Question

我希望能够刷新HTML表格中的值（从SQL数据库中检索），而无需重新加载页面。我希望在没有需要按下的按钮的情况下完成此操作。

enterString baseURl = "http://example.com";
final String privateKey = "private key";**(key is supplied by api provider)**
final int devloperid = userid;  **(devloper id is provider by api provider)**
String uri = "/v2/healthCheck";

public String genrateSignature(final String baseURl, final String privateKey, final String uri, final int devloperid) throws Exception {

    String HMAC_SHA1_ALGORITHIAM = "HmacSHA1";  

    StringBuilder uriWithDevloperId = new StringBuilder().append(uri).append(uri.contains("?") ? "&" : "?").append(devloperid);


    byte[] keyBytes = privateKey.getBytes();
    byte[] uriByte = uriWithDevloperId.toString().getBytes();
    Key signingKey = new SecretKeySpec(keyBytes, HMAC_SHA1_ALGORITHIAM);
    Mac mac = Mac.getInstance(HMAC_SHA1_ALGORITHIAM);
    mac.init(signingKey);
    byte[] signatureBytes = mac.doFinal(uriByte);
    StringBuilder singnature = new StringBuilder(signatureBytes.length * 2);
    for (byte signatureByte : signatureBytes) {

        int intval = signatureByte & 0xff;
        if (intval < 0x10) {
            singnature.append("0");
        }
        singnature.append(Integer.toHexString(intval));
    }
    StringBuilder url = new StringBuilder(baseURl).append(uri).append(uri.contains("?") ? "&" : "?").append("devid=" +devloperid).append("&signature="+singnature.toString().toUpperCase());

    Log.i("hello", String.valueOf(url));
    return url.toString();


}

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
   TrasportTimeTable trasportTimeTable = new TrasportTimeTable();
    try {
        trasportTimeTable.execute(genrateSignature(baseURl,privateKey,uri,devloperid));
    } catch (Exception e) {
        e.printStackTrace();
    }

}

public class TrasportTimeTable extends AsyncTask<String, Void, String> {


    @Override
    protected String doInBackground(String... urls) {
        String result = "";

        URL url;
        HttpURLConnection httpURLConnection;

        try {
            url = new URL(urls[0]);
            httpURLConnection = (HttpURLConnection) url.openConnection();
            InputStream inputStream = httpURLConnection.getInputStream();

            InputStreamReader inputStreamReader = new InputStreamReader(inputStream);

            int data = inputStreamReader.read();

            while (data != -1) {
                char currentCharacter = (char) data;
                result += currentCharacter;

                data = inputStreamReader.read();

            }
            Log.i("result", result);
        } catch (IOException e) {
            e.printStackTrace();
        }

        return null;
    }

    @Override
    protected void onPostExecute(String result) {
        super.onPostExecute(result);


        Toast.makeText(MainActivity.this, "toast recieved", Toast.LENGTH_SHORT).show();
    }
}
}

这是我目前所拥有的，但它没有显示任何内容。

Answer 1

从我所看到的，你没有成功＆＃34;

function refreshPartial() {
$.ajax({
url: "data.php", //where you process the data
dataType: 'script',
format: 'js',
success: function(data)
{
$("#idOfElementWhereYouWantToPutTheData").html(data);
}

})
}

以及＆＃34; url＆＃34;是处理数据的地方，而不是获取数据的地方

自动mySQLi HTML表值刷新

1 个答案: