尽力解释我想要做的事情。
我有两个型号，我和我收到的api响应。当项目api响应进来时，我需要将其映射到我的模型并插入所有项目。这当然很简单。这是问题所在，我需要在不知道自己在做什么的情况下这样做。我的代码将以两个字符串传递，一个是我的模型映射路径，另一个是api响应映射路径。
以下是两条路径
var myPath = "outputModel.items[].uniqueName"
var apiPath = "items[].name"
基本上对于items中的所有apiPath，请items中的myPath并设置为uniqueName
归结为我的代码不知道何时需要映射两个项目，或者即使它们包含数组或简单字段到字段路径。它们甚至可以包含多个数组，如下所示：
********************示例************************* < / p>
var items = [
{
name: "Hammer",
skus:[
{num:"12345qwert"}
]
},
{
name: "Bike",
skus:[
{num:"asdfghhj"},
{num:"zxcvbn"}
]
},
{
name: "Fork",
skus:[
{num:"0987dfgh"}
]
}
]
var outputModel = {
storeName: "",
items: [
{
name: "",
sku:""
}
]
};
outputModel.items[].name = items[].name;
outputModel.items[].sku = items[].skus[].num;
************************以上是
的预期结果var result = {
storeName: "",
items: [
{
name: "Hammer",
sku:"12345qwert"
},
{
name: "Bike",
sku:"asdfghhj"
},
{
name: "Bike",
sku:"zxcvbn"
},
{
name: "Fork",
sku:"0987dfgh" }
]
};
我将获得一组用于映射EACH值的路径。在上面的例子中，我被传递了两组路径，因为我正在映射两个值。它必须遍历两组数组才能在我的模型中创建单个数组。
问题 - 无论两个模型路径是什么样子，我如何动态检测数组并正确移动数据？可能的？
答案 0 :(得分：2)
因此，您已定义了一种语言来定义一些数据寻址和操作规则。让我们考虑一种方法，让您说出
access(apiPath, function(value) { insert(myPath, value); }
access函数会在apiPath中找到所有必需的项目，然后回调insert，将其插入myPath。我们的工作是编写创建access和insert函数的函数;或者，你可以说，将你的小语言“编译”成我们可以执行的函数。
我们将编写名为make_accessor和make_inserter的“编译器”，如下所示：
function make_accessor(program) {
return function(obj, callback) {
return function do_segment(obj, segments) {
var start = segments.shift() // Get first segment
var pieces = start.match(/(\w+)(\[\])?/); // Get name and [] pieces
var property = pieces[1];
var isArray = pieces[2]; // [] on end
obj = obj[property]; // drill down
if (!segments.length) { // last segment; callback
if (isArray) {
return obj.forEach(callback);
} else {
return callback(obj);
}
} else { // more segments; recurse
if (isArray) { // array--loop over elts
obj.forEach(function(elt) { do_segment(elt, segments.slice()); });
} else {
do_segment(obj, segments.slice()); // scalar--continue
}
}
}(obj, program.split('.'));
};
}
我们现在可以通过调用make_accessor('items[].name')来创建一个访问者。
接下来，让我们编写插件：
function make_inserter(program) {
return function(obj, value) {
return function do_segment(obj, segments) {
var start = segments.shift() // Get first segment
var pieces = start.match(/(\w+)(\[\])?/); // Get name and [] pieces
var property = pieces[1];
var isArray = pieces[2]; // [] on end
if (segments.length) { // more segments
if (!obj[property]) {
obj[property] = isArray ? [] : {};
}
do_segment(obj, segments.slice());
} else { // last segment
obj[property] = value;
}
}(obj, program.split('.'));
};
}
现在，您可以将整个逻辑表达为
access = make_accessor('items[].name');
insert = make_inserter('outputModel.items[].uniqueName');
access(apiPath, function(val) { insert(myPath, val); });
答案 1 :(得分：0)
我借用了之前的答案并进行了改进，以便解决您的示例，这应该是通用的。虽然如果你打算用两组输入顺序运行，那么行为就像我在你对原始问题的评论中所概述的那样。
var apiObj = {
items: [{
name: "Hammer",
skus: [{
num: "12345qwert"
}]
}, {
name: "Bike",
skus: [{
num: "asdfghhj"
}, {
num: "zxcvbn"
}]
}, {
name: "Fork",
skus: [{
num: "0987dfgh"
}]
}]
};
var myObj = { //Previously has values
storeName: "",
items: [{
uniqueName: ""
}],
outputModel: {
items: [{
name: "Hammer"
}]
}
};
/** Also works with this **
var myPath = "outputModel.items[].uniqueName";
var apiPath = "items[].name";
*/
var myPath = "outputModel.items[].sku";
var apiPath = "items[].skus[].num";
function make_accessor(program) {
