我有一个复杂的JSON，我需要递归解析。递归的最终结果是Map＆gt;对象的类型，其中键是观众 - 名称值，内部地图是文本键，标题值。 这只是完整JSON的一部分。
"sections": {
"1": {
"1": {
"1": {
"title": "xxx",
"text": "xxx",
"tags": {
"audience": {
"1": {
"name": "xxx",
"title": "xxx",
"id": "xxx"
}
},
"styleHint": {
"1": {
"name": "xxx",
"title": "xxx",
"id": "xxx"
}
}
}
},
"title": "xxx",
"text": "xxx",
"tags": {
"audience": {
"1": {
"name": "xxx",
"title": "xxx",
"id": "xxx"
}
},
"styleHint": {
"1": {
"name": "xxx",
"title": "xxx",
"id": "xxx"
}
}
}
},
"2": {
"title": "xxx",
"text": "xxx",
"tags": {
"audience": {
"1": {
"name": "xxx",
"title": "xxx",
"id": "xxx"
}
},
"styleHint": {
"1": {
"name": "xxx",
"title": "xxx",
"id": "xxx"
}
}
}
},
"title": "xxx",
"text": "xxx",
"tags": {
"audience": {
"1": {
"name": "xxx",
"title": "xxx",
"id": "xxx"
},
"2": {
"name": "xxx",
"title": "xxx",
"id": "xxx"
}
},
"styleHint": {
"1": {
"name": "xxx",
"title": "xxx",
"id": "xxx"
}
}
}
},
"2": {
"title": "xxx",
"text": "xxx",
"tags": {
"audience": {
"1": {
"name": "xxx",
"title": "xxx",
"id": "xxx"
}
},
"styleHint": {
"1": {
"name": "xxx",
"title": "xxx",
"id": "xxx"
}
}
},
"anchor":"xxx"
},
"3": {
"1": {
"title": "xxx",
"text": "xxx",
"tags": {
"audience": {
"tag": {
"name": "xxx",
"title": "xxx",
"id": "xxx"
}
},
"styleHint": {
"tag": {
"name": "xxx",
"title": "xxx",
"id": "xxx"
}
}
}
},
"title": "xxx",
"text": "xxx",
"tags": {
"audience": {
"1": {
"name": "xxx",
"title": "xxx",
"id": "xxxx"
}
},
"styleHint": {
"1": {
"name": "xx",
"title": "xxx",
"id": "xxxx"
}
}
}
}
}
我之前使用JSONObject只是为了很晚才意识到迭代以相反的顺序发生：（
我尝试递归地解析整个结构并将其反转为我的利益。 BUt命令是乱七八糟:( :(主要是因为文本，标题，片段跟随第二个文本，标题，并有2个受众名称。该部分的文本和标题被跳过，因为整个订单被妥协< / p>
请帮忙!!我目前的实施如下
private Map<String, Map<String, String>> parseTextAndTitle(JSONObject json,
Map<String, Map<String, String>> ttMap, String article,
List<String> usrGrp) throws JSONException {
logger.info("Entering method..");
String userGroup = null;
Map<String, String> titleAndText = new LinkedHashMap<String, String>();
Map<String, String> currMap = new LinkedHashMap<String, String>();
Map<String, String> tempMap = new LinkedHashMap<String, String>();
Iterator<String> keys = json.sortedKeys();
while (keys.hasNext()) {
String key = keys.next();
JSONObject value = null;String firstKey = null;
String text = null;String title = null;
int length = 0;
try {
value = json.getJSONObject(key);
if (key.equalsIgnoreCase(STYLEHINT) || key.equalsIgnoreCase(ANCHOR)
|| key.equalsIgnoreCase(INLINE)) {
continue;
}
if (key.equals(TEXT)) {
text = json.getString(key);
text = removeHtmlTag(text);
logger.debug("TEXT RETRIEVED:" + text);
if(text != null) {
titleAndText.put(text, "");
}
else
logger.debug("Text not retrieved!!");
}
if (key.equals(TITLE)) {
title = json.getString(TITLE);
title = appendNewline(title);
logger.debug("TITLE RETRIEVED:" + title);
if (title != null) {
for (Map.Entry<String, String> iter : titleAndText
.entrySet())
firstKey = iter.getKey();
if(firstKey != null) {
titleAndText.put(firstKey, title);
}
else
logger.debug("NO key present in textAndTitle Map!!");
}
}
if (key.equals(AUDIENCE_TAG)) {
try {
length = value.length();
for (int i = 0; i < length; i++) {
userGroup = (String) value.getJSONObject(
String.valueOf(i + 1)).get(NAME);
logger.debug("USERGROUP RETRIEVED:" + userGroup);
usrGrp.add(userGroup);
}
} catch (Exception e) {
userGroup = (String) value.getJSONObject(TAG).get(NAME);
logger.debug("USERGROUP RETRIEVED:" + userGroup);
usrGrp.add(userGroup);
}
}
else{
parseTextAndTitle(value, ttMap, article, usrGrp);
