jquery onchange与php

Question

我是php的新手。请帮我解决一下这个。这是我的HTML代码。

import java.io.*;
import java.security.KeyFactory;
import java.security.PrivateKey;
import java.security.spec.PKCS8EncodedKeySpec;
import android.util.Base64;

public class ReadPKCS8Pem {

    private final static String PRIVATE_KEY = 
            "-----BEGIN PRIVATE KEY-----\n"
            + "MIICdQIBADANBgkqhkiG9w0BAQEFAASCAl8wggJbAgEAAoGBAM7t8Ub1DP+B91NJ\n"
            + "nC45zqIvd1QXkQ5Ac1EJl8mUglWFzUyFbhjSuF4mEjrcecwERfRummASbLoyeMXl\n"
            + "eiPg7jvSaz2szpuV+afoUo9c1T+ORNUzq31NvM7IW6+4KhtttwbMq4wbbPpBfVXA\n"
            + "IAhvnLnCp/VyY/npkkjAid4c7RoVAgMBAAECgYBcCuy6kj+g20+G5YQp756g95oN\n"
            + "dpoYC8T/c9PnXz6GCgkik2tAcWJ+xlJviihG/lObgSL7vtZMEC02YXdtxBxTBNmd\n"
            + "upkruOkL0ElIu4S8CUwD6It8oNnHFGcIhwXUbdpSCr1cx62A0jDcMVgneQ8vv6vB\n"
            + "/YKlj2dD2SBq3aaCYQJBAOvc5NDyfrdMYYTY+jJBaj82JLtQ/6K1vFIwdxM0siRF\n"
            + "UYqSRA7G8A4ga+GobTewgeN6URFwWKvWY8EGb3HTwFkCQQDgmKtjjJlX3BotgnGD\n"
            + "gdxVgvfYG39BL2GnotSwUbjjce/yZBtrbcClfqrrOWWw7lPcX1d0v8o3hJfLF5dT\n"
            + "6NAdAkA8qAQYUCSSUwxJM9u0DOqb8vqjSYNUftQ9dsVIpSai+UitEEx8WGDn4SKd\n"
            + "V8kupy/gJlau22uSVYI148fJSCGRAkBz+GEHFiJX657YwPI8JWHQBcBUJl6fGggi\n"
            + "t0F7ibceOkbbsjU2U4WV7sHyk8Cei3Fh6RkPf7i60gxPIe9RtHVBAkAnPQD+BmND\n"
            + "By8q5f0Kwtxgo2+YkxGDP5bxDV6P1vd2C7U5/XxaN53Kc0G8zu9UlcwhZcQ5BljH\n"
            + "N24cUWZOo+60\n"
            + "-----END PRIVATE KEY-----";

    public static void main(String[] args) throws Exception {
        // Read in the key into a String
        StringBuilder pkcs8Lines = new StringBuilder();
        BufferedReader rdr = new BufferedReader(new StringReader(PRIVATE_KEY));
        String line;
        while ((line = rdr.readLine()) != null) {
            pkcs8Lines.append(line);
        }

        // Remove the "BEGIN" and "END" lines, as well as any whitespace

        String pkcs8Pem = pkcs8Lines.toString();
        pkcs8Pem = pkcs8Pem.replace("-----BEGIN PRIVATE KEY-----", "");
        pkcs8Pem = pkcs8Pem.replace("-----END PRIVATE KEY-----", "");
        pkcs8Pem = pkcs8Pem.replaceAll("\\s+","");

        // Base64 decode the result

        byte [] pkcs8EncodedBytes = Base64.decode(pkcs8Pem, Base64.DEFAULT);

        // extract the private key

        PKCS8EncodedKeySpec keySpec = new PKCS8EncodedKeySpec(pkcs8EncodedBytes);
        KeyFactory kf = KeyFactory.getInstance("RSA");
        PrivateKey privKey = kf.generatePrivate(keySpec);
        System.out.println(privKey);
    }

}

这是我的jquery代码：

<label>Select room type:</label>
<select name="user" id="user ">
  <option>one</option>
  <option>two</option>
  <option>three</option>
</select>
<div id="display"></div>

这是fetch.php：

$(document).ready(function() {
    $("#user").change(function (e) {
      var user = $(this).val();
      $.post("fetch.php",{"user": user},function (data) {
          $('#display').html(data);
      });
    }); 
});

我在运行fetch.php时遇到错误

<?php
   echo $_POST['user'];
?>

请在我出错的地方帮助我

Answer 1

替代方式

jquery代码：

integration.response.body.X-API-VER

<强> fetch.php

$('#user').change(function(){
    var user = $('#user').val();
    $.ajax({url:"fetch.php?user="+user,cache:false,success:function(result){
        $('#display').html(result);
    }});
});