Java 使用jackson创建json对象
如何使用jackson创建一个json数组,如下面的示例所示 我尝试使用ObjectMapper,但这似乎不正确Java 使用jackson创建json对象,java,json,hashmap,jackson,Java,Json,Hashmap,Jackson,如何使用jackson创建一个json数组,如下面的示例所示 我尝试使用ObjectMapper,但这似乎不正确 try (DirectoryStream<Path> ds = Files.newDirectoryStream(path)) { for (Path file : ds) { System.out.println("name:"+file.getFileName()+
try (DirectoryStream<Path> ds = Files.newDirectoryStream(path)) {
for (Path file : ds) {
System.out.println("name:"+file.getFileName()+
"\n"+
"mime:"+Files.probeContentType(file)+
"\n"+
"locked:"+!Files.isWritable(file));
}
} catch (IOException e) {
System.err.println(e);
}
下面是我提供的一个json示例
"files": [
{
"mime": "directory",
"ts": 1334071677,
"read": 1,
"write": 0,
"size": 0,
"hash": "l1_Lw",
"volumeid": "l1_",
"name": "Demo",
"locked": 1,
"dirs": 1
},
{
"mime": "directory",
"ts": 1334071677,
"read": 1,
"write": 0,
"size": 0,
"hash": "l1_Lw",
"volumeid": "l1_",
"name": "Demo",
"locked": 1,
"dirs": 1
},
{
"mime": "directory",
"ts": 1340114567,
"read": 0,
"write": 0,
"size": 0,
"hash": "l1_QmFja3Vw",
"name": "Backup",
"phash": "l1_Lw",
"locked": 1
},
{
"mime": "directory",
"ts": 1310252178,
"read": 1,
"write": 0,
"size": 0,
"hash": "l1_SW1hZ2Vz",
"name": "Images",
"phash": "l1_Lw",
"locked": 1
},
{
"mime": "application\/x-genesis-rom",
"ts": 1310347586,
"read": 1,
"write": 0,
"size": 3683,
"hash": "l1_UkVBRE1FLm1k",
"name": "README.md",
"phash": "l1_Lw",
"locked": 1
}
]
编辑1
Map<String, Object> filesMap = new HashMap<>();
List<Object> files = new ArrayList<Object>();
System.out.println("\nNo filter applied:");
try (DirectoryStream<Path> ds = Files.newDirectoryStream(path)) {
for (Path file : ds) {
Map<String, Object> fileInfo = new HashMap<>();
fileInfo.put("name", file.getFileName().toString());
// Prints Files in Director
// Files.getAttribute(file,"size");
System.out.println("name:" + file.getFileName().toString() +
"\n" +
"mime:" + Files.probeContentType(file) +
"\n" +
"locked:" + !Files.isWritable(file));
ObjectMapper mapper = new ObjectMapper();
String json = mapper.writeValueAsString(fileInfo);
files.add(json);
}
} catch (IOException e) {
System.err.println(e);
}
files.toArray();
filesMap.put("files", files);
ObjectMapper mapper = new ObjectMapper();
String jsonString;
try {
jsonString = mapper.writeValueAsString(filesMap);
} catch (IOException e) {
jsonString = "fail"; //To change body of catch statement use File | Settings | File Templates.
}
最终答案
Map<String, Object> filesMap = new HashMap<>();
List<Object> files = new ArrayList<Object>();
System.out.println("\nNo filter applied:");
try (DirectoryStream<Path> ds = Files.newDirectoryStream(path)) {
for (Path file : ds) {
Map<String, Object> fileInfo = new HashMap<>();
fileInfo.put("name", file.getFileName().toString());
System.out.println("name:" + file.getFileName().toString() +
"\n" +
"mime:" + Files.probeContentType(file) +
"\n" +
"locked:" + !Files.isWritable(file));
files.add(fileInfo);
}
} catch (IOException e) {
System.err.println(e);
}
files.toArray();
filesMap.put("files", files);
ObjectMapper mapper = new ObjectMapper();
String jsonString;
try {
jsonString = mapper.writeValueAsString(filesMap);
} catch (IOException e) {
jsonString = "fail";
}
Map filesMap=newhashmap();
列表文件=新的ArrayList();
System.out.println(“\n未应用过滤器:”);
try(DirectoryStream ds=Files.newDirectoryStream(path)){
用于(路径文件:ds){
Map fileInfo=newhashmap();
fileInfo.put(“name”,file.getFileName().toString());
System.out.println(“名称:”+file.getFileName().toString()+
“\n”+
“mime:+Files.probeContentType(文件)+
“\n”+
锁定:“+!Files.isWritable(file));
添加(fileInfo);
}
}捕获(IOE异常){
系统错误println(e);
}
toArray()文件;
filesMap.put(“文件”,files);
ObjectMapper mapper=新的ObjectMapper();
字符串jsonString;
试一试{
jsonString=mapper.writeValueAsString(fileMap);
}捕获(IOE异常){
jsonString=“失败”;
}
您可以将对象写入json字符串。因此,我希望您的数据位于根据需要定义的类的对象中。以下是如何将该对象转换为json字符串:
//1. Convert Java object to JSON format
ObjectMapper mapper = new ObjectMapper();
String jsonString = mapper.writeValueAsString(yourObject);
有关完整的jackson数据绑定javadoc,请参阅。您需要:
此类具有创建ArrayNode
s、ObjectNode
s、IntNode
s、DecimalNode
s、TextNode
s等的方法ArrayNode
s和ObjectNode
s具有方便的变异方法,可以直接添加大多数JSON原语(非容器)值,而无需经过工厂(在内部,它们引用此工厂,这就是原因)
至于
ObjectMapper
,请注意它既是序列化程序(ObjectWriter
)又是反序列化程序(ObjectReader
)。无需创建POJO并将其转换为JSON即可。我假设您的数据在记录对象中
JsonNode rootNode = mapper.createObjectNode();
ArrayNode childNodes = mapper.createArrayNode();
for (Record record : records) {
JsonNode element = mapper.createObjectNode();
((ObjectNode) element).put("mime": record.getDirectory());
//fill rest of fields which are needed similar to this.
//Also here record.getDirectory() method will should return "directory"
//according to your json file.
childNodes.add(element);
}
((ObjectNode) rootNode).put("files", childNodes);
将JSON对象初始化为单例实例,并构建它:
ObjectNode node=JsonNodeFactory.instance.ObjectNode();//初始化
node.put(“x”,x);//建筑
PS:println use
node.toString()
您的问题似乎缺少一个实际问题。对不起,今晚我遇到了太多的互联网问题,我想我忘了添加这一点:P现在有意义了吗?我很困惑。这里没有明显使用ObjectMapper
。您是否正在尝试将某些内容转换为使用ObjectMapper
或什么?您已经显示了所需的输入和输出,但没有显示您实际尝试的内容:实际转换代码。你看完了吗?我看完了教程,它很有帮助,虽然还不是很好,但是很接近。这让我非常接近,但出于某种原因,我在子对象周围有额外的引号。我在上面添加了我目前拥有的内容。@DirkLachowski Yep:)我将其放在“编辑1”下描述的底部,额外的引号是因为我使用了ObjectMapper两次。删除内部writeValueAsString后,我能够得到预期的结果。谢谢你的帮助。事实上,我会把这个问题作为答案,因为当我给它评分时,另一个答案造成了不可预见的问题。这是我使用的最终答案:)
//1. Convert Java object to JSON format
ObjectMapper mapper = new ObjectMapper();
String jsonString = mapper.writeValueAsString(yourObject);
final JsonNodeFactory factory = JsonNodeFactory.instance;
JsonNode rootNode = mapper.createObjectNode();
ArrayNode childNodes = mapper.createArrayNode();
for (Record record : records) {
JsonNode element = mapper.createObjectNode();
((ObjectNode) element).put("mime": record.getDirectory());
//fill rest of fields which are needed similar to this.
//Also here record.getDirectory() method will should return "directory"
//according to your json file.
childNodes.add(element);
}
((ObjectNode) rootNode).put("files", childNodes);