【发布时间】:2011-09-08 16:27:02
【问题描述】:
我有一个包含多个 activeCheckBoxLists 的页面,这些页面实际上是在 foreach 循环中生成的。我遇到的问题是,每个列表都是使用与其他所有生成列表相同的名称和 ID 生成的。我需要一种方法将迭代或其他内容发送到 id 和 name 中,以便它们是唯一的。
当我使用 jquery 选择列表的第一项时,它会选择每个列表的第一项。
这是生成列表的代码。
// collect all filter titles (level == 0, parent == 0)
$topLevelFilterTitles = $hsf->findAllByAttributes(array('level'=>'0','parent_id'=>'0'));
foreach($topLevelFilterTitles as $filterTitle):
// with each filter title find all children (level == 1, parent == filter title id)
echo "<div class='half menu split'>";
echo "<p class='uppercase-text filter-name'>" . $filterTitle->title . "</p>";
$filterOptions = $hsf->findAllByAttributes(
array(
'level'=>'1',
'parent_id'=>$filterTitle->id,
)
);
$list = CHtml::listData($filterOptions,'filter_name','title');
echo CHtml::activeCheckBoxList(
$hsf, // model
'filter_name',
$list
);
echo "</div>";
endforeach;
生成的列表如下所示
<div class="halfmenu split">
<p class="uppercase-text filter-name">Filter Name R</p>
<input id="ytHardwareSearchFiltering_filter_name" type="hidden" value="" name="HardwareSearchFiltering[filter_name]" />
<input id="HardwareSearchFiltering_filter_name_0" value="r_value0" type="checkbox" name="HardwareSearchFiltering[filter_name][]" />
<label for="HardwareSearchFiltering_filter_name_0">r_name0</label><br/>
<input id="HardwareSearchFiltering_filter_name_1" value="r_value1" type="checkbox" name="HardwareSearchFiltering[filter_name][]" />
<label for="HardwareSearchFiltering_filter_name_1">r_name1</label>
</div>
<div class="halfmenu split">
<p class="uppercase-text filter-name">Filter Name T</p>
<input id="ytHardwareSearchFiltering_filter_name" type="hidden" value="" name="HardwareSearchFiltering[filter_name]" />
<input id="HardwareSearchFiltering_filter_name_0" value="t_value0" type="checkbox" name="HardwareSearchFiltering[filter_name][]" />
<label for="HardwareSearchFiltering_filter_name_0">t_name0</label><br/>
<input id="HardwareSearchFiltering_filter_name_1" value="t_value1" type="checkbox" name="HardwareSearchFiltering[filter_name][]" />
<label for="HardwareSearchFiltering_filter_name_1">t_name1</label><br/>
</div>
我需要(最好)列表之间的 ID 不同,但即使我可以得到不同的名称,这也是一个开始。提前感谢您的帮助。
【问题讨论】:
标签: php activerecord yii