Question

我想迭代Map＆lt;字符串，列表＆gt;＆gt; 在java中：

final JSONObject mappedData = new JSONObject();
final Map<String, List<List<String>>> associatedUsecasesMap = new HashMap<String, List<List<String>>>();
//some code that populates map.
mappedData.put("usecaseDetailsMap",associatedUsecasesMap);
return mappedData;

在jsp中当我用

打印时

<a:text>${usecases.get('usecaseDetailsMap').get('ARRIVED')}</a:text>

我得到了

[["ABC","Realm","patluria","test","1444209946120"],["ABC","Realm","patluria","Abhinav","1444719147503"],["ABC","Realm","patluria","Abhinav","1444719147503"],["ss","Realm","patluria","sa","1443590024585"],["ABC","Realm","patluria","eef","1444382236881"],["passport","Realm","patluria","testing","1444368146430"],["ABC","Realm","patluria","eef","1444382236881"],["ABC","Realm","patluria","as","1444210103588"],["saasa","Realm","patluria","as","1444372196955"],["ABC","Realm","patluria","sasd","1444727640485"],["ABC","Realm","patluria","eef","1444382223024"],["ss","Realm","patluria","sa","1443590006005"],["asa","Realm","patluria","asa","1444386667440"],["ABC","Realm","patluria","dadda","1444639933870"],["asa","Realm","patluria","asa","1444386667440"],["ABC","Realm","patluria","sasd","1444727626874"],["ss","Realm","patluria","sa","1444414479534"],["ABC","Realm","patluria","asa","1444729340806"],["ABC","Realm","patluria","as","1444209965115"],["ss","Realm","patluria","sa","1443590024585"],["saasa","Realm","patluria","as","1444372196955"]]

但我想迭代这个，所以我这样做

<c:forEach items="${usecases.get('usecaseDetailsMap')get('ARRIVED')}" var="usecase" >

但是我收到以下错误：javax.servlet.jsp.JspTagException：不知道如何迭代提供的＆＃34;项目＆＃34;在＆lt; forEach＆gt;

中

我可以知道如何遍历List＆gt;基于指定的密钥。

Answer 1

当您遍历Map时，JSTL forEach将其用作Map.Entry实例。然后你可以使用entry.key＆amp; jsp中的entry.value。因此，您可以直接在JSTL中使用Map，如下所示： -

<c:forEach var="usecaseDetails" items="${usecaseDetailsMap}">
  ${usecaseDetails.key} ${usecaseDetails.value}
</c:forEach>

在jsp上迭代Map <string，list <list <string>＆gt;＆gt;

1 个答案: