問題描述

我有一個(gè)類似于以下內(nèi)容的 JSON 字符串:

I have a JSON string that resembles the following:

{
    "foo" : "bar",
    "id" : 1,
    "children":[
        {
            "some" : "string",
            "id" : 2,
            children : []
        },
        {
            "some" : "string",
            "id" : 2,
            children : []
        }
    ]
}

我對(duì)此字符串進(jìn)行 JSON 解析，然后將所有對(duì)象轉(zhuǎn)換為 HashMap，將所有數(shù)組轉(zhuǎn)換為 HashMap[].我的問題是我需要一個(gè)遞歸函數(shù)來遍歷 Java 中這個(gè) JSON 結(jié)構(gòu)的所有節(jié)點(diǎn).我怎樣才能做到這一點(diǎn)?我在想這樣的事情:

I do a JSON parse of this string, and that turns all objects into HashMaps and all arrays into HashMap[]s. My problem is I need a single recursive function to iterate through all nodes of this JSON structure in Java. How can I do this? I was thinking something like:

public HashMap findNode(boolean isArray, HashMap map, HashMap[] array){
    //array stuff
    if(isArray){
        for(int i=0; i<array.length(); i++){
            Object value = array[i];
            if(value instanceof String)
                System.out.println("value = "+value);
            else if(value instanceof HashMap)
                findNode(false, value, null);
            else if(value instanceof HashMap[])
                findNode(true, null, value);
        }
    //hashmap stuff
    }else{
        for(HashMap.Entry<String, Object> entry : map.entrySet()){
            Object value = entry.getValue();
            if(value instanceof String)
                System.out.println("value = "+value);
            else if(value instanceof HashMap)
                findNode(false, value, null);
            else if(value instanceof HashMap[])
                findNode(true, null, value);
        }
    }
}

推薦答案

假設(shè)你的一個(gè)數(shù)組里面只能有 Maps(而不是其他數(shù)組):

Assuming you an array can only have Maps inside (and not other arrays):

public void findNode(HashMap map) {
    for(HashMap.Entry<String, Object> entry : map.entrySet()){
        Object value = entry.getValue();
        if(value instanceof String)
            System.out.println("value = "+value);
        else if(value instanceof HashMap)
            findNode(value);
        else if(value instanceof HashMap[])
            for(int i=0; i<array.length(); i++){
                findNode(array[i]);
    }
}

或者如果你可以使用 3 個(gè)函數(shù)，你可以讓它變得更簡單

Or you can make it even simpler if you can use 3 functions

public void findNode(HashMap map) {
    for(HashMap.Entry<String, Object> entry : map.entrySet()){
        findNode(entry.getValue());
    }
}

public void findNode(String value) {
    System.out.println("value = "+value);
}

public void findNode(HashMap[] value) {
    for(int i=0; i<array.length(); i++){
        findNode(array[i]);
    }
}

這篇關(guān)于通過深度HashMap遞歸迭代的文章就介紹到這了，希望我們推薦的答案對(duì)大家有所幫助，也希望大家多多支持html5模板網(wǎng)！