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Jestem nowy w PHP, stworzyłem kod do przesyłania obrazów do bazy danych sql i pobierania obrazu za pomocą php.Błąd krytyczny: wywołanie niezdefiniowanej funkcji finfo_open() w php
Oto mój kod:
<html>
<head><title>File Insert</title></head>
<body>
<h3>Please Choose a File and click Submit</h3>
<form enctype="multipart/form-data" action=
"<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input type="hidden" name="MAX_FILE_SIZE" value="10000000" />
<input name="userfile" type="file" />
<input type="submit" value="Submit" />
</form>
<?php
// check if a file was submitted
if(!isset($_FILES['userfile']))
{
echo '<p>Please select a file</p>';
}
else
{
try {
$msg= upload(); //this will upload your image
echo $msg; //Message showing success or failure.
}
catch(Exception $e) {
echo $e->getMessage();
echo 'Sorry, could not upload file';
}
}
// the upload function
function upload() {
include "mysqlconnect.php";
$maxsize = 10000000; //set to approx 10 MB
//check associated error code
if($_FILES['userfile']['error']==UPLOAD_ERR_OK) {
//check whether file is uploaded with HTTP POST
if(is_uploaded_file($_FILES['userfile']['tmp_name'])) {
//checks size of uploaded image on server side
if($_FILES['userfile']['size'] < $maxsize) {
//checks whether uploaded file is of image type
//if(strpos(mime_content_type($_FILES['userfile']['tmp_name']),"image")===0) {
$finfo = finfo_open(FILEINFO_MIME_TYPE);
if(strpos(finfo_file($finfo, $_FILES['userfile']['tmp_name']),"image")===0) {
// prepare the image for insertion
$imgData =addslashes (file_get_contents($_FILES['userfile']['tmp_name']));
// put the image in the db...
// database connection
mysql_connect($host, $user, $pass) OR DIE (mysql_error());
// select the db
mysql_select_db ($db) OR DIE ("Unable to select db".mysql_error());
// our sql query
$sql = "INSERT INTO test_image
(image, name)
VALUES
('{$imgData}', '{$_FILES['userfile']['name']}');";
// insert the image
mysql_query($sql) or die("Error in Query: " . mysql_error());
$msg='<p>Image successfully saved in database with id ='. mysql_insert_id().' </p>';
}
else
$msg="<p>Uploaded file is not an image.</p>";
}
else {
// if the file is not less than the maximum allowed, print an error
$msg='<div>File exceeds the Maximum File limit</div>
<div>Maximum File limit is '.$maxsize.' bytes</div>
<div>File '.$_FILES['userfile']['name'].' is '.$_FILES['userfile']['size'].
' bytes</div><hr />';
}
}
else
$msg="File not uploaded successfully.";
}
else {
$msg= file_upload_error_message($_FILES['userfile']['error']);
}
return $msg;
}
// Function to return error message based on error code
function file_upload_error_message($error_code) {
switch ($error_code) {
case UPLOAD_ERR_INI_SIZE:
return 'The uploaded file exceeds the upload_max_filesize directive in php.ini';
case UPLOAD_ERR_FORM_SIZE:
return 'The uploaded file exceeds the MAX_FILE_SIZE directive that was specified in the HTML form';
case UPLOAD_ERR_PARTIAL:
return 'The uploaded file was only partially uploaded';
case UPLOAD_ERR_NO_FILE:
return 'No file was uploaded';
case UPLOAD_ERR_NO_TMP_DIR:
return 'Missing a temporary folder';
case UPLOAD_ERR_CANT_WRITE:
return 'Failed to write file to disk';
case UPLOAD_ERR_EXTENSION:
return 'File upload stopped by extension';
default:
return 'Unknown upload error';
}
}
?>
</body>
</html>
Teraz mam ten błąd, błąd krytyczny: Zadzwoń do niezdefiniowanej funkcji finfo_open() na tej linii
$finfo = finfo_open(FILEINFO_MIME_TYPE); .
Czy ktoś mógłby mi pomóc rozwiązać ten problem.
Z góry dziękuję.
finfo_open jest dostępny tylko dla php> = 5.3.0 http://www.php.net/manual/en/function.finfo-open.php. Możesz chcieć dwukrotnie sprawdzić, czy twój serwer jest zaktualizowany za pomocą php_info(). Ten link może pomóc: http://stackoverflow.com/questions/21293996/wordpress-plugin-call-to-undefined-function-finfo-open – Jinandra
ok dzięki @sanki –
@sanki: z linku, usunę to plik semicolon php.ini gdzie ta linia "extension = php_fileinfo.dll" ale nadal wyświetla ten błąd –