我正在尝试将已存储的值显示为已选中。
我的问题是我正在尝试从os_hostel_facility的映射表中获取多个便利 但当我试图表明只有第一个值即将出现在控制台中,即使没有显示为选中我怎么能这样做我的代码在底部:
如果问题未达到标准,请建议编辑..!
这是我的剧本:
success: function (response) {
$("#viewhostelfacility").val(response['facility']['id_facility']);
console.log(response['facility']['id_facility']);
这里只有第一个值来了,我想显示整个数组 如选中
这是我的更新表格代码
<ul id="hostel_facility" class="dropdown-menu dropdown-select">
<?php $facility = $conn->query("SELECT * FROM os_facilities ORDER BY id_facility ASC");
while ($facilityresult = $facility->fetch_assoc()) { ?>
<li><a><input type="checkbox" name="hostel_facility[]" id="viewhostelfacility" value="<?php echo $facilityresult['id_facility']; ?>" /><?php echo $facilityresult['facility_name']; ?></a></li>
<?php } ?>
</ul>
这是我的控制器页面,我发送ersponse:
$facilitysearch= $conn->query("SELECT * From os_hostel_facility WHERE id_hostel='".$_POST['hostelId']."'") or die(mysql_error());
$viewfacility=$facilitysearch->fetch_assoc();
$response['facility'] = $viewfacility;
答案 0 :(得分:1)
您可以创建每个函数来获取数组中的所有数据:
success: function (response) {
$.each(response,function(i,e)){
$("#viewhostelfacility").val(e['facility']['id_facility']);
console.log(e['facility']['id_facility']);
}
发给你一个Json对象?如果是,您可以轻松使用数据标记:
<强>对象强>
{"os_hostel":[
{"facility":"Iron","id_facility":"1"},
{"facility":"Landry","id_facility":"2"}
]}
<强>代码强>
success: function (response) {
var hostel = response.os_hostel;
$.each(hostel,function(i,e){
$("#viewhostelfacility").val(e.facility);
console.log(e.facility);
})
<强>段强>
var obj = {"os_hostel":[
{"facility":"Iron","id_facility":"1"},
{"facility":"Landry","id_facility":"2"}
]}
var hostel = obj.os_hostel;
$.each(hostel,function(i,e){
console.log(e.facility);
})
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