return function (obj, callback) {
(function do_segment(obj, segments) {
var start = segments.shift() // Get first segment
var pieces = start.match(/(\w+)(\[\])?/); // Get name and [] pieces
var property = pieces[1];
var isArray = pieces[2]; // [] on end
obj = obj[property]; // drill down
if (!segments.length) { // last segment; callback
if (isArray) {
return obj.forEach(callback);
} else {
return callback(obj);
}
} else { // more segments; recurse
if (isArray) { // array--loop over elts
obj.forEach(function (elt) {
do_segment(elt, segments.slice());
});
} else {
do_segment(obj, segments.slice()); // scalar--continue
}
}
})(obj, program.split('.'));
};
}
function make_inserter(program) {
return function (obj, value) {
(function do_segment(obj, segments) {
var start = segments.shift() // Get first segment
var pieces = start.match(/(\w+)(\[\])?/); // Get name and [] pieces
var property = pieces[1];
var isArray = pieces[2]; // [] on end
if (segments.length) { // more segments
if (!obj[property]) {
obj[property] = isArray ? [] : {};
}
do_segment(obj[property], segments.slice());
} else { // last segment
if (Array.isArray(obj)) {
var addedInFor = false;
for (var i = 0; i < obj.length; i++) {
if (!(property in obj[i])) {
obj[i][property] = value;
addedInFor = true;
break;
}
}
if (!addedInFor) {
var entry = {};
entry[property] = value;
obj.push(entry);
}
} else obj[property] = value;
}
})(obj, program.split('.'));
};
}
access = make_accessor(apiPath);
insert = make_inserter(myPath);
access(apiObj, function (val) {
insert(myObj, val);
});
console.log(myObj);
答案 2 :(得分：0)
（旧解决方案： https://jsfiddle.net/d7by0ywy/）：
当您知道要提前处理的两个对象（此处称为inp和out）时，这是我的新通用解决方案。如果您事先不了解它们，则可以使用旧解决方案中的技巧将=两侧的对象分配给inp和out（{{3} }）。
限制：两边必须有相同数量的数组，数组必须包含对象。如果你想让它变得更复杂，你必须提出一个更好的语法来表达规则（或者为什么你没有功能而不是规则？）。
歧义示例： out.items[].sku=inp[].skus[].num您是否将num的值数组分配给sku，还是使用{分配了一个对象数组？ {1}}属性？
数据：
num程序：
rules = [
'out.items[].name=inp[].name',
'out.items[].sku[].num=inp[].skus[].num'
];
inp = [{
'name': 'Hammer',
'skus':[{'num':'12345qwert','test':'ignore'}]
},{
'name': 'Bike',
'skus':[{'num':'asdfghhj'},{'num':'zxcvbn'}]
},{
'name': 'Fork',
'skus':[{'num':'0987dfgh'}]
}];
答案 3 :(得分：0)
正如评论中所提到的，输入格式没有严格的定义，很难用完美的错误处理和处理所有极端情况。
这是我的冗长实现，适用于您的示例，但在其他一些情况下可能会失败：
function merge_objects(a, b) {
var c = {}, attr;
for (attr in a) { c[attr] = a[attr]; }
for (attr in b) { c[attr] = b[attr]; }
return c;
}
var id = {
inner: null,
name: "id",
repr: "id",
type: "map",
exec: function (input) { return input; }
};
// set output field
function f(outp, mapper) {
mapper = typeof mapper !== "undefined" ? mapper : id;
var repr = "f("+outp+","+mapper.repr+")";
var name = "f("+outp;
return {
inner: mapper,
name: name,
repr: repr,
type: "map",
clone: function(mapper) { return f(outp, mapper); },
exec:
function (input) {
var out = {};
out[outp] = mapper.exec(input);
return out;
}
};
}
// set input field
function p(inp, mapper) {
var repr = "p("+inp+","+mapper.repr+")";
var name = "p("+inp;
return {
inner: mapper,
name: name,
repr: repr,
type: mapper.type,
clone: function(mapper) { return p(inp, mapper); },
exec: function (input) {
return mapper.exec(input[inp]);
}
};
}
// process array
function arr(mapper) {
var repr = "arr("+mapper.repr+")";
return {
inner: mapper,
name: "arr",
repr: repr,
type: mapper.type,
clone: function(mapper) { return arr(mapper); },
exec: function (input) {
var out = [];
