我正在尝试打印字段“term”的所有出现的值..使用我编写的程序它只返回第一次出现..是否有可能完全出现?
var data = {
"_index": "test",
"_type": "news",
"_source": {
"partnerName": "propertyFile 9",
"relatedSources": "null",
"entityCount": "50",
"Categories": {
"Types": {
"Events": [{
"count": 1,
"term": "Time",
"Time": [{
"term": "Dec 9",
"Dec_9": [{
"count": 1,
"term": "2012"
}]
}]
}, {
"count": 4,
"term": "News",
"News": [{
"term": "Germany",
"Germany": [{
"count": 1,
"term": "Election"
}],
"currency": "Euro (EUR)"
}, {
"term": "Egypt",
"Egypt": [{
"count": 1,
"term": "Revolution"
}]
}]
}]
}
}
}};
var findDeepKey = function(obj, key){
var results = [];
if (typeof obj !== 'object') return null;
for (var k in obj) {
if (k === key) return obj[k];
}
for (var k in obj) {
if (typeof obj[k] === 'object') {
if (obj[k].length) {
for (var i = 0, il = obj[k].length; i < il; i++) {
results.push(findDeepKey(obj[k][i], key));
}
} else {
for (var kk in obj[k]) {
if (kk === key) return obj[k][kk];
results.push(findDeepKey(obj[k][kk], key));
}
}
}
}
for (var i = 0, il = results.length; i < il; i++) {
if (results[i] !== null)
return results[i];
}};
alert(findDeepKey(data, 'term'));
答案 0 :(得分:0)
我正在尝试打印字段“term”的所有出现的值
它存在于对象树的任何级别的任何位置?这是一个递归下降搜索:Live example | source
function findProps(name, data, results) {
var key, value;
// If we weren't given a results array, create one
if (Object.prototype.toString.call(results) !== "[object Array]") {
results = [];
}
// Is it an object (including an array)?
if (typeof data === "object") {
// Yes, loop through its properties
for (key in data) {
// Does it have an "own" property with this name?
// (I assume you only want "own" properties, not
// properties inherited from the prototype chain).
if (data.hasOwnProperty(key)) {
// Yes, get the value
value = data[key];
// Is this our property?
if (key === name) {
// Remember it
results.push(value);
}
// Recurse into the value?
if (typeof value === "object") {
// Yes
findProps(name, value, results);
}
}
}
}
// All done
return results;
}
var terms = findProps("term", data);
display("Found 'term' " + terms.length +
" time(s): " + terms.join(", "));