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Php 我想将图像压缩到特定目录_Php - Fatal编程技术网

Php 我想将图像压缩到特定目录

Php 我想将图像压缩到特定目录,php,Php,图像已压缩,但它将存储在where ever脚本文件夹中,但我想存储特定给定目录下的图像 $name = ''; $type = ''; $size = ''; $error = ''; function compress_image($source_url, $destination_url, $quality) { $info = getimagesize($source_url); if ($info['mime'] == 'image/jpeg')

图像已压缩,但它将存储在where ever脚本文件夹中,但我想存储特定给定目录下的图像

$name = ''; $type = ''; $size = ''; $error = '';
function compress_image($source_url, $destination_url, $quality) {

      $info = getimagesize($source_url);

          if ($info['mime'] == 'image/jpeg')
          $image = imagecreatefromjpeg($source_url);

          elseif ($info['mime'] == 'image/gif')
          $image = imagecreatefromgif($source_url);

          elseif ($info['mime'] == 'image/png')
          $image = imagecreatefrompng($source_url);

          imagejpeg($image, $destination_url, $quality);
          return $destination_url;
        }
     if ($_POST) {
            $upload_image = $_FILES["file"][ "name" ];
            if ($_FILES["file"]["error"] > 0)
            {
            $error = $_FILES["file"]["error"];
            }
            else if (($_FILES["file"]["type"] == "image/gif") ||
            ($_FILES["file"]["type"] == "image/jpeg") ||
            ($_FILES["file"]["type"] == "image/png") ||
            ($_FILES["file"]["type"] == "image/pjpeg")) {

            $url = $upload_image;
           $filename = compress_image($_FILES["file"]["tmp_name"], $upload_image, 80);


           // if(move_uploaded_file($image, $target_dir.$url))
            {
              echo "uploaded success";
            }
        }else {
            $error = "Uploaded image should be jpg or gif or png";
        }
        }
替换

imagejpeg($image, $destination_url, $quality);

替换

imagejpeg($image, $destination_url, $quality);


我在下面给出的类似于此的mypath不工作$destination\u url=$\u SERVER['DOCUMENT\u ROOT']./mysite/uploadimages/;imagejpeg($image,$destination\u url,$quality);我在下面给出的类似于此的mypath不工作$destination\u url=$\u SERVER['DOCUMENT\u ROOT']./mysite/uploadimages/;imagejpeg($image,$destination\u url,$quality);如果你用流利的英语提问,并用代码另外解释问题,你的问题更有可能得到回答。sure@Blackam。如果你用流利的英语提问,并用代码另外解释问题,你的问题更有可能得到回答。sure@Blackam。
imagejpeg($image, "uploads/".$destination_url, $quality);