for (var i=0; i<input.length; i++) {
out.push(mapper.exec(input[i]));
}
return out;
}
};
}
function combine(m1, m2) {
var type = (m1.type == "flatmap" || m2.type == "flatmap") ? "flatmap" : "map";
var repr = "combine("+m1.repr+","+m2.repr+")";
return {
inner: null,
repr: repr,
type: type,
name: "combine",
exec:
function (input) {
var out1 = m1.exec(input);
var out2 = m2.exec(input);
var out, i, j;
if (m1.type == "flatmap" && m2.type == "flatmap") {
out = [];
for (i=0; i<out1.length; i++) {
for (j=0; j<out2.length; j++) {
out.push(merge_objects(out1[i], out2[j]));
}
}
return out;
}
if (m1.type == "flatmap" && m2.type != "flatmap") {
out = [];
for (i=0; i<out1.length; i++) {
out.push(merge_objects(out1[i], out2));
}
return out;
}
if (m1.type != "flatmap" && m2.type == "flatmap") {
out = [];
for (i=0; i<out2.length; i++) {
out.push(merge_objects(out2[i], out1));
}
return out;
}
return merge_objects(out1, out2);
}
};
}
function flatmap(mapper) {
var repr = "flatmap("+mapper.repr+")";
return {
inner: mapper,
repr: repr,
type: "flatmap",
name: "flatmap",
clone: function(mapper) { return flatmap(mapper); },
exec:
function (input) {
var out = [];
for (var i=0; i<input.length; i++) {
out.push(mapper.exec(input[i]));
}
return out;
}
};
}
function split(s, t) {
var i = s.indexOf(t);
if (i == -1) return null;
else {
return [s.slice(0, i), s.slice(i+2, s.length)];
}
}
function compile_one(inr, outr) {
inr = (inr.charAt(0) == ".") ? inr.slice(1, inr.length) : inr;
outr = (outr.charAt(0) == ".") ? outr.slice(1, outr.length) : outr;
var box = split(inr, "[]");
var box2 = split(outr, "[]");
var m, ps, fs, i, j;
if (box == null && box2 == null) { // no array!
m = id;
ps = inr.split(".");
fs = outr.split(".");
for (i=0; i<fs.length; i++) { m = f(fs[i], m); }
for (j=0; j<ps.length; j++) { m = p(ps[j], m); }
return m;
}
if (box != null && box2 != null) { // array on both sides
m = arr(compile_one(box[1], box2[1]));
ps = box[0].split(".");
fs = box[0].split(".");
for (i=0; i<fs.length; i++) { m = f(fs[i], m); }
for (j=0; j<ps.length; j++) { m = p(ps[j], m); }
return m;
}
if (box != null && box2 == null) { // flatmap
m = flatmap(compile_one(box[1], outr));
ps = box[0].split(".");
for (j=0; j<ps.length; j++) { m = p(ps[j], m); }
return m;
}
return null;
}
function merge_rules(m1, m2) {
if (m1 == null) return m2;
if (m2 == null) return m1;
if (m1.name == m2.name && m1.inner != null) {
return m1.clone(merge_rules(m1.inner, m2.inner));
} else {
return combine(m1, m2);
}
}
var input = {
store: "myStore",
items: [
{name: "Hammer", skus:[{num:"12345qwert"}]},
{name: "Bike", skus:[{num:"asdfghhj"}, {num:"zxcvbn"}]},
{name: "Fork", skus:[{num:"0987dfgh"}]}
]
};
var m1 = compile_one("items[].name", "items[].name");
var m2 = compile_one("items[].skus[].num", "items[].sku");
var m3 = compile_one("store", "storeName");
var m4 = merge_rules(m3,merge_rules(m1, m2));
var out = m4.exec(input);
alert(JSON.stringify(out));
答案 4 :(得分：0)
嗯，一个有趣的问题。以编程方式constructing nested objects from a property accessor string（或the reverse）不是问题，即使是multiple descriptors in parallel也是如此。它变得复杂的是阵列，需要迭代;当它在setter和getter方面以及并行的多个描述符字符串达到不同的级别时，它就不再那么有趣了。
首先，我们需要区分脚本中每个访问者描述的数组级别，并解析文本：
function parse(script) {
return script.split(/\s*[;\r\n]+\s*/g).map(function(line) {
var assignment = line.split(/\s*=\s*/);
return assignment.length == 2 ? assignment : null; // console.warn ???