}
} catch (Exception e) {
logger.debug("value not a JSON Object..rather an element");
// Extract the text values
if (key.equals(TEXT)) {
text = json.getString(key);
text = removeHtmlTag(text);
logger.debug("TEXT RETRIEVED:" + text);
if(text != null) {
titleAndText.put(text, "");
}
else
logger.debug("Text not retrieved!!");
}
if (key.equals(TITLE)) {
title = json.getString(TITLE);
title = appendNewline(title);
logger.debug("TITLE RETRIEVED:" + title);
if (title != null) {
for (Map.Entry<String, String> iter : titleAndText
.entrySet())
firstKey = iter.getKey();
if(firstKey != null) {
titleAndText.put(firstKey, title);
}
else
logger.debug("NO key present in textAndTitle Map!!");
}
}
}
if (!(usrGrp.isEmpty()) && !(titleAndText.isEmpty())
&& title != null) {
if(usrGrp.size() > 1)
{
for(int i=0;i<usrGrp.size();i++)
{
//If user group already present, extract current text,title map
//If not put usergroup as key, text,title map as value
if (ttMap.containsKey(usrGrp.get(i))) {
currMap = ttMap.get(usrGrp.get(i));
if (currMap.isEmpty()) {
ttMap.put(usrGrp.get(i), titleAndText);
} else {
currMap = ttMap.get(usrGrp.get(i));
for (Map.Entry<String, String> entry : currMap
.entrySet()) {
tempMap.put(entry.getKey(),
(String) entry.getValue());
}
for (Map.Entry<String, String> ttEntry : titleAndText
.entrySet()) {
tempMap.put(ttEntry.getKey(),
(String) ttEntry.getValue());
}
ttMap.put(usrGrp.get(i),tempMap);
// titleAndText = new LinkedHashMap<String, String>();
tempMap = new LinkedHashMap<String, String>();
}
}
else {
ttMap.put(usrGrp.get(i), titleAndText);
}
}
titleAndText.clear();
}
else
{
if (ttMap.isEmpty())
{
tempMap = titleAndText;
ttMap.put(usrGrp.get(0), tempMap);
}
else {
currMap = ttMap.get(usrGrp.get(0));
if (currMap.isEmpty()) {
ttMap.put(usrGrp.get(0), titleAndText);
}else {
currMap = ttMap.get(usrGrp.get(0));
for (Map.Entry<String, String> entry : currMap
.entrySet()) {
tempMap.put(entry.getKey(),
(String) entry.getValue());
}
for (Map.Entry<String, String> ttEntry : titleAndText
.entrySet()) {
tempMap.put(ttEntry.getKey(),
(String) ttEntry.getValue());
}
ttMap.put(usrGrp.get(0),tempMap);
titleAndText.clear();
}
}
}
usrGrp.clear();
}
}
logger.info("Exiting method..");
return ttMap;
}
答案 0 :(得分：1)
package Test.json;
import java.util.Iterator;
import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;
public class App {
public static void main(String[] args) {
String str = "{\"a\":\"1\", \"b\":\"2\", \"c\":[{\"d\":\"4\"},{\"e\":\"5\"},{\"f\":[{\"g\":\"6\"},{\"h\":\"7\"}]}], \"i\":8}";
try {
loopThroughJson(new JSONObject(str));
} catch (JSONException e) {
e.printStackTrace();
}
}
public static void loopThroughJson(Object input) throws JSONException {
if (input instanceof JSONObject) {
Iterator<?> keys = ((JSONObject) input).keys();
while (keys.hasNext()) {
String key = (String) keys.next();
if (!(((JSONObject) input).get(key) instanceof JSONArray))
System.out.println(key + "=" + ((JSONObject) input).get(key));
else
loopThroughJson(new JSONArray(((JSONObject) input).get(key).toString()));
}
}
if (input instanceof JSONArray) {
for (int i = 0; i < ((JSONArray) input).length(); i++) {
JSONObject a = ((JSONArray) input).getJSONObject(i);
Object key = a.keys().next().toString();
if (!(a.opt(key.toString()) instanceof JSONArray))
System.out.println(key + "=" + a.opt(key.toString()));
else
loopThroughJson(a.opt(key.toString()));
}
}
}
}
Output:
a=1
b=2
d=4
e=5
g=6
h=7
i=8
答案 1 :(得分：1)
修改后的@sklimkovitch代码以使其在某些复杂的Json结构中工作...
