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Android的HTTP POST + Web服务PHP

本文介绍了Android的HTTP POST + Web服务PHP的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我在PHP返回一个字符串的Web服务，他收到
2参数应用，身份证和Te。
我测试过它与mozzila插件使用的海报，所以我决定用它为我的Andr​​oid应用。

这是我的Andr​​oid code：

 最后的查询字符串= NULL;AsyncHttpClient客户端=新AsyncHttpClient（）;
RequestParams RP =新RequestParams（）;
rp.put（ID，NUM）;
rp.put（TE，标签）;
Log.i（HTTP，送之前\\ n）;
client.post（http://appdomain.hol.es/webService.php，RP，新JsonHtt presponseHandler（）{    公共无效的onSuccess（字符串jObject）
    {
        query.replace（查询，jObject）;
Log.i（HTTP，recived：+ jObject +\\ n）;    }
    公共无效onFailure处（Throwable的为arg0）
    {
Log.i（HTTP，失败）;
    }
}）;

我调试蒙山log.i，我已经可以看到，它并不显示既不recived没有失败。
任何人都可以helpl我？

PD：我留下最相关的WebService 

 的$ id = $ _ POST [ID];
$ TE = $ _ POST [TE];
$查询=SELECT`preg`，`respA`​​，`respB`，`respC`，`respD`，``respV`FROM$忒`其中`id` =$ ID。;$ resultado =的mysql_query（$查询，$链接）;
$ arraySalida =阵列（）;
而（$ registro = mysql_fetch_assoc（$ resultado））：
    $卡德纳= \"{$registro['$p$pg']};{$registro['respA']};{$registro['respB']};{$registro['respC']};{$registro['respD']};{$registro['respV']}\";
    $ arraySalida [] = $卡德纳;ENDWHILE;
回声破灭（：，$ arraySalida）;

@jaimin作品的解决办法，但编译器说：类型不匹配：不能从AsyncTask的转换为字符串在

（！）

这是code：

 公共字符串BBDD（INT NUM，串片）
    {
        HttpAsyncTask httpAsyncTask =新HttpAsyncTask（将String.valueOf（NUM）选项卡）;
        （！）/ * * /字符串resul = httpAsyncTask.execute（http://opofire.hol.es/webServiceOpoFire.php）;        返回resul;
}


解决方案

有关HTTP POST我会建议你使用的AsyncTask＆LT;>将在用户界面上单独的线程中运行
这里是code两个月以来我使用和它的做工精细

 私有类HttpAsyncTask扩展的AsyncTask＆LT;弦乐，太虚，字符串＆GT; {
 私人字符串ID，忒;
 公共HttpAsyncTask（ID字符串，字符串TE）{
        this.id = ID;
        this.te = TE;    }        @覆盖
        保护字符串doInBackground（字符串的URL ...）{            返回POST（网址[0]）;
        }
        // onPostExecute显示的AsyncTask的结果。
        @覆盖
        保护无效onPostExecute（字符串结果）{
            Toast.makeText（getBaseContext（），发送的数据！，Toast.LENGTH_LONG）.show（）;
       }
    }
公共静态字符串POST（字符串URL）{
    为InputStream的InputStream = NULL;
    字符串结果=;
    尝试{        // 1.创建HttpClient的
        HttpClient的HttpClient的=新DefaultHttpClient（）;        // 2.化妆POST请求到指定的URL
        HttpPost httpPost =新HttpPost（URL）;
       以这种方式//传递参数       清单＆LT;＆的NameValuePair GT; namevaluepairs中=新的ArrayList＆LT;＆的NameValuePair GT;（2）;
    nameValuePairs.add（新BasicNameValuePair（ID，价值））;
    nameValuePairs.add（新BasicNameValuePair（TE，价值））;    //添加数据
    httppost.setEntity（新UrlEn codedFormEntity（namevaluepairs中））;        // 8.执行POST请求到指定的URL
        HTT presponse HTT presponse = httpclient.execute（httpPost）;        // 9.收到响应为InputStream的
        的InputStream = HTT presponse.getEntity（）的getContent（）。        // 10.转换的InputStream串
        如果（的InputStream！= NULL）
            结果= convertInputStreamToString（InputStream的）;
        其他
            结果=没有工作！    }赶上（例外五）{
        Log.d（InputStream的，e.getLocalizedMessage（））;
    }
 // 11返回结果
    返回结果;}私人静态字符串convertInputStreamToString（的InputStream的InputStream）抛出IOException
    // TODO自动生成方法存根    的BufferedReader的BufferedReader =新的BufferedReader（新的InputStreamReader（InputStream的））;
    串线=;
    字符串结果=;
    而（（行= bufferedReader.readLine（））！= NULL）
        结果+ =行;    inputStream.close（）;
    返回结果;}

传递值，您可以创建构造这样。

形成你的活动做到这一点。

  HttpAsyncTask httpAsyncTask =新HttpAsyncTask（ID，TE）; //这会传递变量的值
 字符串ResultfromServer = httpAsyncTask.execute（urlStr）; //字符串ResultfromServer是你的响应字符串

和中的AsyncTask我已经创建构造

这是非常有帮助，我希望它会帮助你太

I have a web service in PHP which returns a String, he recives two parametres, id and te.I've tested its using with the mozzila addon poster, so i decided to use it for my android app.

