我和我的朋友在将图片从Android应用程序上传到php中的Codeigniter框架时遇到了问题。问题出现了(我们认为),因为它不是图像文件,因此不能用于图像操作。
一个重要的通知是图像上传在使用html页面进行测试时有效。所有其他数据(如电子邮件)都提供并在Android应用程序中工作。唯一不起作用的是文件上传。但它确实与另一个php文件一起工作,但我的问题是我如何使用最后的PHP脚本工作并将其转换为codeigniter?我想使用这个框架,但到目前为止我们还没有。
以下是上传的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);
}
上传的codeigniter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 $short_url;
}
运行较旧的php脚本,请注意注释的代码不起作用。因为它可能会产生与CI相同的效果。
<?php
/*if ((($_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";
}*/
?>
请帮助我们修改CI功能或java代码,以便我们上传我们心爱的图像。感谢您的建议和更好的智慧!
答案 0 :(得分:4)
答案是在java代码中。我需要添加的是文件体的更多参数。
multipartContent.addPart("userfile", new FileBody(new File(filename), filename, "image/jpeg", "utf-8"));
这完美无瑕。 有关filebody构造函数的信息,您可以在此处找到:http://hc.apache.org/httpcomponents-client-ga/httpmime/apidocs/org/apache/http/entity/mime/content/FileBody.html