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php - Android 应用程序发送一个字符串以在网页上显示

转载 作者:行者123 更新时间:2023-11-30 00:24:21 25 4
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几个月来我一直在尝试创建一个 Android 应用程序和网页,应用程序向网站发送一个简单的字符串,网站显示该文本。我试过很多教程,但它们都不完整或过时。

解决方案

因为在编码方面我只是一个菜鸟,所以我无法判断这些建议是完整的故事还是故事的一部分。原来我只知道了一部分。

我的目标是:创建一个 Android 应用程序和网页,其中应用程序向网站发送一个简单的字符串,网站显示该文本供所有人查看。我在问题中没有提到最后一部分。因此,询问完整有助于获得正确答案。

我必须采取几个步骤:

  1. 注册一个域名(我做的)
  2. 签约托管服务(我做过)
  3. 将带有哈希代码(见下文)的 php 文件上传到该网页顶层目录中的网页文件管理器,确保没有内容管理器正在为该域或子域运行(这里我犯了一个错误:我让 Parallels Presence Builder 在目标域上运行。因为它总是使用模板,所以我无法发布“干净”版本的 php 文件。这导致我的网页返回以 <!DOCTYPE 等开头的垃圾信息。我创建了一个没有内容管理的新子域,并使用 Hash 的 php 代码上传了一个 php 文件。它运行良好)
  4. 然后我使用 Hash 代码创建了 Android 应用程序(见下文,不要忘记 .jar 文件并且不要忘记将 .jar 文件添加为库!)。
  5. 由于我的问题不完整,Hash 的代码没有处理网页必须显示文本供所有人查看的部分。由于 php 对每个用户都是个人的,因此我必须对代码进行补充。

看这里:

<?php
$msg = $_POST['msg'];
if($_SERVER['REQUEST_METHOD']=='POST'){
$myfile = fopen("yourfilename.txt", "a") or die("Unable to open file!");
fwrite($myfile, $msg . "\n");
fclose($myfile);
}

$myfile = fopen("yourfilename.txt", "r") or die("Unable to open file!");
echo fread($myfile,filesize("yourfilename.txt"));
fclose($myfile);
?>

最佳答案

好吧,让我给你一个非常简单的方法来连接你的应用程序和 PHP 服务页面......首先你必须确保你已经有这个库:1- http-core-4.1.jar2- httpclient-4.0.3.jar

然后创建一个名为“JSONParser”的新类并粘贴此代码:

import android.util.Log;

import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.client.utils.URLEncodedUtils;
import org.apache.http.impl.client.DefaultHttpClient;
import org.json.JSONException;
import org.json.JSONObject;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.UnsupportedEncodingException;
import java.util.List;

public class JSONParser {

static InputStream is = null;
static JSONObject jObj = null;
static String json = "";

// constructor
public JSONParser() {

}

// function get json from url
// by making HTTP POST or GET mehtod
public JSONObject makeHttpRequest(String url, String method,
List<NameValuePair> params) {

// Making HTTP request
try {

// check for request method
if(method == "POST"){
// request method is POST
// defaultHttpClient
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url);
httpPost.setEntity(new UrlEncodedFormEntity(params));

HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();

}else if(method == "GET"){
// request method is GET
DefaultHttpClient httpClient = new DefaultHttpClient();
String paramString = URLEncodedUtils.format(params, "utf-8");
url += "?" + paramString;
HttpGet httpGet = new HttpGet(url);

HttpResponse httpResponse = httpClient.execute(httpGet);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
}

} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}

try {
BufferedReader reader = new BufferedReader(new InputStreamReader(
is, "iso-8859-1"), 8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
json = sb.toString();
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}

// try parse the string to a JSON object
try {
jObj = new JSONObject(json);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}

// return JSON String
return jObj;

}
}

好吧,现在转到你的主要 Activity 并按如下方式进行:

public class MainActivity extends AppCompatActivity {
JSONParser jparser = new JSONParser();
ProgressDialog pdialog;

@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);

String msg = "Hello";

new connector().execute(msg); //This will create a new thread to connect with your service page.


}

//The AsyncClass connector..



class connector extends AsyncTask<String,String,String> {
@Override
protected void onPreExecute(){
pdialog = new ProgressDialog(MainActivity.this);
pdialog.setMessage("Sending Hi...");
pdialog.setCancelable(false);
pdialog.setIndeterminate(false);
pdialog.show();
}
//ArrayList to carry your values
List<NameValuePair> data = new ArrayList<>();

@Override
protected String doInBackground(String... params) {

try {

data.add(new BasicNameValuePair("msg",params[0])); //here we sit the key "msg" to carry your value which you passed on thread execute.

//Then create the JSONObject to make the rquest and pass the data(ArrayList).
JSONObject json = jparser.makeHttpRequest("Http://example.com/mypage.php","POST",data);

/*In case you want to receive a value from your page you should do something like this..
String getString = json.getString("key1");
Int getInt = json.getInt("key2");
key1,key2 are keys that you send from your page.
*/





} catch (Exception e) {
/* Any Exception here */
}

return null;

}

@Override
protected void onPostExecute(String file_url){
pdialog.dismiss();
// This After the thread did it's operate
/* For Example you Toast your key1,key2 values :D*/

}
}



}

在您的 PHP 页面中,您应该执行以下操作..

<?php
//In case of receive..
$msg = $_POST['msg'];
echo $msg;

//In case of send..
$values = array();
$values["key1"] = "Hello this is key1";
$values["key2"] = "Hello this is key2";
echo json_encode($values); //this will make a json encode which you can get it back in your application ;)

?>

希望对您有所帮助 :),抱歉我的英语不好 :v

关于php - Android 应用程序发送一个字符串以在网页上显示,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/45663685/

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