Question

我正在使用Web API集成到我的应用程序中。我同样使用.net服务器API，但问题是当我使用URL传递参数时，如果API的任何参数包含空格，它将引发错误。当我传递字符串值时，由于空间原因会引发错误，如果我删除空间，则一切都非常完美。请帮忙!!!

 private void getEnterLoadDetail() {

    pBar.setVisibility(View.VISIBLE);
    pref = getContext().getSharedPreferences(PREF_NAME, Context.MODE_PRIVATE);

    String userName = pref.getString(KEY_USER_NAME, "username");
    Log.e("TAG", "USERNAME" + userName);
    tripId = txtTripId.getText().toString();
    materialCode = txtMaterialCode.getText().toString();
    vendorCode = txtVendorCode.getText().toString();
    String url = "http://18.216.239.216:8086/api/Mobile?username="+userName+"&tripid=" + tripId + "&VehicleNo=" + vehicleNo + "&MaterialName=" + materialname + "&MaterialCode=" + materialCode + "&VendorName=" + vendorName + "&VendorCode=" + vendorCode + "&MaterialRate=" + rate + "&Remarks="+remarks+"";


    StringRequest stringRequest = new StringRequest(Request.Method.POST, url,
            new Response.Listener<String>() {
                @Override
                public void onResponse(String response) {

                    Log.e("response", response.toString());

                    try {
                        if (response.contains("success")) {
                            Toast.makeText(getActivity(), R.string.load_detail_submitted, Toast.LENGTH_LONG).show();
                            edtVehicleNo.setText("");
                            edtRate.setText("");
                            edtRemarks.setText("");
                            txtMaterialCode.setText("Material Code");
                            txtVendorCode.setText("Vendor Code");
                            spinnerVendorName.setSelection(0);
                            spinnerMaterialName.setSelection(0);
                            txtTripId.setText("Trip Id");
                        }
                    } catch (Exception e) {
                        e.printStackTrace();
                        Activity activity = getActivity();
                        if (activity != null && isAdded()) {
                            Toast.makeText(activity, R.string.some_error_occured, Toast.LENGTH_LONG).show();

                        }

                    }
                    pBar.setVisibility(View.GONE);
                }


            }, new Response.ErrorListener() {

        @Override
        public void onErrorResponse(VolleyError error) {
            VolleyLog.e("Error: " + error.getMessage());
            Activity activity = getActivity();
            if (activity != null && isAdded()) {
                Toast.makeText(activity, R.string.some_error_occured, Toast.LENGTH_LONG).show();
                pBar.setVisibility(View.GONE);
            }

        }
    });

    RequestQueue requestQueue = Volley.newRequestQueue(getActivity());

    //adding the string request to request queue
    requestQueue.add(stringRequest);
}

-------- 材料名称，所有其他变量均包含空格，并且API引发错误。

Answer 1

尝试

String url = "http://18.216.239.216:8086/api/Mobile?username="+userName+"&tripid=" + tripId + "&VehicleNo=" + vehicleNo + "&MaterialName=" + materialname + "&MaterialCode=" + materialCode + "&VendorName=" + vendorName + "&VendorCode=" + vendorCode + "&MaterialRate=" + rate + "&Remarks="+remarks+"";
String convertedUrl = URLEncoder.encode(url, "utf-8");

参数中的空格导致android中的错误

1 个答案: