HtmlUnit 503服务不可用

Question

我想用HtmlUnit抓取Google搜索结果。我的代码工作了一段时间，但现在我得到了这个例外：

  

com.gargoylesoftware.htmlunit.ScriptException：503 https://ipv4.google.com/sorry/IndexRedirect?continue=的服务不可用......

它还会打印出我认为无效的JavaScript：

function () {
      if (1 != c.readyState) {
          var e = !1;
          try {
              e = 0 == c.status && 4 == c.readyState;
          }
          catch (k) {
              e = !0;
          }
          var f, g = d;
          e ? f = 21 : QS_kea(c.readyState, c.status) && 0 > (c.getResponseHeader("Content-Type") || "").indexOf("application/json") ? (f = 12, g = {response: c.responseText, url: d}) : QS_lea(c.status, 400, 500) ? f = 25 : QS_lea(c.status, 500, 600) && (f = 1);
          if (void 0 !== f) {
              QS_gea(a, f, null, g), QS_mea(a, b);
          } else {
              if (3 == c.readyState && a.ya && !a.ka) {
                  b.RJ = QS_hea(a, c.responseText, b.RJ, d);
              } else {
                  if (4 == c.readyState && !b.complete) {
                      b.complete = !0;
                      var h = QS_d(function (b) {
                          a.ka || (b.RJ = QS_hea(a, b.ov.responseText, b.RJ, b.url, !0));
                          if (a.ka) {
                              var c = QS_d(a.Aa, a, h, 0);
                              a.ra.push(window.requestAnimationFrame(c));
                          } else {
                              QS_mea(a, b);
                          }
                      }, a, b);
                      200 == c.status ? h() : QS_mea(a, b);
                  }
              }
          }
      }
  }

如果我压制脚本错误和状态代码错误，我无法得到任何结果，我猜是因为页面加载不正确？

我也想到，也许Google会检测到刮擦并阻止我的代码。那会很奇怪，因为它会立即检测到第一次尝试。