Java 如何将数据发送到php服务器并将数据接收到android

Java 如何将数据发送到php服务器并将数据接收到android,java,android,Java,Android,我是android编程的新手,我想将数据发送到php服务器并接收数据以在android中显示 但我不知道如何创建它,我不知道如何使用库?可以给我举一些例子来练习 对不起,我的英语不好 示例Myphp“xxx.xxx.x.x/api.php” 我想从edittext向php服务器发送关键字和索引,并接收json数据以显示listview。HttpURLConnection可用于获取、放置、发布和删除请求: try { // Construct the URL

我是android编程的新手,我想将数据发送到php服务器并接收数据以在android中显示

但我不知道如何创建它,我不知道如何使用库?可以给我举一些例子来练习

对不起,我的英语不好

示例Myphp“xxx.xxx.x.x/api.php”


我想从edittext向php服务器发送
关键字
索引
,并接收json数据以显示listview。

HttpURLConnection可用于获取、放置、发布和删除请求:

try {
                // Construct the URL
    URL url = new URL("http://xxxxxxx/webservice/&query="
    +keyword.getText().toString()
    +"&index="+index.getText().toString());
    // Create the request 
            urlConnection = (HttpURLConnection) url.openConnection();
            urlConnection.setRequestMethod("GET");
            urlConnection.connect();
    // Read the input stream into a String
            InputStream inputStream = urlConnection.getInputStream();
            StringBuffer buffer = new StringBuffer();
            if (inputStream == null) {
                // Nothing to do.
                return null;
            }
            reader = new BufferedReader(new InputStreamReader(inputStream));

            String line;
            while ((line = reader.readLine()) != null) {
                // Since it's JSON, adding a newline isn't necessary (it won't affect parsing)
                // But it does make debugging a *lot* easier if you print out the completed
                // buffer for debugging.
                buffer.append(line + "\n");
            }

            if (buffer.length() == 0) {
                // Stream was empty.  No point in parsing.

            }
            JsonStr = buffer.toString(); //result
} catch (IOException e) {
 }

其中
关键字
索引
将是您的编辑文本变量。

哦!对不起,我忘了告诉php文件名
“111.11.11/api.php”
,无法使用
http://xxxxxxx/webservice/&query=“…
ok然后将url替换为111.11.11/api.php?&query=“+关键字.getText().toString()+”&index=“+index.getText().toString()))我收到错误行
返回null
无法从具有无效结果类型的方法返回值。
connect()
是多余的。输入流不能为null,因此对此进行测试毫无意义。异常应该传播到调用方,而不是被吞没。@EJP您能给我举个例子吗?或者告诉我该怎么办
try {
                // Construct the URL
    URL url = new URL("http://xxxxxxx/webservice/&query="
    +keyword.getText().toString()
    +"&index="+index.getText().toString());
    // Create the request 
            urlConnection = (HttpURLConnection) url.openConnection();
            urlConnection.setRequestMethod("GET");
            urlConnection.connect();
    // Read the input stream into a String
            InputStream inputStream = urlConnection.getInputStream();
            StringBuffer buffer = new StringBuffer();
            if (inputStream == null) {
                // Nothing to do.
                return null;
            }
            reader = new BufferedReader(new InputStreamReader(inputStream));

            String line;
            while ((line = reader.readLine()) != null) {
                // Since it's JSON, adding a newline isn't necessary (it won't affect parsing)
                // But it does make debugging a *lot* easier if you print out the completed
                // buffer for debugging.
                buffer.append(line + "\n");
            }

            if (buffer.length() == 0) {
                // Stream was empty.  No point in parsing.

            }
            JsonStr = buffer.toString(); //result
} catch (IOException e) {
 }