图像大小调整PHP [英] Image resize PHP

问题描述

可能重复：

任何人都可以建议最好的图像调整脚本在PHP？

我仍然是一个PHP的图像处理或文件处理的新手。

以下

我使用简单的html表单发布图像文件并通过php上传。
当我尝试和改变我的代码，以适应更大的文件（即调整大小）我得到一个错误。
已经在网上搜索，但无法找到任何真正简单的东西。

$ $ $ $ $ $ $ $ $ $ $文件格式（$ _ FILES ['image' ] [ 'tmp_name的值']）;

//将大小与我们定义的最大大小进行比较，如果大于
则打印错误if（$ size == FALSE）
{
$ errors = 1;
} else if（$ size [0]> 300）{//如果width大于300px
$ aspectRatio = 300 / $ size [0];
$ newWidth = round（$ aspectRatio * $ size [0]）;
$ newHeight = round（$ aspectRatio * $ size [1]）;
$ imgHolder = imagecreatetruecolor（$ newWidth，$ newHeight）;
}

$ newname = ROOTPATH.LOCALDIR。/ images /\".$ image_name; // image_name生成

$ copy = imagecopyresized（$ imgHolder，$ _FILES ['image'] ['tmp_name']，0，0，0，0，$ newWidth，$ newHeight，$ size [0]，$ size [1]）;
move_uploaded_file（$ copy，$ newname）; //我要将文件移动到$ newname

的位置，我得到的错误是：
$ b

imagecopyresized（）：提供的参数
不是中的有效图像资源

在此先感谢

---

感谢您的所有意见，改变它到这个

  $ oldImage = imagecreatefromstring（file_get_contents（$ _ FILES ['image'] ['tmp_name']））; 
 $ copy = imagecopyresized（$ imgHolder，$ oldImage，0，0，0，$ newWidth，$ newHeight，$ size [0]，$ size [1]）; 
 if（！move_uploaded_file（$ copy，$ newname））{
 $ errors = 1; 
 
 
 
 $ b 
没有得到一个PHP日志错误，但没有保存：（
 
 
 
有什么想法？ 
 
 
再次感谢 
 
 
 
 
 
 结果  
 
 
以下作品 
 
 
  $ oldImage = imagecreatefromjpeg（$ img）; 
 $ imageHolder = imagecreatetruecolor（$ newWidth，$ newHeight）; 
 imagecopyresized（$ imageHolder，$ oldImage，0，0， 0，$ newWidth，$ newHeight，$ width，$ height）; 
 imagejpeg（$ imageHolder，$ newname，100）; 
  
感谢大家的帮助 
 
解决方案
  imagecopyresized  imagecreatefromFILETYPE 来创建一个文件名，例如，如果它是一个JPEG，使用 imagecreatefromjpeg 并传递文件名 - 这将返回一个图像资源。
 
 
 
如果您不知道文件类型，则全部不会丢失。您可以以字符串的形式读取文件，并使用 imagecreatefromstring （自动检测文件类型）来加载它，如下所示： 
 
 
  $ oldImage = imagecreatefromstring（file_get_contents（$ _ FILES ['image'] ['tmp_name']））; 
  
 

  
Possible Duplicate:

  Can anybody suggest the best image resize script in php?  
I'm still a newbie regarding image handling or file handling for that matter in PHP.


Would appreciate any input regarding the following

I post an image file using a simple html form and upload it via php.
When i try and alter my code to accomodate larger files (i.e. resize) I get an error.
Have been searching online but cant find anything really simple.
$size = getimagesize($_FILES['image']['tmp_name']);

//compare the size with the maxim size we defined and print error if bigger
if ($size == FALSE)
{
    $errors=1;
}else if($size[0] > 300){   //if width greater than 300px
    $aspectRatio = 300 / $size[0];
    $newWidth = round($aspectRatio * $size[0]);
    $newHeight = round($aspectRatio * $size[1]);
    $imgHolder = imagecreatetruecolor($newWidth,$newHeight);
}

$newname= ROOTPATH.LOCALDIR."/images/".$image_name; //image_name is generated

$copy = imagecopyresized($imgHolder, $_FILES['image']['tmp_name'], 0, 0, 0, 0, $newWidth, $newHeight, $size[0], $size[1]);
move_uploaded_file($copy, $newname); //where I want to move the file to the location of $newname
The error I get is:

  
imagecopyresized(): supplied argument
  is not a valid Image resource in
Thanks in advance



Thanks for all your input, i've changed it to this
$oldImage = imagecreatefromstring(file_get_contents($_FILES['image']['tmp_name']));
$copy = imagecopyresized($imgHolder, $oldImage, 0, 0, 0, 0, $newWidth, $newHeight, $size[0], $size[1]);
if(!move_uploaded_file($copy, $newname)){
    $errors=1;
}
Not getting a PHP log error but its not saving :(

Any ideas?

Thanks again



Result

Following works.
$oldImage = imagecreatefromjpeg($img);
$imageHolder = imagecreatetruecolor($newWidth, $newHeight);
imagecopyresized($imageHolder, $oldImage, 0, 0, 0, 0, $newWidth, $newHeight, $width, $height);
imagejpeg($imageHolder, $newname, 100);
Thanks for everyones help
解决方案
imagecopyresized takes an image resource as its second parameter, not a file name. You'll need to load the file first. If you know the file type, you can use imagecreatefromFILETYPE to load it. For example, if it's a JPEG, use imagecreatefromjpeg and pass that the file name - this will return an image resource.

If you don't know the file type, all is not lost. You can read the file in as a string and use imagecreatefromstring (which detects file types automatically) to load it as follows:
$oldImage = imagecreatefromstring(file_get_contents($_FILES['image']['tmp_name']));


                        
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