Android Firebase正在更新错误的子项

Android Firebase正在更新错误的子项,android,firebase,Android,Firebase,我希望更新数据库中的users节点以及presentUsersByPhoneNumber节点 但是,当使用下面概述的代码时: // Create in users table: final Firebase usersRef = myFeastFirebaseNewUser.child("users"); Map<String, String> newUserAttributes = new HashMap&l

我希望更新数据库中的
users
节点以及
presentUsersByPhoneNumber
节点

但是,当使用下面概述的代码时:

            // Create in users table:
            final Firebase usersRef = myFeastFirebaseNewUser.child("users");

            Map<String, String> newUserAttributes = new HashMap<String, String>();
            newUserAttributes.put("username", mUsername.getText().toString());
            newUserAttributes.put("password", mPassword.getText().toString());
            newUserAttributes.put("phoneNumber", mPhoneNum.getText().toString());
            newUserAttributes.put("email", mEmail.getText().toString());

            Map<String, Object> createNewUser = new HashMap<String, Object>();
            createNewUser.put(result.get("uid").toString(), newUserAttributes);
            usersRef.updateChildren(createNewUser);

            // Create in presentUsersByPhoneNumber:
            final Firebase userByPhoneNumber = myFeastFirebasePhoneSearch.child("presentUsersByPhoneNumber");

            Map<String, String> newUserID = new HashMap<String, String>();
            newUserID.put("uid", result.get("uid").toString());

            Map<String, Object> createNewUsesPhoneRef = new HashMap<String, Object>();
            createNewUser.put("1" + mPhoneNum.getText().toString(), newUserID);
            userByPhoneNumber.updateChildren(createNewUsesPhoneRef);
//在用户表中创建:
final Firebase usersRef=myFeastFirebaseNewUser.child(“用户”);
Map newUserAttributes=newhashmap();
newUserAttributes.put(“用户名”,mUsername.getText().toString());
newUserAttributes.put(“密码”,mPassword.getText().toString());
newUserAttributes.put(“phoneNumber”,mPhoneNum.getText().toString());
newUserAttributes.put(“email”,mEmail.getText().toString());
Map createNewUser=newhashmap();
createNewUser.put(result.get(“uid”).toString(),newUserAttributes);
usersRef.updateChildren(createNewUser);
//在presentUsersByPhoneNumber中创建:
final Firebase userByPhoneNumber=myFeastFirebasePhoneSearch.child(“presentUsersByPhoneNumber”);
Map newUserID=newhashmap();
newUserID.put(“uid”),result.get(“uid”).toString();
Map createNewUsesPhoneRef=newHashMap();
createNewUser.put(“1”+mPhoneNum.getText().toString(),newUserID);
userByPhoneNumber.updateChildren(createNewUsesPhoneRef);
结果是:

我想做的是在保存时设置节点
presentUsersByPhoneNumber
,内容为红色


如何实现这一点?

Android代码应该是:

            Map<String, Object> createNewUsesPhoneRef = new HashMap<String, Object>();
            createNewUsesPhoneRef.put("1" + mPhoneNum.getText().toString(), newUserID);
            userByPhoneNumber.updateChildren(createNewUsesPhoneRef);
Map createNewUsesPhoneRef=newHashMap();
createNewUsesPhoneRef.put(“1”+mPhoneNum.getText().toString(),newUserID);
userByPhoneNumber.updateChildren(createNewUsesPhoneRef);

不是对
createNewUser

的引用问题中包含了JSON树的图片。请将其替换为实际的JSON作为文本,您可以通过单击Firebase数据库中的导出按钮轻松获得该文本。将JSON作为文本使其具有可搜索性,允许我们轻松地使用它来测试您的实际数据,并在我们的答案中使用它,这通常是一件好事。谢谢,但问题在于Android代码。在更新电话号码引用时,我使用了以下行上的对象:
createNewUser.put(“1”+mPhoneNum.getText().toString(),newUserID)这是不正确的,不能相信我错过了它…感谢更新。但下一次,请包含实际的JSON,而不是JSON的屏幕截图。StackOverflow推特。让我们先关闭这个或自我回答,因为它不太可能对任何人有任何用处。除此之外:你不需要在无关的问题上给我贴标签。我要么去你的,要么不去。