Question

如果code = 1，则插入成功。如果code = 0，则插入是不成功。我不能这样。我想在android屏幕上看到这些信息。我很抱歉我的英语不好。如果你能提供帮助我会很高兴。

php代码：

<?php
    $host='123.45.67.89';
    $uname='username';
    $pwd='password';
    $db="database";

    $con = mysql_connect($host,$uname,$pwd) or die("connection failed");
    mysql_select_db($db,$con) or die("db selection failed");

    $name=$_REQUEST['name'];
    $surname=$_REQUEST['surname'];
    $age=$_REQUEST['age'];

    $flag['code']=0;

    if($r=mysql_query("insert into table values('$name','$surname','$age') ",$con))
    {
        $flag['code']=1;
        echo"hi";
    }

    print(json_encode($flag));
    mysql_close($con);
?>

Java代码：

insert.setOnClickListener(new OnClickListener() {

        @Override
        public void onClick(View arg0) {


            name=editText1.getText().toString();
            surname=editText2.getText().toString();
            age=editText3.getText().toString();

            String data[]={name,surname,age};

            new DoThatThingBro().execute(data);


        }
    });

}


    protected JSONObject addPost(String name, String surname, String age) throws ClientProtocolException, IOException, JSONException{

        String URL = "http://website.com/insert.php?name="+name+"&surname="+surname+"&age="+age;
        HttpClient httpclient = new DefaultHttpClient();
        HttpGet get = new HttpGet(URL);
        HttpResponse response = httpclient.execute(get);
        StatusLine status = response.getStatusLine();
        s = status.getStatusCode();

        if(s == 200){

            HttpEntity e = response.getEntity();
            String data = EntityUtils.toString(e);
            JSONArray posts = new JSONArray(data);
            JSONObject result = posts.getJSONObject(0);
            return result;


        }

        return null;
    }

    public class DoThatThingBro extends AsyncTask<String, String, String>{

        @Override
        protected String doInBackground(String... params) {
            // TODO Auto-generated method stub

            try {
                json = addPost(params[0],params[1],params[2]);
                String data = json.getString("reuslt");

                return data;
            } catch (ClientProtocolException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            } catch (IOException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            } catch (JSONException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }

            return null;
        }


    }

}

Answer 1

在你的php中试试这样的代码。当你想看表时叫它。

$result = mysqli_query($con,"SELECT * FROM tablename");

将表存储在$ result变量中。

while($row = mysqli_fetch_array($result)) {
echo $row['firstcolumn'] . " " . $row['secondcolumn'];
echo "<br>";
}

这会将表中的所有数据作为数组获取并打印所有结果，直到没有其他内容可以打印。希望这有帮助。让我知道。

Answer 2

编辑：评论后，问题更加明确。

将你的php设置为：

$r = mysql_query("insert into table values('$name','$surname','$age') ",$con)
if (!$r) {
    echo "Insert failed: " . mysql_error());
} else {
    $flag['code']=1;
    echo "Successfully inserted " . mysql_affected_rows() . " rows";
}

你的java：

String URL = "http://website.com/insert.php?name="+name+"&surname="+surname+"&age="+age;
HttpClient httpclient = new DefaultHttpClient();
HttpGet get = new HttpGet(URL);
try {
    String response = httpclient.execute(get, new BasicResponseHandler());
    Toast.makeText(this, response, Toast.LENGTH_LONG);
} catch (IOException e) {
    e.printStackTrace();
}

将数据插入表格时，如何知道插入是否成功？

2 个答案: