我想编写一个函数来检查对象atrribute数组中是否包含关键字。如果在该数组中找到关键字，则应该返回该对象。
我尝试了下面的代码，但它没有用。我希望有一个聪明的人可以告诉我在这段代码中我做错了什么:)
我的目标是功能：ImageObjects.DE.brumi返回var ImageObjects = function() {
var DE = {
brumi : {
name:"Brummi-Werk",
keywords: ["brumi","auto", "LKWs", "lkws","lkw-werkstatt","die Brummis"]
},
medien : {
name: "medien",
keywords:["vier","nummer vier", "springer", "neue medien", "neue medien ag", "die neue medien ag"]
},
mautmaxe : {
name: "mau",
keywords:["fünf","nummer fünf", "mautmaxe", "maut", "currywust", "currywurst stand"]
}
};
return {
DE: DE
}
}();
function searchObj(obj, query) {
var data ='';
for (var property in obj) {
if (obj.hasOwnProperty(property)) {
if (typeof obj[property] === "object") {
data = searchObj(obj[property], query);
if(data !='') return data;
}
else {
if (obj["keywords"].indexOf(query) >= 0) {
return obj;
}
else {
return null;
}
}
}
}
return data;
}
var Projects = function() {
function detectProjectByKeyword(project) {
var foundObject = null;
for (var x in ImageObjects.DE){
if ((searchObj(ImageObjects.DE[x], project))!=null){
foundObject = searchObj(ImageObjects.DE[x], project)
}
}
return foundObject;
}
return {
detectProjectByKeyword: detectProjectByKeyword
}
}();
或许有更简单的方法来实现目标？我非常感谢任何帮助，因为我已经被困在这里一个多星期了:(
{{1}}
答案 0 :(得分：1)
如果结构永远不变，IE; ImageObjects.LANGUAGE.project.keywords，这就足够了：
function detectProjectByKeyword(keyword, obj, language) {
for(var a in obj[language]) {
if(obj[language].hasOwnProperty(a)) {
if(obj[language][a].keywords.indexOf(keyword) >= 0) return obj[language][a]
}
}
}
console.log(detectProjectByKeyword('brumi', ImageObjects, 'DE'));
console.log(detectProjectByKeyword('lkws', ImageObjects, 'DE'));
console.log(detectProjectByKeyword('mautmaxe', ImageObjects, 'DE'));
答案 1 :(得分：1)
我已经冒昧地稍微改变了您的数据结构，但如果我正确理解您正在尝试做的事情，那么我希望以下内容能够发挥作用：
var ImageObjects = {
DE: {
brumi : {
name: "Brummi-Werk",
keywords: [
"brumi","auto", "LKWs", "lkws","lkw-werkstatt","die Brummis"
]
},
medien : {
name: "medien",
keywords: [
"vier","nummer vier", "springer", "neue medien", "neue medien ag", "die neue medien ag",
/* added to demonstrate retrieving projects using a keyword shared across projects */, "brumi"
]
},
mautmaxe : {
name: "mau",
keywords: [
"fünf","nummer fünf", "mautmaxe", "maut", "currywust", "currywurst stand"
]
}
}
};
// Returns an array containing all projects with the keyword
// If only one project has a keyword, a one-element array will be returned
function findAllProjectsByKeyword(keyword) {
var results = [];
for (var country in ImageObjects) {
for (var project in ImageObjects[country]) {
var projectKeywords = ImageObjects[country][project].keywords;
if ( projectKeywords && projectKeywords.indexOf(keyword) != -1 ) {
results.push(ImageObjects[country][project]);
}
}
}
return results;
}
// Returns a single project with keyword
// If multiple projects have a keyword, only one will be returned
// The one that is returned may vary even for the same data, because the order
// of for-in loop enumeration may vary across implementations/runs
function findProjectByKeyword(keyword) {
for (var country in ImageObjects) {
for (var project in ImageObjects[country]) {
var projectKeywords = ImageObjects[country][project].keywords;
if ( projectKeywords && projectKeywords.indexOf(keyword) != -1 ) {
return ImageObjects[country][project];
}
}
}
}
// Example where a keyword is shared by two projects
var brumiProjects = findAllProjectsByKeyword('brumi');
console.log(brumiProjects.length); // <= 2
console.log(brumiProjects[0].name); // <= Brummi-Werk
console.log(brumiProjects[1].name); // <= medien
// Only one project is returned with findProjectByKeyword, even if multiple projects have that keyword
console.log(findProjectByKeyword('brumi').name); // <= Brummi-Werk (could be medien), depending on how the object is enumerated
// Example where only one project has a keyword
console.log(findAllProjectsByKeyword('vier')[0].name); // <= medien
console.log(findProjectByKeyword('vier').name); // <= medien
ImageObjects结构有三个级别，country（我假设），项目和项目信息（名称和关键字）。 findAllProjectsByKeyword循环前两个级别，以便搜索所有国家/地区的所有项目。对于每个项目，它在项目对象的keywords数组中搜索关键字（作为findAllProjectsByKeyword的参数传递）;如果找到关键字，则Array.indexOf返回索引 - 否则返回-1。如果找到关键字，则找到的项目对象将被推送到结果数组中。给定单个关键字，findAllProjectsByKeyword将找到多个项目。 findProjectByKeyword使用相同的方法，但只返回为给定关键字找到的第一个项目。
您可能还需要考虑其他细微之处;例如，Array.indexOf区分大小写，因此如果您搜索关键字＆＃39; Brumi＆＃39;，您将无法获得任何结果。
答案 2 :(得分：0)
如何通过多个关键字查找对象并使用现代ECMA-Script 2015及更高版本的工具和精彩内容呢？
请参阅以下代码段中的注释，以获取有关理解整个解决方案的见解：
var ImageObjects = {
DE: {
brumi: {
name: "Brummi-Werk",
keywords: [
"brumi", "auto", "LKWs", "lkws", "lkw-werkstatt", "die Brummis"
]
},
medien: {
name: "medien",
keywords: [
"vier", "nummer vier", "springer", "neue medien", "neue medien ag", "die neue medien ag", "mautmaxe"
]
},
mautmaxe: {
name: "mau",
keywords: [
"fünf", "nummer fünf", "mautmaxe", "maut", "currywust", "currywurst stand", "springer"
]
}
}
};
// #1 Second parameter is a rest parameter. Second and any other parameter
// will be captured as an array called "keywords"
//
// #2 Object.keys returns own given object's properties (you don't need to
// filter them out with x.hasOwnProperty("blah"))
//
// #3 For each property of a given language object, filter those
// that contain the whole given keywords. The "...keywords" will
// call Array.prototype.includes as many times as keywords you've
// provided, and the operation will be done like using && (AND) logical
// operator: all keywords should be present or it will be false.
//
// #4 Finally, Array.prototype.map gets the whole found property and returns
// the object held by that property!
var findByKeywords = (language, ...keywords) =>
Object.keys(ImageObjects[language])
.filter(property => ImageObjects[language][property].keywords.includes(...keywords))
.map(property => ImageObjects[language][property]);
var result = findByKeywords("DE", "mautmaxe", "springer");
console.log(result);
进一步阅读：