Php - passing variable inside an array

Question

my script work fine if this function is written :

$extension_allowed =array(".pdf",".jpg",".jpeg",".png",".txt",".doc",".docx", ".odt", ".rtf", ".xls", ".xlsx", ".ods");

But, if I want config the extensions from application config and extract this extension from a query, how to pass the extensions inside an array?

// extensions
$extensionsUploadFileQ = mysql_query("
SELECT 
    *
FROM table_config
WHERE field1 = 'EXTENSIONS_UPLOAD'
");
$extensionsUploadFileR=mysql_fetch_array($extensionsUploadFileQ);
$extensionUploadFileParameter=$extensionsUploadFileR['value_extensions'];

the result of query, passing at the variable $extensionUploadFileParameter but not work by this array:

$extension_allowed =array($extensionUploadFileParameter);

How to fix this?

Thanks

If using array(".pdf",".jpg",".jpeg",".png",".txt",".doc",".docx", ".odt", ".rtf", ".xls", ".xlsx", ".ods"); the script responde with trust message. But if using $extension_allowed =array($extensionUploadFileParameter); the script responde with failure message. — Frankie
– Frankie, Commented Oct 23, 2016 at 10:23
How you are storing this parameter? If it is stored as php code, you can eval this array (php.net/manual/pl/function.eval.php), or you can store it as a json (functions json_decode and json_encode) — Skamielina
– Skamielina, Commented Oct 23, 2016 at 10:33

Subscriberius · Accepted Answer · 2016-10-23 10:57:54Z

1

If I understood well in you MySQL DB you have your "EXTENSIONS_UPLOAD" as a string like ".pdf,.jpg,.jpeg,.png,.txt,.doc,.docx" and you want just to convert it to the PHP array type.

If yes then try to use explode PHP function:

// extensionUploadFileParameter = ".pdf,.jpg,.jpeg";
$extension_allowed = explode(",", $extensionUploadFileParameter);

/* 
   $extension_allowed will have:

   array(3)
   (
     [0] => string(4) ".pdf"
     [1] => string(4) ".jpg"
     [2] => string(4) ".jpeg"
   )
*/

OR

// extensionUploadFileParameter = '".pdf",".jpg",".jpeg"';
preg_match_all('/(.\w+)/', $extensionUploadFileParameter, $matches);  
$extension_allowed = $matches[0];

/*
 $extension_allowed will have:

 Array
 (
   [0] => .pdf
   [1] => .jpg
   [2] => .jpeg
 )
*/

I hope it helps.

Good luck.

P.S. Try to avoid using "eval" function:

Caution The eval() language construct is very dangerous because it allows execution of arbitrary PHP code. Its use thus is discouraged. If you have carefully verified that there is no other option than to use this construct, pay special attention not to pass any user provided data into it without properly validating it beforehand.

edited Oct 23, 2016 at 10:57

answered Oct 23, 2016 at 10:35

Subscriberius

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4 Comments

Frankie Over a year ago

in mySQL filed i have this value string: ".pdf",".jpg",".jpeg",".png",".txt",".doc",".docx", ".odt", ".rtf", ".xls", ".xlsx", ".ods" I think: when this tring is extract from a query, was passing at the variable $extensionUploadFileParameter

Paul T. Rawkeen Over a year ago

array_explode() ?!

Frankie Over a year ago

your first solution $extension_allowed = explode(",", $extensionUploadFileParameter); not work: response with an error message: File not uploaded.

Frankie Over a year ago

your secon solution preg_match_all('/(.\w+)/', $extensionUploadFileParameter, $matches); $extension_allowed = $matches[0]; WORK FINE!!!!!!!!!!!

Skamielina · Accepted Answer · 2016-10-23 10:51:30Z

1

You wrote:

in mySQL filed i have this value string: ".pdf",".jpg",".jpeg",".png",".txt",".doc",".docx", ".odt", ".rtf", ".xls", ".xlsx", ".ods" I think: when this tring is extract from a query, was passing at the variable $extensionUploadFileParameter – Frankie

You can use eval function

eval('$extension_allowed = array(' . $extensionUploadFileParameter . ');');

edited Oct 23, 2016 at 10:51

answered Oct 23, 2016 at 10:42

Skamielina

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3 Comments

Frankie Over a year ago

I try your solution, but displayed this error: Parse error: syntax error, unexpected end of file in upload.php(452) : eval()'d code on line 1

Skamielina Over a year ago

I have fixed my solution. Pleas try this.

Frankie Over a year ago

not work: Parse error: syntax error, unexpected end of file .. eval()'d code on line 1

Frankie · Accepted Answer · 2016-10-23 11:01:48Z

0

Thanks at all. The script work fine with this:

 preg_match_all('/(.\w+)/', $extensionUploadFileParameter, $matches);  
$extension_allowed = $matches[0];

eval() is dangerous: The eval() language construct is very dangerous because it allows execution of arbitrary PHP code

answered Oct 23, 2016 at 11:01

Frankie

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