当我在选项中选择一个聚会名称时,我正在从表中获取acno。到目前为止,我已经很累了,我从表中获取了acno,但是它不在选项框中。
我的控制器代码:
public function get_states2()
{
$name = $this->input->post('name');
$result = $this->db->query("SELECT TAcNo FROM tipup_payment LEFT OUTER JOIN parmaster on parmaster.pcode = tipup_payment.TName WHERE PName='$name' ")->result_array();
echo json_encode($result);
}
我的查看页面代码:
<div class="col-md-6">
<div class="form-group form-group-xs">
<div class="col-lg-9">
Party Name:
<select class="form-control countries" name="City">
<option></option>
<?php foreach ($PName as $row ): ?>
<option value="<?php echo trim($row['PName']); ?>"><?php echo trim($row['PName']); ?></option><?php endforeach ?>
</select>
</div>
</div>
<div class="form-group form-group-xs">
<div class="col-lg-9">
AcNo:
<select multiple="multiple" style="height: 85px;" id="Name" class="form-control states">
<option value=""></option>
</select>
<?php echo form_error('Area', '<div class="text-danger">', '</div>'); ?>
</div>
</div>
<div id="item">
<input type="checkbox" name="item">With Details</center></div>
</div>
</div>
我的脚本代码:
<script type="text/javascript">
$(document).ready(function(){
$('.countries').change(function(){
var name = $('.countries').val();
$.ajax({
type: "POST",
url: "<?php echo base_url();?>Tieup/get_states2",
data:{name:name},
datatype: 'json',
success: function (data) {
/*get response as json */
alert(data);
var result = jQuery.parseJSON(data);
var no = result.TAcNo;
$("#Name").val(no);
/*ends */
}
});
});
});
</script>
答案 0 :(得分:1)
解决方案
您需要像这样触发更改以更新选择值
$(“#Name”)。val(no).change();
答案 1 :(得分:1)
为您的下拉菜单提供一个类或ID ur html
<select class="product">
</select>
您的jquery代码
遍历所有ur值并将其一一设置为ur选项值,最后使用.html()
var value = [{"TAcNo":"341"}]
var options = '<option value="">Select</option>';
$(value).each((index, item) => { //loop through your elements
console.log(item)
options += '<option value="'+item.TAcNo+'">'+item.TAcNo+'</option>';
});
$('.product').html(options);
希望有帮助