在这种情况下如何将设置转换为列表

时间:2015-08-17 16:10:05

标签: java spring jsp

这是我的控制者:

private void setUniqueSignSize(List<TemplateMasterModel> templateTypeList, final Map<String, Object> model) {

    Set<String> signSizeList = new HashSet<String>();

    for(TemplateMasterModel templateType: templateTypeList) {
        signSizeList.add(templateType.getTemplateGrpCode());
    }

    model.put("signSizeList", signSizeList);

    //getPaperTypes(templateTypeList, signSizeList, model);

}

这是我的jsp:

<table style="width:500px; margin-bottom: 10px;margin-left:50px;">
    <tr>
        <td class="requiredFieldIndicator">Sine Size</td>
        <td>
            <input type="hidden" name="tempTypeName" id="tempTypeName" value="">

            <select name="signTypeDesc" id="signTypeDesc" onchange="setTemplChangeValue();" style="width: 200px">
            <%
                String changeY = "";
                if(request.getAttribute("changeY") != null){
                    changeY = String.valueOf(request.getAttribute("changeY"));
                }

                List<TemplateMasterModel> templateTypeList = (List) request.getAttribute("signSizeList");
                if(templateTypeList != null && templateTypeList.size() != 0){
                    if(templateTypeList.size() == 1)
                    {
                        if(changeY.equals("No"))
                        {
                            %><option value="Select">None Selected</option><%   
                        }
                        %>
                        <option value="<%=templateTypeList.get(0).getTemplateGrpCode()%>"><%=templateTypeList.get(0).getTemplateGrpCode()%></option>
                        <%
                    }

如何将集合signSizeList转换为列表?

1 个答案:

答案 0 :(得分:0)

要么你可以去

List templateTypeList  = new ArrayList((HashSet)request.getAttribute("signSizeList"));

或使用List的.addAll()方法将set的所有元素添加到新的ArrayList。