user2671532 this will always result in in_array returning false when you try to match it with $mimes. — Maximus2012, Aug 12 '13 at 01:31
@Maximus2012 I realize that, and I tried having csvFile without a temp name and then imploding it, still got false even though var_dump gave me "csv.csv" name — Gjorg, Aug 12 '13 at 01:32
you need to get the mime type for the file. See if this helps: http://stackoverflow.com/questions/134833/how-do-i-find-the-mime-type-of-a-file-with-php — Maximus2012, Aug 12 '13 at 01:32

score 1 · Accepted Answer · edited May 23 '17 at 10:25

1

$_FILES['uploadfile']['tmp_name'];

you should be looking in 'type' not 'tmp_name'

$_FILES['uploadfile']['type'];

However, there is an SO discussion here about why a csv file can have the application/octet-stream mime-type. It suggests that you need to verify the file-type yourself, and not rely on ['type']. The best way would be to use fgetcsv() to try to parse it as csv. If this fails then assume it is not a csv.

edited May 23 '17 at 10:25

Community

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answered Aug 12 '13 at 01:32

Andy G

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exact same thing happens – Gjorg Aug 12 '13 at 01:33
Print or echo the ['type'] to see what it is. – Andy G Aug 12 '13 at 01:34
application/octet-stream – Gjorg Aug 12 '13 at 01:36
That indicates it is a binary file. Is it really a csv? Open it in Notepad++ or similar. The filename isn't something.csv.csv or possibly csv.csv.txt is it? – Andy G Aug 12 '13 at 01:38
The name is "csv.csv", it's made in OpenOffice.org Calc and saved as "CSV file", under the file it says "CSV file" – Gjorg Aug 12 '13 at 01:41
You are welcome. csv's are not as easy to discover as, for example, an image. – Andy G Aug 12 '13 at 01:53

if(in_array) mime check always returns FALSE

1 Answers1