我正在尝试在用户填写表单时创建应用程序,然后单击“发送”按钮,通过URL参数发送信息,如:
localhost / my.php?param1 = Name&param2 = SurName
我已经通过互联网阅读了它,但是无法得到它,我找到了一些例子但他们告诉我如何通过Json,我只想通过链接发送。 (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);
}
任何帮助将不胜感激。
我建议看看OkHttp。它是一个简单的库,可以以您喜欢的任何方式发送HTTP请求。在Vogella上,您还可以找到一个使用其他参数发送GET请求的示例。基本上它看起来像这样
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();
之后你需要在OkHttpClient
的帮助下发送你的请求(异步)
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
}
}
(代码来自Vogella的任务)