This is my android code:

final String query = null;

AsyncHttpClient client = new AsyncHttpClient();
RequestParams rp = new RequestParams();
rp.put("id", num);
rp.put("te", tab);
Log.i("http","before send\n");
client.post("http://appdomain.hol.es/webService.php",rp, new JsonHttpResponseHandler(){

    public void onSuccess(String jObject)
    {
        query.replace(query, jObject);
Log.i("http","recived: "+jObject+"\n");

    }
    public void onFailure(Throwable arg0)
    {
Log.i("http","fail");
    }
});

I'm debugging whith log.i and i've could seen that it doesn't show neither recived neither fail.can anyone helpl me?

PD: i leave the most relevant of webService

$id = $_POST["id"];
$te = $_POST["te"];
$query = "SELECT `preg` , `respA` , `respB` , `respC` , `respD` , `respV`FROM `".$te."` WHERE `id` =".$id;

$resultado= mysql_query($query,$link);
$arraySalida = array();
while($registro = mysql_fetch_assoc ($resultado) ):
    $cadena = "{$registro['preg']};{$registro['respA']};{$registro['respB']};{$registro['respC']};{$registro['respD']};{$registro['respV']}";
    $arraySalida[]= $cadena;

endwhile;
echo implode(":",$arraySalida);

the solution of @jaimin works but the compiler says: Type mismatch: cannot convert from AsyncTask to String in (!)

this is the code:

public String BBDD(int num, String tab)
    {
        HttpAsyncTask httpAsyncTask = new HttpAsyncTask(String.valueOf(num),tab);
        /*(!)*/String resul = httpAsyncTask.execute("http://opofire.hol.es/webServiceOpoFire.php");

        return resul;
}
解决方案

for Http Post i'll suggest you to use AsyncTask<> which will run in separate thread from UIhere is the code i am using since two months and its working fine

 private class HttpAsyncTask extends AsyncTask<String, Void, String> {
 private String id,te;
 public  HttpAsyncTask(String id,String te){
        this.id = id;
        this.te = te;

    }

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



            return POST(urls[0]);
        }
        // onPostExecute displays the results of the AsyncTask.
        @Override
        protected void onPostExecute(String result) {
            Toast.makeText(getBaseContext(), "Data Sent!", Toast.LENGTH_LONG).show();
       }
    }
public static String POST(String url){
    InputStream inputStream = null;
    String result = "";
    try {

        // 1. create HttpClient
        HttpClient httpclient = new DefaultHttpClient();

        // 2. make POST request to the given URL
        HttpPost httpPost = new HttpPost(url);
       // pass parameters in this way

       List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
    nameValuePairs.add(new BasicNameValuePair("id", "value "));
    nameValuePairs.add(new BasicNameValuePair("te", "value"));

    //add data
    httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

        // 8. Execute POST request to the given URL
        HttpResponse httpResponse = httpclient.execute(httpPost);

        // 9. receive response as inputStream
        inputStream = httpResponse.getEntity().getContent();

        // 10. convert inputstream to string
        if(inputStream != null)
            result = convertInputStreamToString(inputStream);
        else
            result = "Did not work!";

    } catch (Exception e) {
        Log.d("InputStream", e.getLocalizedMessage());
    }
 // 11. return result
    return result;

}

private static String convertInputStreamToString(InputStream inputStream) throws IOException {
    // TODO Auto-generated method stub

    BufferedReader bufferedReader = new BufferedReader( new InputStreamReader(inputStream));
    String line = "";
    String result = "";
    while((line = bufferedReader.readLine()) != null)
        result += line;

    inputStream.close();
    return result;

}

for passing values you can create constructor like this.

form your activity do this

 HttpAsyncTask httpAsyncTask = new HttpAsyncTask(id,te);//this will pass variables values
 String  ResultfromServer = httpAsyncTask.execute(urlStr);// String ResultfromServer is your response string

and in AsyncTask I've created constructor

this is very helpful for me hope it will help you too

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08-19 18:05
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