}).filter(Boolean).map(function(as) {
as = as.map(function(accessor) {
var parts = accessor.split("[]").map(function(part) {
return part.split(".");
});
for (var i=1; i<parts.length; i++) {
// assert(parts[i][0] == "")
var prev = parts[i-1][parts[i-1].length-1];
parts[i][0] = prev.replace(/s$/, ""); // singular :-)
}
return parts;
});
if (as[0].length == 1 && as[1].length > 1) // getter contains array but setter does not
as[0].unshift(["output"]); // implicitly return array (but better throw an error)
return {setter:as[0], getter:as[1]};
});
}
这样，文本输入可以变成可用的数据结构，现在看起来像这样：
[{"setter":[["outputModel","items"],["item","name"]],
"getter":[["items"],["item","name"]]},
{"setter":[["outputModel","items"],["item","sku"]],
"getter":[["items"],["item","skus"],["sku","num"]]}]
getter已经很好地转换为嵌套循环，如
for (item of items)
for (sku of item.skus)
… sku.num …;
这正是我们要去的地方。这些规则中的每一个都相对容易处理，复制对象的属性和迭代数组的数组，但这是我们最关键的问题：我们有多个规则。我们处理迭代多个数组时的基本解决方案是创建他们的cartesian product，这确实是我们需要的。但是，我们希望对此进行大量限制 - 而不是在输入中创建所有name和所有num的所有组合，我们希望按照它们来的item对它们进行分组从
为此，我们将为输出结构构建某种前缀树，其中包含对象的生成器，每个recursivley都是相应输出子结构的树。
function multiGroupBy(arr, by) {
return arr.reduce(function(res, x) {
var p = by(x);
(res[p] || (res[p] = [])).push(x);
return res;
}, {});
}
function group(rules) {
var paths = multiGroupBy(rules, function(rule) {
return rule.setter[0].slice(1).join(".");
});
var res = [];
for (var path in paths) {
var pathrules = paths[path],
array = [];
for (var i=0; i<pathrules.length; i++) {
var rule = pathrules[i];
var comb = 1 + rule.getter.length - rule.setter.length;
if (rule.setter.length > 1) // its an array
array.push({
generator: rule.getter.slice(0, comb),
next: {
setter: rule.setter.slice(1),
getter: rule.getter.slice(comb)
}
})
else if (rule.getter.length == 1 && i==0)
res.push({
set: rule.setter[0],
get: rule.getter[0]
});
else
console.error("invalid:", rule);
}
if (array.length)
res.push({
set: pathrules[0].setter[0],
cross: product(array)
});
}
return res;
}
function product(pathsetters) {
var groups = multiGroupBy(pathsetters, function(pathsetter) {
return pathsetter.generator[0].slice(1).join(".");
});
var res = [];
for (var genstart in groups) {
var creators = groups[genstart],
nexts = [],
nests = [];
for (var i=0; i<creators.length; i++) {
if (creators[i].generator.length == 1)
nexts.push(creators[i].next);
else
nests.push({path:creators[i].path, generator: creators[i].generator.slice(1), next:creators[i].next});
}
res.push({
get: creators[0].generator[0],
cross: group(nexts).concat(product(nests))
});
}
return res;
}
现在，我们的规则集group(parse(script))如下所示：
[{
"set": ["outputModel","items"],
"cross": [{
"get": ["items"],
"cross": [{
"set": ["item","name"],
"get": ["item","name"]
}, {
"get": ["item","skus"],
"cross": [{
"set": ["item","sku"],
"get": ["sku","num"]
}]
}]
}]
}]
这是一个我们可以实际使用的结构，因为它现在清楚地传达了如何将所有嵌套数组及其中的对象匹配在一起的意图。 让我们动态地解释这一点，为给定的输入构建输出：
function transform(structure, input, output) {
for (var i=0; i<structure.length; i++) {
output = assign(output, structure[i].set.slice(1), getValue(structure[i], input));
}
return output;
}
function retrieve(val, props) {
return props.reduce(function(o, p) { return o[p]; }, val);
}
function assign(obj, props, val) {
if (!obj)
if (!props.length) return val;
else obj = {};
for (var j=0, o=obj; j<props.length-1 && o!=null && o[props[j]]; o=o[props[j++]]);
obj[props[j]] = props.slice(j+1).reduceRight(function(val, p) {
var o = {};
o[p] = val;
return o;
}, val);
return obj;
}
function getValue(descriptor, input) {
if (descriptor.get) // && !cross
return retrieve(input, descriptor.get.slice(1));
var arr = [];
