无法从以前的数据中更新数据

时间:2018-09-04 07:13:43

标签: java android

我有3个活动,分别是AddAds,ViewAds和EditAds。用户添加新添加的内容后,它将在ViewAds中显示/检索,当用户单击UpdateButton时,它将转到EditAds。在这里,我无法弄清楚为什么不能从以前的数据更新数据?有人可以帮助我,我在哪里错过我的编码?预先感谢...

UpdateButton编码:-

UpdateAdvertisementButton.setOnClickListener(new View.OnClickListener() {
    @Override
    public void onClick(View v) {
        SendUserToViewAdsActivity();

        String id = databaseAdvertisement.push().getKey();
        String tuitionimage = image;
        String providername = EditProviderName.getText().toString();
        String providergender = EditProviderGender.getSelectedItem().toString();;

        if (TextUtils.isEmpty(tuitionimage)) {
            Toast.makeText(EditAdsActivity.this, "Insert tuition image.", Toast.LENGTH_SHORT).show();
        } else if (TextUtils.isEmpty(providername)) {
            Toast.makeText(EditAdsActivity.this, "Enter tuition provider name.", Toast.LENGTH_SHORT).show();
        } else if(EditProviderGender.getSelectedItemPosition() == 0) {
           Toast.makeText(EditAdsActivity.this, "Select the options listed, either male or female.", Toast.LENGTH_SHORT).show();
        } else {
            loadingBar.setTitle("Updating Tuition Centre");
            loadingBar.setMessage("Please wait while we are updating your ads in our system.");
            loadingBar.show();
            loadingBar.setCanceledOnTouchOutside(true);

            uploadImageAndGetURL(databaseAdvertisement.push().getKey());

            updateAds(id, tuitionimage, providername, providergender);
        }
    }
});

updateAds()方法编码:-

private boolean updateAds(String id, String tuitionimage, String providername, String providergender) {
    DatabaseReference databaseReference = FirebaseDatabase.getInstance().getReference("Advertisement").child(id);

    Advertisement advertisement = new Advertisement(id, tuitionimage, providername, providergender);

    databaseReference.setValue(advertisement);

    Toast.makeText(EditAdsActivity.this, "Advertisement has been updated successfully!", Toast.LENGTH_SHORT).show();

    return true;
}

0 个答案:

没有答案