大家好,我当前的 Android 应用程序中有一个 Activity ,其中有一个 Web View 。我想知道如何从该网站上的 php 获取变量并将其存储到我的 android 应用程序上的变量中(是的,我控制该网站,并且具有完整的编辑功能)。
最佳答案
我会解释一般需要做的事情。首先,在你的 Android 应用程序端,应该有类似这样的代码。
1)在你的android端
String url = "www.yoururl.com/yourphpfile.php";
List<NameValuePair> parmeters = new ArrayList<NameValuePair>();
parameters.add(new BasicNameValuePair("category",category));
parameters.add(new BasicNameValuePair("subcategory",subcategory));// These are the namevalue pairs which you may want to send to your php file. Below is the method post used to send these parameters
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url); // You can use get method too here if you use get at the php scripting side to receive any values.
httpPost.setEntity(new UrlEncodedFormEntity(parameters));
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
BufferedReader reader = new BufferedReader(new InputStreamReader(
is, "iso-8859-1"), 8); // From here you can extract the data that you get from your php file..
StringBuilder builder = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
builder.append(line + "\n");
}
is.close();
json = sb.toString(); // Here you are converting again the string into json object. So that you can get values with the help of keys that you send from the php side.
String message = json.getString("message");
int success = json.getInt("success"); // In this way you again convert json object into your original data.
2) 在你的 php 端
$response = array();
// check for post fields
if (isset($_POST['category']) && isset($_POST['subcategory'])) {
$category = $_POST['category'];
$subcategory = $_POST['subcategory'];
// include db connect class
require_once __DIR__ . '/db_connect.php';
// connecting to db
$db = new DB_CONNECT();
// mysql inserting a new row
$result = mysql_query("INSERT INTO categorytable(category, subcategory) VALUES('$category', '$subcategory')");
// check if row inserted or not
if ($result) {
// successfully inserted into database
$response["success"] = 1;
$response["message"] = "Data inserted into database.";
// echoing JSON response
echo json_encode($response);// Here you are echoing json response. So in the inputstream you would get the json data. You need to extract it over there and can display according to your requirement.
}
正如您所说,您对 php 相当精通,您可能已经发现 php 中的上述内容变得尽可能简单,并且不安全。因此,您可以遵循一些 pdo 或其他一些安全方法在 php 端进行编码。确保将 Android 端的代码包含在 asynctask 中,以便在需要的地方在单独的线程中运行。希望这会有所帮助。
关于php - 如何将变量从 php 发送到 Android 应用程序?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/13635395/