上传图像文件

Question

我想知道是否有人可以帮助我。

通过使用一些在线教程和“反复试验”，我将下面的脚本放在一起，允许用户上传图像文件。

<?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>

我很欣赏这可能是一个真正的初学者问题，但我在保存用户提供的“标题”方面遇到了问题，因为它是识别图像文件的一种形式。如果用户没有提供“标题”，则单词'（无标题提供）'将保存为文件名，我可以使其正常工作，但如果我填写'标题'文本字段，则值为n不会拉动并插入文件名。

我知道答案就在这段代码中：

$title = 'title'; 

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

但我已经尝试过各种各样的方法来试图克服这个没有任何运气的方法。我只是想知道是否有人可以看看这个，让我知道我哪里出错了。

非常感谢和亲切的问候。

Answer 1

您需要从文本字段中获取信息：

$title = $_POST['title'];

最好使用empty()检查字段是否为空：

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

Answer 2

感谢@ThatOtherPerson，我意识到我没有捕获'Title'值。解决方案是

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

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