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ファイルが既に存在するかどうかを確認しようとしていますが、動作しません LINE 6 

if(isset($_POST['upload']) && $_FILES['userfile']['size'] > 0)
{

$fileName = $_FILES['userfile']['name'];
$tmpName  = $_FILES['userfile']['tmp_name'];
$fileSize = $_FILES['userfile']['size'];
$fileType = $_FILES['userfile']['type'];
if (file_exists($_FILES['userfile']['name']))
      {
      echo $_FILES['userfile"]["name'] . " already exists. ";
      }else{
$fp= fopen($tmpName, 'r');
$content = fread($fp, filesize($tmpName));
$content = addslashes($content);
fclose($fp);

if(!get_magic_quotes_gpc())
{
    $fileName = addslashes($fileName);
}
//include 'library/config.php';
//include 'library/opendb.php';

$query = "INSERT INTO upload (name, size, type, content )
VALUES ('$fileName', '$fileSize', '$fileType', '$content')";

mysql_query($query) or die('Error, query failed'); 
//include 'library/closedb.php';

echo "<br>File $fileName uploaded<br>";
} 
}
4

1 に答える 1

0

$_FILES['userfile"]["name']

開始の引用符と終了の引用符が一致していることを確認してください。有効な文字列の例は、たとえば"this"andです。'this'

echo $_FILES['userfile']['name'] . " already exists. ";
echo $_FILES["userfile"]["name"] . " already exists. ";
// etc.
于 2012-10-16T04:41:35.403 に答える