我是Bootstrap在IE11中选择用户.
当我用ajax填充数据并打开它然后它自动向下移动病房,当我选择任何值时,它没有被选中.如何删除此错误?
jQuery.ajax({
url: base_url + "UserBusinesses/ajaxState/" + countryId,
dataType: 'json',
beforeSend: function () {
$('#UserBusinessStateId').html('<option value="">Loding... </option>');
$('#UserBusinessStateId').selectpicker('refresh');
$('#UserBusinessCityId').html('<option value="">Select city</option>');
$('#UserBusinessCityId').selectpicker('refresh');
},
success: function (obj) {
var addHtml = '';
addHtml += '<option value="">Select state</option>';
$.each(obj, function (key, value) {
var selected = "";
if (selectedState == key) {
var selected = "selected='selected'";
}
addHtml += "<option value='" + key + "' " + selected + ">" + value + "</option>";
});
jQuery("#UserBusinessStateId").html(addHtml);
$('#UserBusinessStateId').selectpicker('refresh');
}
});
Run Code Online (Sandbox Code Playgroud) 我正在使用paypal rest api,我不知道如何创建client_id和clientSecret.如果您有任何想法,请告诉我...
谢谢
$array1 = array("a"=>1, "b"=>2, "c"=>3, "d"=>4, "e"=>5);
print_r(array_filter($array1, "odd"));
function odd($var)
{
// returns whether the input integer is odd//
return($var & 1);
}
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在返回值中有什么操作&符?它如何返回奇数
我在MongoDB中嵌入文档我怎么能在那里的pin码只删除一个地址:140901和更新其中的pin码为:152364
db.add_fun.insert({
"_id" : ObjectId("5a82e6dc1139b572569fa785"),
"name" : "Vikas",
"salary" : 72.0,
"address" : [
{
"address_id" : ObjectId("5a82f0e51139b572569fa78c"),
"address" : "Mullanpur ",
"pincode" : "140901",
"country" : "India"
},
{
"address_id" : ObjectId("5a82f0e51139b572569fa78d"),
"address" : "mohali ",
"pincode" : "152364",
"country" : "India"
}
]
})
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我试试这个但不行
db.add_fun.update({},
{
$pull: {
address: {
$elemMatch: {
pincode: "140901"
}
}
}
},
{
multi:true
}
)
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想要删除这个
{
"address_id" : ObjectId("5a82f0e51139b572569fa78c"),
"address" : "Mullanpur ",
"pincode" : "140901",
"country" : "India" …Run Code Online (Sandbox Code Playgroud) 我有 2 个查询,我想知道哪个查询最好
Model::select('id')->where('status','1')->first();
Model::lists('id')->where('status','1')->first();
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请告诉我