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我有以下的有效JSON(我测试了jsonlint.com):如何解析简单的JSON我JAVA
{
"Structure": {
"paper": "Paper",
"solid_reviewer": "Solid Reviewer",
"second_reviewer": "2. Reviewer",
"third_reviewer": "3. Reviewer"
},
"ReviewerGroup": {
"3457": {
"paper": "3457",
"solid_reviewer": {
"druidkey": "168",
"match": "Interest match"
},
"second_reviewer": {
"druidkey": "192",
"match": "Interest match"
},
"third_reviewer": {
"druidkey": "155",
"match": "Interest match"
}
},
"3458": {
"paper": "3458",
"solid_reviewer": {
"druidkey": "229",
"match": "Interest match"
},
"second_reviewer": {
"druidkey": "145",
"match": "Interest match"
},
"third_reviewer": {
"druidkey": "123",
"match": "Interest match"
}
},
"275": {
"paper": "275",
"solid_reviewer": {
"druidkey": "999",
"match": "Interest match"
},
"second_reviewer": {
"druidkey": "412",
"match": "Interest match"
},
"third_reviewer": {
"druidkey": "713",
"match": "Interest match"
}
}
},
"failedPapers": {}
}
在PHP中,解码此我只是做:
$response = json_decode($response);
和我可以遍历$ response对象:
foreach ($response->ReviewerGroup as $r){
$paperRev1 = $paperReviewDAO->find(**$r->paper**, 1);
$revConf = $revConfDAO->findRevConfFromReviewer($confId,
**$r->solid_reviewer->druidkey**);
.
.
}
但是它在Java中似乎要复杂得多。
我见过很多关于如何读取一个对象或对象数组的例子。
但是,如何获取所有'ReviewerGroup'(RG)对象并遍历它们?我需要从每个对象中提取信息,并将数据插入数据库(我知道该怎么做......)作为每个RG对象的实体。
我不在乎使用什么JSON库。我已经看过GSON和杰克逊。
感谢您的任何意见和想法,...
/ 金
这和Jackson很相似,除了它是“readValue”和“ObjectMapper” –