繁体   English   中英

使用php和JSON将android连接到mySQL

[英]Connecting android to mySQL using php and JSON

我正在尝试创建一个通过php和JSON连接到mySQL数据库的android应用程序。 这是我在服务器上的php文件:

<?php
$host="my_host";
$username="my_name";
$password="my_password";
$db_name="my_db_name";

$con=mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");
$sql = "SELECT    UserID, DisplayName
FROM      User
WHERE     (UserName LIKE '%it%') OR (DisplayName LIKE '%it%')"; 
$result = mysql_query($sql);
$json = array();

if(mysql_num_rows($result)){
while($row=mysql_fetch_assoc($result)){
$json['UserRes'][]=$row;
}
}
mysql_close($con);
echo json_encode($json); 
?> 

这是我的JSON Java类代码:

public class JSONClass extends AsyncTask<String,Void,String>{

public HashMap<String, String> tbl = new HashMap<String, String>();
private Context context;

public JSONClass(Context context) {
    this.context = context;
}

@Override
protected String doInBackground(String... arg0) {
    try{
        String link = "some url";
        URL url = new URL(link);
        HttpClient client = new DefaultHttpClient();
        HttpGet request = new HttpGet();
        request.setURI(new URI(link));
        HttpResponse response = client.execute(request);
        BufferedReader in = new BufferedReader (new InputStreamReader(response.getEntity().getContent()));
        StringBuffer sb = new StringBuffer("");
        String line="";
        while ((line = in.readLine()) != null) {
            sb.append(line);
            break;
        }
        in.close();
        return sb.toString();
    }catch(Exception e){
        return new String("Exception: " + e.getMessage());
    }
}
}

我是新来的,只想从那里开始学习基础知识和工作。 php文件应该返回一行作为答案,但是据我所知,该函数属于try-catch部分。 我知道结果应该是HashMap。 有人可以告诉我我做错了什么,还是给我小费以得到结果? 先感谢您!

这是您从url获取数据的方式,其中以下url是您的PHP脚本的网址:

HttpClient client = new DefaultHttpClient();
HttpPost post = new HttpPost(url);
post.addHeader("Content-Type", "application/x-www-form-urlencoded");
HttpResponse response = client.execute(post);
HttpEntity entity = response.getEntity();
String result = EntityUtils.toString(entity, "utf-8");
JSONArray ja = new JSONArray(result);
for(int i = 0 ; i < ja.length() ; i++){
  String name = ja.getJSONObject(i).getString("name"); //write name of column
}

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM