이미지 파일 업로드

Question

누군가 제발 저를 도울 수 있는지 궁금합니다.

온라인 자습서와 '시행 착오'를 사용하여 사용자가 이미지 파일을 업로드 할 수 있도록 아래 스크립트를 작성했습니다.

<?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> 

나는이 진짜 초보자 질문이 될 수 있음을 인식,하지만 난 이미지 파일을 식별하는 형태로 사용자가 제공 한 '제목을'절약에 문제가 있습니다. 사용자가 '제목'을 제공하지 않으면 '제공되지 않음'이라는 단어가 파일 이름으로 저장되고 올바르게 작동 할 수 있지만 '제목'텍스트 필드를 완성하면 값은 isn이 아닙니다. 파일 이름으로 가져 와서 삽입했습니다.

나는 대답은이 코드 조각 내에있는 것을 알고 :

$title = 'title'; 

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

그러나 나는 시도하고 운이없이이를 극복 할 수있는 방법의 모든 방법을 시도했습니다. 나는 누군가가 이걸 좀 봐줄 수 있는지 궁금해하고 내가 잘못한 곳을 알려주지.

많은 감사와 친절한 인사.

Answer 1

덕분에 @ThatOtherPerson 덕분에 'Title'값을 캡처하지 못하고 있음을 깨달았습니다. 이 솔루션은 사용자가 텍스트 필드에서 정보를 잡아 필요

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

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

Answer 2

입니다 :

$title = $_POST['title']; 

필드가 비어 된 경우 확인하기 위해 empty()를 사용하는 것이 더 나을 수 있습니다

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