Android 中通过 URL 参数发送发送数据
Sending send data through URL parameters in Android
我正在尝试制作一个应用程序,当用户填写表单并单击发送按钮时,通过 url 参数发送信息,例如:
localhost/my.php?param1=Name¶m2=SureName
我已经通过互联网阅读了它,但无法理解,我找到了一些示例,但它们告诉了如何通过 Json,我只想通过 link 发送。 (我认为这是最简单的方法)
我的php代码是:
<?php
require "connection.php";
if (!isset($_GET['param1'])) {
echo '<p align="center"> No data passed!</p>"';
}
if (strcmp($_GET['param1'],'FirstParam')== 0)
{
$sql= "Query HERE TO ADD INTO DB";
mysqli_query($connection,$sql);
if ($connection->query($sql) === TRUE) {
echo "Record updated successfully";
} else {
echo "Error updating record: " . $connection->error;
}
mysqli_close($connection);
}
我们将不胜感激。
我建议您查看 OkHttp. It is an easy library to send HTTP requests in whatever way you like. On Vogella 您还可以找到发送带有附加参数的 GET 请求的示例。基本上是这样的
HttpUrl.Builder urlBuilder = HttpUrl.parse("https://api.github.help").newBuilder();
urlBuilder.addQueryParameter("v", "1.0");
urlBuilder.addQueryParameter("user", "vogella");
String url = urlBuilder.build().toString();
Request request = new Request.Builder()
.url(url)
.build();
然后你需要在 OkHttpClient
的帮助下(异步地)发送你的请求,就像这样
client.newCall(request).enqueue(new Callback() {
@Override
public void onFailure(Call call, IOException e) {
e.printStackTrace();
}
@Override
public void onResponse(Call call, final Response response) throws IOException {
if (!response.isSuccessful()) {
throw new IOException("Unexpected code " + response);
} else {
// do something wih the result
}
}
(代码取自 Vogella)
我正在尝试制作一个应用程序,当用户填写表单并单击发送按钮时,通过 url 参数发送信息,例如:
localhost/my.php?param1=Name¶m2=SureName
我已经通过互联网阅读了它,但无法理解,我找到了一些示例,但它们告诉了如何通过 Json,我只想通过 link 发送。 (我认为这是最简单的方法)
我的php代码是:
<?php
require "connection.php";
if (!isset($_GET['param1'])) {
echo '<p align="center"> No data passed!</p>"';
}
if (strcmp($_GET['param1'],'FirstParam')== 0)
{
$sql= "Query HERE TO ADD INTO DB";
mysqli_query($connection,$sql);
if ($connection->query($sql) === TRUE) {
echo "Record updated successfully";
} else {
echo "Error updating record: " . $connection->error;
}
mysqli_close($connection);
}
我们将不胜感激。
我建议您查看 OkHttp. It is an easy library to send HTTP requests in whatever way you like. On Vogella 您还可以找到发送带有附加参数的 GET 请求的示例。基本上是这样的
HttpUrl.Builder urlBuilder = HttpUrl.parse("https://api.github.help").newBuilder();
urlBuilder.addQueryParameter("v", "1.0");
urlBuilder.addQueryParameter("user", "vogella");
String url = urlBuilder.build().toString();
Request request = new Request.Builder()
.url(url)
.build();
然后你需要在 OkHttpClient
的帮助下(异步地)发送你的请求,就像这样
client.newCall(request).enqueue(new Callback() {
@Override
public void onFailure(Call call, IOException e) {
e.printStackTrace();
}
@Override
public void onResponse(Call call, final Response response) throws IOException {
if (!response.isSuccessful()) {
throw new IOException("Unexpected code " + response);
} else {
// do something wih the result
}
}
(代码取自 Vogella)