从Android的插入多个条目 - ＆GT; PHP - ＆GT; MYSQL [英] INSERT multiple entries from Android -> PHP -> MYSQL

问题描述

我想从一个Android应用程序的外部Mysql数据库中插入多个（1-50）项。我完全得到了一个PHP脚本为单个INSERT查询工作。但我没有，到目前为止，使这项工作有关条目的整个阵列，很可能是由于我有限的PHP的理解。

I am trying to insert multiple (1-50) entries from an Android application to an external Mysql database. I perfectly got a PHP script to work for single INSERT queries. But I am failing so far to make this work for a whole array of entries, most likely due to my limited understanding of PHP.

Android的code：

Android code:

List<NameValuePair> upload_array = new ArrayList<NameValuePair>();
upload_array.add(new BasicNameValuePair("mFirstname[0]", "FirstName 1"));
upload_array.add(new BasicNameValuePair("mFirstname[1]", "FirstName 2"));
upload_array.add(new BasicNameValuePair("mLastname[0]", "LastName 1"));
upload_array.add(new BasicNameValuePair("mLastname[1]", "LastName 2"));
upload_array.add(new BasicNameValuePair("mNickname[0]", "NickName 1"));
upload_array.add(new BasicNameValuePair("mNickname[1]", "NickName 2"));

HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://url/script.php");
HttpResponse response = null;
try {
    httppost.setEntity(new UrlEncodedFormEntity(upload_array));
    response = httpclient.execute(httppost);
} catch (Exception e) {
    e.printStackTrace();
}

和在PHP中：

<?php

$connect = mysqli_connect("***","***","***", "***");

if(mysqli_connect_errno($connect))
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
else
{
    echo "success";
}

$query = mysqli_prepare("INSERT INTO `namelist` (`firstname`,`lastname`,`nickname`)
    VALUES(?,?,?)");

$mFirstname = $_POST['mFirstname'];
$mLastname = $_POST['mLastname'];
$mNickname = $_POST['mNickname'];

foreach($mFirstname as $key as $key => $value) {
    $query->bind_param('sss',$value["mFirstname"],$value["mLastname"],$value["mNickname"];
    $query->execute();
}

mysqli_close($connect);
?>

时的错误在code的Andr​​oid部分已经发生或做到这一点PHP脚本只是没有读我送正确的数据？任何有识之士将是非常欢迎的。

Is the mistake happening in the Android part of the code already or does this PHP script just not read the data I sent properly? Any insight would be very welcome.

推荐答案

好吧，我做了使用JSON数组这项工作。如果任何人使用它，这里是怎么一回事呢：

Ok, I made this work using a JSON Array. In case anyone has use for it, here's how it goes:

Android的，创建JSON字符串：

Android, create JSON String:

//Create JSON string start
String json_string ="{\"upload_fishes\":[";

//Repeat and loop this until all objects are added (and add try+catch)
JSONObject obj_new = new JSONObject();
obj_new.put("fish_id", your_looped_string_1[i]);
obj_new.put("fish_lat", your_looped_string_2[i]);
obj_new.put("fish_lon", your_looped_string_3[i]);
json_string = json_string + obj_new.toString() + ",";

//Close JSON string
json_string = json_string.substring(0, json_string.length()-1);
json_string += "]}";

Android的数据发送到PHP（加试捕+）：

Android send data to PHP (add try+catch):

HttpParams httpParams = new BasicHttpParams();
HttpConnectionParams.setConnectionTimeout(httpParams, TIMEOUT_MILLISEC);
HttpConnectionParams.setSoTimeout(httpParams, TIMEOUT_MILLISEC);
HttpClient client = new DefaultHttpClient(httpParams);

String url = "http://yourserver.com/script.php";

HttpPost request = new HttpPost(url);
request.setEntity(new ByteArrayEntity(json_string.getBytes("UTF8")));
request.setHeader("json", json_string);
HttpResponse response = client.execute(request);

Log.d("FISHY", response.getStatusLine().toString());

PHP脚本：

<?php

//CONNECT TO THE DATABASE
 $DB_HOST = 'yourhost.com';
 $DB_USER = 'user';
 $DB_PASS = 'password';
 $DB_NAME = "db_name";

 $mysqli = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);

 if(mysqli_connect_errno())
{
//    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
else
{
//    echo "Connected to MySQL";
}


   $postdata = file_get_contents("php://input"); 
   $data = json_decode($postdata, true);

   if (is_array($data['upload_fishes'])) {
      foreach ($data['upload_fishes'] as $record) {
        $fid = $record['fish_id'];
        $flat = $record['fish_lat'];
    $flon = $record['fish_lon'];

        mysqli_query($mysqli,"INSERT INTO `fishes`(`fish_type_id`, `fish_lat`, `fish_lon`) VALUES ($fid, $flat, $flon)");
      }
   }


mysqli_close($mysqli);
?>

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