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画像の名前をデータベースにアップロードしています!

問題は、名前がデータベースにあり、名前の前に空白があることです!

問題はどこだ？

$original_name = strtolower(trim($arquivo['name']));
$caracteres = array("ç","~","^","]","[","{","}",";",":","´",",",">",
                   "<","-","/","|","@","$","%","ã","â","á","à","é",
                  "è","ó","§","ò","+","=","*","&","(",")","!","#","?",
                  "`","ã"," ","©","£");

$original_name = str_replace(' ', '', $original_name);
$final_name = str_replace($caracteres,"",$original_name);
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1 に答える 1

2

次のような許可された文字のホワイトリストを使用することをお勧めします[a-zA-Z0-9_]

そして使用：

$final_name = preg_replace("#[^a-z0-9_]+#i", "", $arquivo['name']);
于 2013-07-21T21:23:27.743 に答える