public void loopThroughJson(Object input) throws JSONException {
if (input instanceof JSONObject) {
Iterator<?> keys = ((JSONObject) input).keys();
while (keys.hasNext()) {
String key = (String) keys.next();
if (!(((JSONObject) input).get(key) instanceof JSONArray))
if (((JSONObject) input).get(key) instanceof JSONObject) {
loopThroughJson(((JSONObject) input).get(key));
} else
System.out.println(key + "=" + ((JSONObject) input).get(key));
else
loopThroughJson(new JSONArray(((JSONObject) input).get(key).toString()));
}
}
if (input instanceof JSONArray) {
for (int i = 0; i < ((JSONArray) input).length(); i++) {
JSONObject a = ((JSONArray) input).getJSONObject(i);
loopThroughJson(a);
}
}
}
答案 2 :(得分：0)
而不是
while (keys.hasNext()) {
<blah blah>
if (key.equalsIgnoreCase(STYLEHINT) || key.equalsIgnoreCase(ANCHOR)
|| key.equalsIgnoreCase(INLINE)) {
continue;
}
if (key.equals(TEXT)) {
<blah blah>
}
if (key.equals(TITLE)) {
....
可以简单地编码：
text = json.getString(TEXT);
<deal with text>
title = json.getString(TITLE);
<etc>
如果某些键值可能不存在，只需在获取它们之前使用has测试它们的缺失。
由于STYLEHINT，ANCHOR和INLINE被忽略，因此无法获取它们。
要处理JSON的棘手布局，请执行以下操作：
if (json.has("title")) {
<extract title/text/tags/stylehint as described above>
}
else {
Iterator<String> keys = json.sortedKeys();
while (keys.hasNext()) {
// Note that "key" must be "1", "2", "3"...
String key = keys.next();
value = json.getJSONObject(key);
<recursively call method using "value">
}
}
答案 3 :(得分：0)
找到了ordered..ditched JSONObject API的解决方案，并改为使用gson JsonObject
private Map<String, List<String>> parseJsonSection(
Map<String, List<String>> retTextMap, JsonObject jsonObject,
String lastKey, StringBuffer tt, List<String> ttext)
throws ParseException, JSONException {
for (Entry<String, JsonElement> entry : jsonObject.entrySet()) {
String key = entry.getKey();
Object value = entry.getValue();
logger.debug("Key:" + key + "\n" + value.toString());
if (key.equalsIgnoreCase(STYLEHINT) || key.equalsIgnoreCase(INLINE)
|| key.equalsIgnoreCase(ANCHOR))
continue;
if (key.equalsIgnoreCase(TEXT)) {
tt.append(value.toString());
ttext.add(tt.toString());
}
if (key.equalsIgnoreCase(TITLE) && tt.length() == 0) {
tt = new StringBuffer();
tt.append(value.toString() + "-");
}
if (key.equalsIgnoreCase(NAME)) {
logger.debug("Value of usergrp:" + value.toString());
String usrGrp = value.toString();
if (retTextMap.isEmpty()) {
if (tt.toString() != null) {
List<String> temp = new ArrayList<String>();
temp = ttext;
retTextMap.put(usrGrp, temp);
}
return retTextMap;
} else if (retTextMap.get(usrGrp) != null) {
List<String> temp = retTextMap.get(value.toString());
if (!temp.contains(tt.toString()))
temp.add(tt.toString());
retTextMap.put(usrGrp, temp);
} else if (retTextMap.get(usrGrp) == null) {
if (tt != null) {
List<String> temp = new ArrayList<String>();
temp.add(tt.toString());
retTextMap.put(usrGrp, temp);
return retTextMap;
}
}
}
if (value instanceof JsonObject) {
parseJsonSection(retTextMap, (JsonObject) value, key, tt, ttext);
}
}
return retTextMap;
}