Javascript Firebase.update 失败：第一个参数在属性中包含未定义
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Firebase.update failed : first argument contains undefined in property
提问by jonathanhuo11
I have a simple Firebase function that updates some data. However, the interpreter says that the first argument contains "undefined" in property 'users.tester1'. Can somebody help me please?
我有一个简单的 Firebase 函数来更新一些数据。但是，解释器说第一个参数在属性“users.tester1”中包含“undefined”。有人可以帮我吗？
var objify = function() {
var rv = {};
for (var i = 0; i < arguments.length; ++i)
rv[arguments[i]] = rv[arguments[i+1]];
return rv;
}
addUser("tester1", []);
var addUser = function(name, edges){
if(!checkIfUsernameExists(name) && !checkIfNodeNameExists(name) && !checkIfEdgeNameExists(name)){
var time = Firebase.ServerValue.TIMESTAMP;
//HERE: I think the error is on this line
refs.users.update(objify(name, "filler"));
refs.users.child(name).set({
"id" : time,
"edges" : "filler"
});
refs.users.child(name).child("edges").update({
"to" : "filler",
"from" : "filler"
});
addNode(new Node(name, time, name));
for(var e in edges){
refs.users.child(name).child("edges/to").update(objify(edges[e].name, true));
addEdge(new Edge(edges[e].name, time, edges[e].to, edges[e].arrows));
//TODO add a "from" edge so that you know who wants to eat you
}
refs.users.child(name).child("edges/to").set({"filler" : null});
} else {
alert("user/node/edge name taken.");
}
};
回答by Frank van Puffelen
When you pass an object to Firebase, the values of the properties can be a value or null (in which case the property will be removed). They can not be undefined, which is what you're passing in according to the error.
当您将对象传递给 Firebase 时，属性的值可以是一个值或 null（在这种情况下，该属性将被删除）。它们不能是undefined，这是您根据错误传递的内容。
Simply running this snippet in isolation shows the problem:
简单地单独运行此代码段会显示问题：
var objify = function() {
var rv = {};
for (var i = 0; i < arguments.length; ++i)
rv[arguments[i]] = rv[arguments[i+1]];
return rv;
}
objify("name", "filler")
Results in:
结果是：
{name: undefined, filler: undefined}
{名称：未定义，填充物：未定义}
My best bet is that you want to pass key/value pairs into objifyas even/odd parameters. In that case you want to change the function to:
我最好的选择是您希望将键/值对objify作为偶数/奇数参数传递。在这种情况下，您要将函数更改为：
var objify = function() {
var rv = {};
for (var i = 0; i < arguments.length; i+=2)
rv[arguments[i]] = arguments[i+1];
return rv;
}
objify("name", "filler")
Results in:
结果是：
{name: "filler"}
{名称：“填充物”}
回答by Jeremy Souffir
to make sure your object does not contain any undefined props use this simple trick:
要确保您的对象不包含任何未定义的道具，请使用以下简单技巧：
JSON.parse( JSON.stringify(YourJsonData ) )
For more info take a look at this codePen: http://codepen.io/ajmueller/pen/gLaBLX
有关更多信息，请查看此 codePen：http://codepen.io/ajmueller/pen/gLaBLX
回答by garrettmac
Like said above, you need all undefinedvalues to be nullif you want them to save as empty values in firebase.
如上所述，如果您希望它们在 firebase 中保存为空值，则需要所有undefined值null。
I take this approach to correct all nested values.
我采用这种方法来纠正所有嵌套值。
//to search and replace
const replaceAll =(s="",f="",r="")=> s.replace(new RegExp(f.replace(/[-\/\^$*+?.()|[\]{}]/g, '\$&'), 'g'), r)
//to save
firebase.database().ref(`path`).update(JSON.parse(replaceAll(JSON.stringify(val),"undefined","null")))
