PHP If状态-仅当后台有Joomla模块文本字段的文本输出时才呈现HTML - PHP If state - Render HTML only if there is text output from Joomla module text field int the backend

PHP If state - Render HTML only if there is text output from Joomla module text field int the backend

我尝试在Joomla 3中构建一个简单的FAQ模块。

我在模块后端有4个字段。问题1 &;问题2 & &;回答2"。

我的php和Html是这样构建的:

<div id="accordion">
  <h3 class="question"> <?php echo $params->get('question1'); ?></h3>
  <div  class="question">
  <?php echo $params->get('answer1'); ?>
  </div>
  <h3  class="question"> <?php echo $params->get('question2'); ?></h3>
  <div  class="question">
  <?php echo $params->get('answer2'); ?>
  </div>
</div>

我喜欢添加一些php的if语句，这样代码块只呈现，如果文本字段在后端，如果他们是由用户填写:

(如果字段"question3"有任何文本，显示此代码)

<?php if ($this->params->get('question3')) : ?>  //This is the line of code i do not know how to do.
  <h3  class="question"> <?php echo $params->get('question2'); ?></h3>
  <div  class="question">
  <?php echo $params->get('answer2'); ?>
  </div>
</div>
<?php endif; ?>

XML是这样的:

                <!-- Question 1 -->                 
                <field
                    label="Question 1"
                    default=""
                    name="question1"
                    description="Fill in the question"
                    type="text"
                    size="60"   
                />
                <!-- End of Question 1 -->
                <!-- Answer 1 -->                   
                <field
                    label="Answer1" 
                    default="Fill in the answer"
                    name="answerswer1"
                    description="Fill in the answer"
                    type="editor"
                    width="200"
                    height="100"
                    hide="readmore,pagebreak" 
                    filter="safehtml" 
                    size="260"  
                />
                <!-- End of Answer 1 -->
                <field type="spacer" name="questionspacer2" label="&lt;b&gt;Question and Answer 2 1&lt;/b&gt;" />

                    <!-- Question 2 -->                 
                <field
                    label="Question 2"
                    default=""
                    name="question2"
                    description="Fill in the question"
                    type="text"
                    size="60"   
                />
                <!-- End of Question 2 -->
<!-- Answer 2 -->                   
                <field
                    label="Answer2" 
                    default=""
                    name="answerswer2"
                    description="Fill in the answer"
                    type="editor"
                    width="200"
                    height="100"
                    hide="readmore,pagebreak" 
                    filter="safehtml" 
                    size="260"  
                />
                <!-- End of Answer 2 -->

任何提示都会很棒!我已经找了一晚上了

根据您如何设置参数，如果没有填写，您需要检查输出如下:

$answer2 = $params->get('answer2'));
// (Applies if the not filled data returns NULL etc.)
if ( isset($answer2) ) { YOUR CODE } 
// (Applies if the not filled data returns an empty string)
if ( !empty($answer2) ) { YOUR CODE }

没有测试，但应该可以工作;)

编辑:就像下面的评论中讨论的那样，让它看起来像这样:)