我是jquery的新手。我做了一半的工作。但我找不到任何解决方案。
在这里你看到图像有四个复选框钉,头发,皮肤护理,按摩。我希望当用户点击沙龙,移动美容师和两个上面的指甲复选框时将勾选,当用户点击选择你的服务指甲复选框将被取消选中,所以这适用于所有。我已经这样做但问题是,当我第二次从指甲中选择项目时,不会选中复选框。它的工作只有一次。请帮我。我在这里有代码: -
HTML: -
<div class="one-row">
<?php foreach ($services as $key => $allservices) {
if ($key <= 3) {
if (!empty($data['services'])) {
if (in_array($allservices['id'], $data['services'])) {
$checked = "checked";
} else {
$checked = "";
}
} else {
$checked = "";
} ?>
<div class="div_img_part-2">
<span class="img_part_class-2"><img src="{{ asset('images/ServiceImages/'. $allservices['image'])}}">
</span>
<span class="text_part_class-2">
<p class="custom-checkbox firstpart">
<input class="firstdisable" type="checkbox" id="{{ $key }}" name="services[]"
value="{{ $allservices['id'] }}" <?= $checked; ?>/>
<label for="{{ $key }}">{{$allservices['name']}}</label>
/p>
</span>
</span>
<select name="service_type[<?php echo $allservices['name']; ?>]" class="selectpicker">
<option value="">Select Your Sevice</option>
<option value="Salon" <?php if (!empty($data['service_type'][$allservices['name']])) {
if ($data['service_type'][$allservices['name']] == "Salon") { ?> selected
<?php }
} ?> >Salon
</option>
<option value="Mobile beautician" <?php if (!empty($data['service_type'][$allservices['name']])) {
if ($data['service_type'][$allservices['name']] == "Mobile beautician") { ?> selected
<?php }
} ?> >Mobile beautician
</option>
<option value="Both" <?php if (!empty($data['service_type'][$allservices['name']])) {
if ($data['service_type'][$allservices['name']] == "Both") { ?> selected
<?php }
} ?>>Both
</option>
</select>
</div>
<?php }
} ?>
</div>
我正在使用Laravel框架 这是我的jQuery代码: -
$('.selectpicker').selectpicker('refresh');
$(".selectpicker").on('change', function() {
var value = $(this).parents(".div_img_part-2").find(".selectpicker").val();
alert(value);
if (value == "") {
$(this).parents(".div_img_part-2").find("input[type='checkbox']").attr('checked', false);
if ($("input:checked").length == 0) {
$('.disable').prop('disabled', false);
$('.selectpicker').selectpicker('refresh');
}
} else {
$(this).parents(".div_img_part-2").find("input[type='checkbox']").attr('checked', true);
}
});
答案 0 :(得分:0)
使用nearest()和prop()代替parent()和attr()。
试试这个:
$(".selectpicker").on('change', function () {
var value = $(this).val();
if (value == "") {
$(this).closest(".div_img_part-2").find("input[type='checkbox']").prop('checked', false);
if ($("input:checked").length == 0) {
$('.disable').prop('disabled', false);
$('.selectpicker').selectpicker('refresh');
}
} else {
$(this).closest(".div_img_part-2").find("input[type='checkbox']").prop('checked', true);
}
});