画像ファイルのアップロード

Question

誰かが私を喜ばせることができるかどうか疑問に思う。

いくつかのオンラインチュートリアルと「試行錯誤」を使用して、私は以下のスクリプトをまとめて、ユーザーが画像ファイルをアップロードできるようにしました。

<?php 
//define a maxim size for the uploaded images 
//define ("MAX_SIZE","100"); 
// define the width and height for the thumbnail 
// note that theese dimmensions are considered the maximum dimmension and are not fixed, 
// because we have to keep the image ratio intact or it will be deformed 
define ("WIDTH","150"); 
define ("HEIGHT","100"); 

// this is the function that will create the thumbnail image from the uploaded image 
// the resize will be done considering the width and height defined, but without deforming the image 
function make_thumb($img_name,$filename,$new_w,$new_h) 
{ 
//get image extension. 
$ext=getExtension($img_name); 
//creates the new image using the appropriate function from gd library 
if(!strcmp("jpg",$ext) || !strcmp("jpeg",$ext)) 
$src_img=imagecreatefromjpeg($img_name); 

if(!strcmp("png",$ext)) 
$src_img=imagecreatefrompng($img_name); 

//gets the dimmensions of the image 
$old_x=imageSX($src_img); 
$old_y=imageSY($src_img); 

// next we will calculate the new dimmensions for the thumbnail image 
// the next steps will be taken: 
// 1. calculate the ratio by dividing the old dimmensions with the new ones 
// 2. if the ratio for the width is higher, the width will remain the one define in WIDTH variable 
// and the height will be calculated so the image ratio will not change 
// 3. otherwise we will use the height ratio for the image 
// as a result, only one of the dimmensions will be from the fixed ones 
$ratio1=$old_x/$new_w; 
$ratio2=$old_y/$new_h; 
if($ratio1>$ratio2) { 
$thumb_w=$new_w; 
$thumb_h=$old_y/$ratio1; 
} 
else { 
$thumb_h=$new_h; 
$thumb_w=$old_x/$ratio2; 
} 

// we create a new image with the new dimmensions 
$dst_img=ImageCreateTrueColor($thumb_w,$thumb_h); 

// resize the big image to the new created one 
imagecopyresampled($dst_img,$src_img,0,0,0,0,$thumb_w,$thumb_h,$old_x,$old_y); 

// output the created image to the file. Now we will have the thumbnail into the file named by $filename 
if(!strcmp("png",$ext)) 
imagepng($dst_img,$filename); 
else 
imagejpeg($dst_img,$filename); 

//destroys source and destination images. 
imagedestroy($dst_img); 
imagedestroy($src_img); 
} 

// This function reads the extension of the file. 
// It is used to determine if the file is an image by checking the extension. 
function getExtension($str) { 
$i = strrpos($str,"."); 
if (!$i) { return ""; } 
$l = strlen($str) - $i; 
$ext = substr($str,$i+1,$l); 
return $ext; 
} 

// This variable is used as a flag. The value is initialized with 0 (meaning no error found) 
//and it will be changed to 1 if an error occurs. If the error occures the file will not be uploaded. 
$errors=0; 
// checks if the form has been submitted 
if(isset($_POST['Submit'])) 
{ 

// cleaning title field 
$title = 'title'; 

if ($title == '') // if title is not set 
$title = '(No Title Provided)';// use (empty title) string 

//reads the name of the file the user submitted for uploading 
$image=$_FILES['image']['name']; 
// if it is not empty 
if ($image) 
{ 
// get the original name of the file from the clients machine 
$filename = stripslashes($_FILES['image']['name']); 

// get the extension of the file in a lower case format 
$extension = getExtension($filename); 
$extension = strtolower($extension); 
// if it is not a known extension, we will suppose it is an error, print an error message 
//and will not upload the file, otherwise we continue 
if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png")) 
{ 
echo '<b> Error! </b> - The image that you attempted to upload is not in the correct format. The file format <b> must </b> be one of the following: <b> "jpg", "jpeg" </b> or <b> "png" </b>. Please try again.'; 
$errors=1; 
} 
else 
{ 
// get the size of the image in bytes 
// $_FILES[\'image\'][\'tmp_name\'] is the temporary filename of the file in which the uploaded file was stored on the server 
$size=getimagesize($_FILES['image']['tmp_name']); 
$sizekb=filesize($_FILES['image']['tmp_name']); 

//compare the size with the maxim size we defined and print error if bigger 
if ($sizekb > 1150000) 
{ 
echo '<b> Error! </b> - The file that you are attempting to upload is greater than the prescribed <b> 1MB </b> limit. Please try again.'; 
$errors=1; 
} 

//we will give an unique name, for example the time in unix time format 
$image_name=$title.'.'.$extension; 
//the new name will be containing the full path where will be stored (images folder) 
$newname="images/".$image_name; 
$copied = copy($_FILES['image']['tmp_name'], $newname); 
//we verify if the image has been uploaded, and print error instead 
if (!$copied) 
{ 
echo '<b> Error! </b> Your file has not been loaded'; 
$errors=1; 
} 
else 
{ 
// the new thumbnail image will be placed in images/thumbs/ folder 
$thumb_name='images/thumbs/'.$image_name; 
// call the function that will create the thumbnail. The function will get as parameters 
//the image name, the thumbnail name and the width and height desired for the thumbnail 
$thumb=make_thumb($newname,$thumb_name,WIDTH,HEIGHT); 
}} }} 

//If no errors registred, print the success message and show the thumbnail image created 
if(isset($_POST['Submit']) && !$errors) 
{ 
echo '<br><b> Success! </b> - Your image has been uploaded</br>'; 
echo '<img src="'.$thumb_name.'">'; 
} 

?> 
<!-- next comes the form, you must set the enctype to "multipart/form-data" and use an input type "file" --> 

<form name="newad" method="post" enctype="multipart/form-data" action=""> 
<table> 
<tr><td><input type="text" name="title" ></td></tr> 
<tr><td><input type="file" name="image" ></td></tr> 
<tr><td><input name="Submit" type="submit" value="Upload image"></td></tr> 
</table> 
</form> 

私は、これは本当の初心者の質問であり得ることを認めるが、私は、画像ファイルを特定の形式として、ユーザが提供する「タイトル」を保存中に問題を抱えています。ユーザーが 'Title'を指定しないと、 '（No Title Provided）'がファイル名として保存されますが、正しく動作するようになりますが、 'Title'テキストフィールドを入力すると値isnファイル名として引っ張られて挿入されました。

は、私の答えは、コードのこの部分内にあることを知っている：

$title = 'title'; 

if ($title == '') // if title is not set 
$title = '(No Title Provided)';// use (empty title) string 

しかし、私は試してみて、任意の運なしでこれを克服する方法のすべての方法を試してみました。私はちょうど誰かがこれを見て、私が間違っている場所を教えてもらえるかどうか疑問に思った。

多くのありがとうと親切に感謝します。

Answer 1

@ThatOtherPersonのおかげで、私は 'タイトル'の値をキャプチャしていないことに気付きました。解決策は、あなたがテキストフィールドから情報を取得する必要があり

$title = ($_POST['title']); 

if ($title == '') // if title is not set 
$title = '(No Title Provided)';// use (empty title) string 

Answer 2

です：

$title = $_POST['title']; 

フィールドを空のままにされている場合は、チェックするためempty()を使用するほうが良いことがあります

if (empty($title)) 
$title = "(No Title Provided)";