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Spring 如何正确地使用FreeMarker迭代并将1到31的月份打印到select中?_Spring_Spring Mvc_Freemarker_Template Engine - Fatal编程技术网

Spring 如何正确地使用FreeMarker迭代并将1到31的月份打印到select中?

Spring 如何正确地使用FreeMarker迭代并将1到31的月份打印到select中?,spring,spring-mvc,freemarker,template-engine,Spring,Spring Mvc,Freemarker,Template Engine,我是一个全新的FreeMarker开发者,我正在开发一个使用这个模板引擎的Spring MVC应用程序 我有一个非常简单的问题要用FreeMarker解决 在一个表单中,我有一个标签,用户可以在其中选择一个月的哪一天,如下所示: <select id="birth-day" name="birthDay" class="form-control selectpicker select-mm" title="Giorno" data-validation="[NOTEMPTY, INTEG

我是一个全新的FreeMarker开发者,我正在开发一个使用这个模板引擎的Spring MVC应用程序

我有一个非常简单的问题要用FreeMarker解决

在一个表单中,我有一个标签,用户可以在其中选择一个月的哪一天,如下所示:

<select id="birth-day" name="birthDay" class="form-control selectpicker select-mm" title="Giorno"  data-validation="[NOTEMPTY, INTEGER]" data-validation-label="giorno di nascita" aria-label="seleziona il giorno di nascita" aria-required="true" tabindex="30">
    <option value="01">01</option>
    <option value="02">02</option>
    <option value="03">03</option>
    ...........................................
    ...........................................
    <option value="31">31</option>
</select>

01
02
03
...........................................
...........................................
31
实际上,这些选项是静态创建的,但我希望通过1到31之间的迭代来创建它,避免将所有这些选项都写入select标记中

查看FreeMarker文档,我觉得它只显示了如何在列表上迭代:


但我必须这样做,这更像是一个标准循环(for(I=1;I由我自己解决,这是代码:

<select id="birth-day" name="birthDay" class="form-control selectpicker select-mm" title="Giorno"  data-validation="[NOTEMPTY, INTEGER]" data-validation-label="giorno di nascita" aria-label="seleziona il giorno di nascita" aria-required="true" tabindex="30">

    <#assign x = 31>
    <#list 1..x as i>
        <#if i < 10>
            <option value="0${i}">0${i}</option>
        <#else>
            <option value="${i}">${i}</option>
        </#if>
    </#list>

</select>

0${i}
${i}

如果你想摆脱
如果
那么你可以使用并非所有的月份都有31天。。。