從Android的Java應用程序上傳圖片Codeigniter

Question

我和我的朋友有一個問題，當涉及到從Android應用程序上傳到Codeigniter框架的圖片在PHP。問題出現了（我們認爲），因爲它不是圖像文件，因此不能用於圖像操作。

一個很大的問題是，使用html頁面進行測試時，圖片上傳工作正常。並且所有其他數據（如電子郵件）都在android應用程序中提供並運行。唯一不起作用的是文件上傳。但它可以與另一個php文件一起工作，但是我的問題是如何將最後一個php腳本運行並將其轉換爲codeigniter？我想用這個框架，但到目前爲止我們還沒有做到。

下面是上傳的javacode - >

HttpClient httpClient = new DefaultHttpClient(); 
HttpContext httpContext = new BasicHttpContext(); 
HttpPost httpPost = new HttpPost("http://url.com/controller-name"); 

try 
{ 
    CustomMultiPartEntity multipartContent = new CustomMultiPartEntity(new ProgressListener() 
    { 
    @Override 
    public void transferred(long num) 
    { 
     publishProgress((int) ((num/(float) totalSize) * 100)); 
    } 
    }); 

    // We use FileBody to transfer an image 
    multipartContent.addPart("userfile", new FileBody(new File(filename))); 
    totalSize = multipartContent.getContentLength(); 
    multipartContent.addPart("email", new StringBody("email@email.com")); 
    multipartContent.addPart("submit", new StringBody("upload")); 

    // Send it 
    httpPost.setEntity(multipartContent); 
    HttpResponse response = httpClient.execute(httpPost, httpContext); 
    String serverResponse = EntityUtils.toString(response.getEntity()); 

    Log.i("SERVER", "Response: " + response.toString()); 
    return serverResponse; 
} 

catch (Exception e) 
{ 
    System.out.println(e); 
} 

，用來上傳的笨方法：

function do_upload() { 
    $image_name = time() . $this->get_random_string(); 
    $config = array(
     'file_name' => $image_name, 
     'allowed_types' => 'jpg|jpeg|gif|png', 
     'upload_path' => $this->gallery_path 

    ); 
    $this->load->library('upload', $config); 
    $this->upload->do_upload(); 
    $image_data = $this->upload->data(); 
    $config = array(
     'source_image' => $image_data['full_path'], 
     'new_image' => $this->gallery_path . '/thumbs', 
     'maintain_ratio' => false, 
     'width' => 300, 
     'height' => 300 
    ); 

    $this->load->library('image_lib', $config); 
    $this->image_lib->resize(); 
    echo $this->image_lib->display_errors(); 

    $image_name = $image_name . $image_data['file_ext']; 
    //Insert into the database 
    $data = array(
     'image_name' => $image_name, 
     'upload_email' => $this->input->post('email'), 
     'ip' => $this->input->ip_address() 
    ); 

    // Insert the new data of the image! 
    $insert = $this->db->insert('table', $data); 
    $short_url = $this->alphaID($this->db->insert_id()); 

    return $short_url; 

} 

年長的PHP腳本的作品，注意的是，註釋代碼不起作用。因爲它可能會產生與CI相同的效果。

<?php 
/*if ((($_FILES["file"]["type"] == "image/gif") 
|| ($_FILES["file"]["type"] == "image/jpeg") 
|| ($_FILES["file"]["type"] == "image/jpg") 
|| ($_FILES["file"]["type"] == "image/png") 
|| ($_FILES["file"]["type"] == "application/octet-stream") 
|| ($_FILES["file"]["type"] == "image/pjpeg")) 
&& ($_FILES["file"]["size"] < 20000)) 
{*/ 
    if ($_FILES["file"]["error"] > 0) 
    { 
     echo "Return Code: " . $_FILES["file"]["error"] . "<br />"; 
    } 
    else 
    { 
     echo "Upload: " . $_FILES["file"]["name"] . "<br />"; 
     echo "Type: " . $_FILES["file"]["type"] . "<br />"; 
     echo "Size: " . ($_FILES["file"]["size"]/1024) . " Kb<br />"; 
     echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br />"; 

     if (file_exists("images/" . $_FILES["file"]["name"])) 
     { 
      echo $_FILES["file"]["name"] . " already exists. "; 
     } 
     else 
     { 
      move_uploaded_file($_FILES["file"]["tmp_name"], 
       "images/" . $_FILES["file"]["name"]); 
      echo "Stored in: " . "images/" . $_FILES["file"]["name"]; 
     } 
    } 
/*} 
else 
{ 
    echo "Invalid file"; 
}*/ 
?> 

請幫助我們修改CI函數或java代碼，以便我們可以上傳我們心愛的圖像。感謝您的建議和更好的智慧！

Answer 1

答案是在java代碼中。我需要添加的是filebody的一些參數。

multipartContent.addPart("userfile", new FileBody(new File(filename), filename, "image/jpeg", "utf-8")); 

這工作完美無瑕。 有關您可以在這裏找到的filebody構造函數的信息：http://hc.apache.org/httpcomponents-client-ga/httpmime/apidocs/org/apache/http/entity/mime/content/FileBody.html