descriptor.cross.reduce(function horror(next, d) {
if (descriptor.set)
return function (inp, cb) {
next(inp, function(res){
cb(assign(res, d.set.slice(1), getValue(d, inp)));
});
};
else // its a crosser
return function(inp, cb) {
var g = retrieve(inp, d.get.slice(1)),
e = d.cross.reduce(horror, next)
for (var i=0; i<g.length; i++)
e(g[i], cb);
};
}, function innermost(inp, cb) {
cb(); // start to create an item
})(input, function(res) {
arr.push(res); // store the item
});
return arr;
}
这确实适用于
var result = transform(group(parse(script)), items); // your expected result
但我们可以做得更好，性能更高：
function compile(structure) {
function make(descriptor) {
if (descriptor.get)
return {inputName: descriptor.get[0], output: descriptor.get.join(".") };
var outputName = descriptor.set[descriptor.set.length-1];
var loops = descriptor.cross.reduce(function horror(next, descriptor) {
if (descriptor.set)
return function(it, cb) {
return next(it, function(res){
res.push(descriptor)
return cb(res);
});
};
else // its a crosser
return function(it, cb) {
var arrName = descriptor.get[descriptor.get.length-1],
itName = String.fromCharCode(it);
var inner = descriptor.cross.reduce(horror, next)(it+1, cb);
return {
inputName: descriptor.get[0],
statement: (descriptor.get.length>1 ? "var "+arrName+" = "+descriptor.get.join(".")+";\n" : "")+
"for (var "+itName+" = 0; "+itName+" < "+arrName+".length; "+itName+"++) {\n"+
"var "+inner.inputName+" = "+arrName+"["+itName+"];\n"+
inner.statement+
"}\n"
};
};
}, function(_, cb) {
return cb([]);
})(105, function(res) {
var item = joinSetters(res);
return {
inputName: item.inputName,
statement: (item.statement||"")+outputName+".push("+item.output+");\n"
};
});
return {
statement: "var "+outputName+" = [];\n"+loops.statement,
output: outputName,
inputName: loops.inputName
};
}
function joinSetters(descriptors) {
if (descriptors.length == 1 && descriptors[0].set.length == 1)
return make(descriptors[0]);
var paths = multiGroupBy(descriptors, function(d){ return d.set[1] || console.error("multiple assignments on "+d.set[0], d); });
var statements = [],
inputName;
var props = Object.keys(paths).map(function(p) {
var d = joinSetters(paths[p].map(function(d) {
var names = d.set.slice(1);
names[0] = d.set[0]+"_"+names[0];
return {set:names, get:d.get, cross:d.cross};
}));
inputName = d.inputName;
if (d.statement)
statements.push(d.statement)
return JSON.stringify(p) + ": " + d.output;
});
return {
inputName: inputName,
statement: statements.join(""),
output: "{"+props.join(",")+"}"
};
}
var code = joinSetters(structure);
return new Function(code.inputName, code.statement+"return "+code.output+";");
}
所以这就是你最终会得到的：
> var example = compile(group(parse("outputModel.items[].name = items[].name;outputModel.items[].sku = items[].skus[].num;")))
function(items) {
var outputModel_items = [];
for (var i = 0; i < items.length; i++) {
var item = items[i];
var skus = item.skus;
for (var j = 0; j < skus.length; j++) {
var sku = skus[j];
outputModel_items.push({"name": item.name,"sku": sku.num});
}
}
return {"items": outputModel_items};
}
> var flatten = compile(group(parse("as[]=bss[][]")))
function(bss) {
var as = [];
for (var i = 0; i < bss.length; i++) {
var bs = bss[i];
for (var j = 0; j < bs.length; j++) {
var b = bs[j];
as.push(b);
}
}
return as;
}
> var parallelRecords = compile(group(parse("x.as[]=y[].a; x.bs[]=y[].b")))
function(y) {
var x_as = [];
for (var i = 0; i < y.length; i++) {
var y = y[i];
x_as.push(y.a);
}
var x_bs = [];
for (var i = 0; i < y.length; i++) {
var y = y[i];
x_bs.push(y.b);
}
return {"as": x_as,"bs": x_bs};
}
现在您可以轻松地将输入数据传递给动态创建的函数，并且可以非常快速地进行转换： - ）