Question

我想使用模板片段创建最多三个项目的列表。无论是否有项目，都会始终显示三个项目空间，因此它看起来像这样。

<div>
  <div th:if="${#lists.size(block.children) > 0}"
    th:insert="code-block :: block(${block.children[0]})" 
    th:remove="tag">
  </div>
</div>;
<div>
  <div th:if="${#lists.size(block.children) > 1}"
    th:insert="code-block :: block(${block.children[1]})" 
    th:remove="tag">
  </div>
</div>;
<div>
  <div th:if="${#lists.size(block.children) > 2}"
    th:insert="code-block :: block(${block.children[2]})" 
    th:remove="tag">
  </div>
</div>

但是，即使th:if语句使用空列表求值为false，它仍会尝试执行`th：include语句，给出以下错误：

Caused by: org.springframework.expression.spel.SpelEvaluationException:
EL1025E:(pos 14): The collection has '0' elements, index '0' is invalid

如何让if语句优先于片段执行？

Answer 1

是的，不幸的是，自include has precedence over if以来，你必须将if向上移动。最简单的方法是将它放在th:block中，如下所示：

<div>
  <th:block th:if="${#lists.size(block.children) > 0}">
    <div th:insert="code-block :: block(${block.children[0]})" th:remove="tag" />
  </th:block>
</div>;
<div>
  <th:block th:if="${#lists.size(block.children) > 1}">
    <div th:insert="code-block :: block(${block.children[1]})" th:remove="tag" />
  </th:block>
</div>;
<div>
  <th:block th:if="${#lists.size(block.children) > 2}">
    <div th:insert="code-block :: block(${block.children[2]})" th:remove="tag" />
  </th:block>
</div>

你也可以简化你的代码，看看更像这样的东西：

<th:block th:each="i: ${#numbers.sequence(0, 2)}">
  <th:block th:if="${#lists.size(block.children) > i}">
    <div th:insert="code-block :: block(${block.children[i]})" th:remove="tag" />
  </th:block>

  <th:block th:unless="${i == 2}">;</th:block>
</th:block>

条件性Thymeleaf片段

1 个答案: