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javascript - 使用一个 div 获取多个值(动态)时,jssor slider 会出现问题

转载 作者:行者123 更新时间:2023-12-03 11:06:46 25 4
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这里我使用jssor slider,我需要在php中使用一个div(slider1_container)动态获取值,

这可能吗?

我尝试了下面的代码,

while($row = $result->fetch_assoc()) {
?>
<div class="category-box1">
<div id="slider1_container" class="common" style="position: relative; top: 0px; left: 0px; width:220px; height: 147px;">
<!-- Slides Container -->
<div u="slides" style="cursor: move; position: absolute; overflow: hidden; left: 0px; top: 0px; width:220px; height:147px; cursor:pointer">
<div class="imagecommon">
<img class="img-responsive center-block" alt="tour" src="<?php echo $this->getBaseUrl() ?>media/images/category/thumbs/<?php echo $row['thumbnail']?>" />
</div>
<div class="imagecommon">
<img class="img-responsive center-block" alt="tour1" src="<?php echo $this->getBaseUrl() ?>media/images/category/thumbs/<?php echo $row['image']?>" />
</div>
</div>
</div>
<div class="col-md-12 letter"><?php echo $row['title']?> </div>

</div>
<?php }
} ?>

<script>
$(window).bind("load", function() {
$(document).ready(function ($) {
var options = { $AutoPlay: true };
var jssor_slider1 = new $JssorSlider$('slider1_container', options);
});
});
</script>

注意:在此代码中获取值很好,但问题是只有一个 slider 正在工作,我需要所有 slider 才能工作,可以的话如何获得呢?提前致谢

最佳答案

不同的 slider 使用不同的名称,如果您将第一个 slider 命名为“slider1”,请将第二个 slider 命名为“slider2”。

while($row = $result->fetch_assoc()) {
?>
<div class="category-box1">
<div id="slider2_container" class="common" style="position: relative; top: 0px; left: 0px; width:220px; height: 147px;">
<!-- Slides Container -->
<div u="slides" style="cursor: move; position: absolute; overflow: hidden; left: 0px; top: 0px; width:220px; height:147px; cursor:pointer">
<div class="imagecommon">
<img class="img-responsive center-block" alt="tour" src="<?php echo $this->getBaseUrl() ?>media/images/category/thumbs/<?php echo $row['thumbnail']?>" />
</div>
<div class="imagecommon">
<img class="img-responsive center-block" alt="tour1" src="<?php echo $this->getBaseUrl() ?>media/images/category/thumbs/<?php echo $row['image']?>" />
</div>
</div>
</div>
<div class="col-md-12 letter"><?php echo $row['title']?> </div>

</div>
<?php }
} ?>

<script>
$(window).bind("load", function() {
$(document).ready(function ($) {
var options = { $AutoPlay: true };
var jssor_slider2 = new $JssorSlider$('slider2_container', options);
});
});
</script>

关于javascript - 使用一个 div 获取多个值(动态)时,jssor slider 会出现问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/27812608/

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