如何查看給定對象是否與firebase返回的對象之一匹配

Question

我試圖測試給定的object匹配存儲並從firebase返回的stored objects 。 如果是，則返回true或在變量中存儲true。

以下是給定對象和firebase返回數據的示例

給定對象：

{name: "John", age: "32"}

存儲對象：

{ 
    '-JrYqLGTNLfa1zEkTG5J': { name: "John", age: "32" },
    '-JrkhWMKHhrShX66dSWJ': { name: "Steve", age: "25" },
    '-JrkhtHQPKRpNh0B0LqJ': { name: "Kelly", age: "33" },
    '-JrkiItisvMJZP1fKsS8': { name: "Mike", age: "28" },
    '-JrkiPqA8KyAMj2R7A2h': { name: "Liz", age: "22" } 
}

Answer 1

var storedObject ={ 
    '-JrYqLGTNLfa1zEkTG5J': { name: "John", age: "32" },
    '-JrkhWMKHhrShX66dSWJ': { name: "Steve", age: "25" },
    '-JrkhtHQPKRpNh0B0LqJ': { name: "Kelly", age: "33" },
    '-JrkiItisvMJZP1fKsS8': { name: "Mike", age: "28" },
    '-JrkiPqA8KyAMj2R7A2h': { name: "Liz", age: "22" } 
};
var givenObject = {name: "John", age: "32"};

這是你如何檢查對象是否相等的方法。 你只需要循環對象

Javascript方式

for(key in storedObject){
    if(JSON.stringify(storedObject[key]) === JSON.stringify(givenObject)){
     alert('equal'+givenObject['name'] +givenObject['age']);
    }
};

jQuery方式

使用$.each()函數

$.each(storedObject,function(key,value){
    if(JSON.stringify(value) === JSON.stringify(givenObject)){
     alert('equal'+givenObject['name'] +givenObject['age']);
    }
});

查看FIDDLE LINK另外，有關您的信息，請查看JavaScript LINK中的此SO 對象比較

Answer 2

為了完整起見，這是iOS / MacOS中的解決方案

//query for all users
FQuery *allUsersRef = [usersRef queryOrderedByChild:@"name"];

//filter that query with a query for users named "John"
FQuery *justJohnRef = [allUsersRef queryEqualToValue:@"John"]; 

//read in all of the resulting users named John
[justJohnRef observeSingleEventOfType:FEventTypeValue withBlock:^(FDataSnapshot *snapshot) {

    NSArray *johnArray = [snapshot.value allObjects];

    NSPredicate *predicate = [NSPredicate predicateWithFormat:@"age == %@", @"32"];
    NSArray *result = [johnArray filteredArrayUsingPredicate:predicate];

    NSLog(@"%@", result);

}];

這里的概念是我們使用廣泛的查詢來引入我們需要的一系列數據，在這種情況下，所有名為John的用戶，然后在代碼中為所有返回的32歲的John用戶過濾該數據。

請注意，如果有多個John年齡為32歲，則結果數組將包含所有這些數據。 應該向NSPredicate提供更詳細的信息，以獲得您正在尋找的確切John（即SSN = x或駕駛執照= y）

Answer 3

以下是使用可重用代碼（需要符合ES5環境）的通用示例，因此它不僅限於您提供的數據。

 // some generic reuseable code (function () { 'use strict'; function $isPrimitive(inputArg) { var type = typeof inputArg; return type === 'undefined' || inputArg === null || type === 'boolean' || type === 'string' || type === 'number' || type === 'symbol'; } function $isFunction(inputArg) { return typeof inputArg === 'function'; } function $isDate(inputArg) { return Object.prototype.toString.call(inputArg) === '[object Date]'; } function $isRegExp(inputArg) { return Object.prototype.toString.call(inputArg) === '[object RegExp]'; } function $isString(inputArg) { return Object.prototype.toString.call(inputArg) === '[object String]'; } function $isArguments(inputArg) { return Object.prototype.toString.call(inputArg) === '[object Arguments]'; } function $getItem(object, index) { var item; if ($isString(object)) { item = object.charAt(index); } else { item = object[index]; } return item; } var de = function (a, b, circ) { if (a === b) { return true; } var aType, bType, aIsArgs, bIsArgs, aIsPrim, bIsPrim, aCirc, bCirc, ka, kb, length, index, it; if ($isDate(a) && $isDate(b)) { return a.getTime() === b.getTime(); } if ($isRegExp(a) && $isRegExp(b)) { return a.source === b.source && a.global === b.global && a.multiline === b.multiline && a.lastIndex === b.lastIndex && a.ignoreCase === b.ignoreCase && a.sticky === b.sticky; } aIsPrim = $isPrimitive(a); bIsPrim = $isPrimitive(b); if ((aIsPrim || $isFunction(a)) && (bIsPrim || $isFunction(b))) { /*jslint eqeq: true */ return a == b; } if (aIsPrim || bIsPrim) { return a === b; } if (a.prototype !== b.prototype) { return false; } if (circ.a.indexOf(a) === -1) { circ.a.push(a); } else { aCirc = true; } if (circ.b.indexOf(b) === -1) { circ.b.push(b); } else { bCirc = true; } if (aCirc && bCirc) { circ.cnt += 1; } else { circ.cnt = 0; } if (circ.cnt > 200) { throw new RangeError('Circular reference limit exceeded'); } aIsArgs = $isArguments(a); bIsArgs = $isArguments(b); if ((aIsArgs && !bIsArgs) || (!aIsArgs && bIsArgs)) { return false; } if (aIsArgs) { return de(Array.prototype.slice.call(a), Array.prototype.slice.call(b), circ); } ka = Object.keys(a); kb = Object.keys(b); length = ka.length; if (length !== kb.length) { if (Array.isArray(a) && Array.isArray(b)) { if (a.length !== b.length) { return false; } } else { return false; } } else { ka.sort(); kb.sort(); for (index = 0; index < length; index += 1) { if (ka[index] !== kb[index]) { return false; } } } for (index = 0; index < length; index += 1) { it = ka[index]; if (!de($getItem(a, it), $getItem(b, it), circ)) { return false; } } aType = typeof a; bType = typeof b; return aType === bType; }; if (!Object.prototype.deepEqual) { Object.defineProperty(Object.prototype, 'deepEqual', { enumerable: false, configurable: true, writable: true, value: function (b) { var a = this; return de(a, b, { a: [], b: [], cnt: 0 }); } }); } if (!Object.prototype.forKeys) { Object.defineProperty(Object.prototype, 'forKeys', { enumerable: false, configurable: true, writable: true, value: function (fn, thisArg) { var object = Object(this), keys, length, val, index, it; if (!$isFunction(fn)) { throw new TypeError('Argument is not a function: ' + fn); } keys = Object.keys(object); length = keys.length; val = false; for (index = 0; index < length; index += 1) { it = keys[index]; val = !!fn.call(thisArg, $getItem(object, it), it, object); if (val) { break; } } return val; } }); } }()); // example of use with your data var stored = { '-JrYqLGTNLfa1zEkTG5J': { name: "John", age: "32" }, '-JrkhWMKHhrShX66dSWJ': { name: "Steve", age: "25" }, '-JrkhtHQPKRpNh0B0LqJ': { name: "Kelly", age: "33" }, '-JrkiItisvMJZP1fKsS8': { name: "Mike", age: "28" }, '-JrkiPqA8KyAMj2R7A2h': { name: "Liz", age: "22" } }, given = { name: "John", age: "32" }, found = stored.forKeys(function (item) { return item.deepEqual(this); }, given); document.getElementById('out').textContent = 'given was matched in stored: ' + found; 

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