创建不适合我的动态下拉列表 [英] Create dynamic dropdown list not working for me

查看:49
本文介绍了创建不适合我的动态下拉列表的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我想在javascript函数调用中创建动态下拉列表,并在ajax的帮助下用数据库填充drodownlist



我尝试过:



i want create dynamic dropdownlist on function call in javascript and fill the drodownlist with database with help of ajax

What I have tried:

function dropdownlistbind(cn, dropdownid, tablename) {
    var obj = {};
    obj.coluname = cn;
    obj.tablename = tablename;
    $.ajax({
        url:         "webmethod.aspx/dropdownbine",
        data:        JSON.stringify(obj),
        dataType:    "json",
        type:        "POST",
        contentType: "application/json; charset=utf-8",
        success: function(r) {
            var data = new Array();
            var len = r.d.length;
            for (var i = 0; i < len; i++) {
                data[i] = r.d[i];
            }
            alert(data);
            return data;
        },
        error: function(XMLHttpRequest, textStatus, errorThrown) {
            alert('Got an Error ');
        }
    });
}

function addSelect(divname, columname, dropid, tablename) {
    var newDiv = document.createElement('div');
    var strin = "ffgggffggf";
    var html = '<select>', filter = dropdownlistbind(columname, dropid, tablename), i;
    alert();
    for (i = 0; filter.length; i++) {
        html += "<option value='" + filter[i] + "'>" + filter[i] + "</option>";
    }
    html += '</select>';
    newDiv.innerHTML = html;
    document.getElementById(divname).appendChild(newDiv);
}

推荐答案

.ajax({
url: webmethod.aspx / dropdownbine
data: JSON .stringify(obj ),
dataType: json
类型: POST
contentType: < span class =code-string> application / json; charset = utf-8,
success: function (r){
var data = new 数组 ();
var len = rdlength;
for var i = 0 ;我< LEN; i ++){
data [i] = r.d [i];
}
alert(数据);
返回数据;
},
错误: function (XMLHttpRequest,textStatus,errorThrown){
alert(' 收到错误');
}
});
}

function addSelect(divname,columname,dropid,tablename){
var newDiv = document .createElement(' DIV');
var strin = ffgggffggf;
var html = ' < select> ',filter = dropdownlistbind(columname,dropid,tablename),i;
alert();
for (i = 0 ; filter.length; i ++){
html + = < option value =' + filter [i] + '> + filter [i] + < /选项>中;
}
html + = ' < / select>';
newDiv.innerHTML = html;
document .getElementById(divname).appendChild(newDiv);
}
.ajax({ url: "webmethod.aspx/dropdownbine", data: JSON.stringify(obj), dataType: "json", type: "POST", contentType: "application/json; charset=utf-8", success: function(r) { var data = new Array(); var len = r.d.length; for (var i = 0; i < len; i++) { data[i] = r.d[i]; } alert(data); return data; }, error: function(XMLHttpRequest, textStatus, errorThrown) { alert('Got an Error '); } }); } function addSelect(divname, columname, dropid, tablename) { var newDiv = document.createElement('div'); var strin = "ffgggffggf"; var html = '<select>', filter = dropdownlistbind(columname, dropid, tablename), i; alert(); for (i = 0; filter.length; i++) { html += "<option value='" + filter[i] + "'>" + filter[i] + "</option>"; } html += '</select>'; newDiv.innerHTML = html; document.getElementById(divname).appendChild(newDiv); }


你不能从成功的回调函数返回 ajax call ,而你应该在成功回调中写下你所有的逻辑。



You cant return the value from Success callback function of ajax call, instead you should write all your logic inside success callback.

function addSelect(divname, columname, dropid, tablename) {

           var newDiv = document.createElement('div');
           var html = '<select>'
           var obj = {};
           obj.coluname = cn;
           obj.tablename = tablename;


.ajax({
url: webmethod.aspx / dropdownbine
data: JSON .stringify(obj),
dataType: json
type: POST
contentType: application / json; charset = utf-8
成功: function (r){

< span class =code-keyword> var result = rd;
for (i = 0 ; i< result.length; i ++)
html + = < option value =' + result [i] + '> + result [i] + < / option>;

html + = ' < / select>';
newDiv.innerHTML = html;
document .getElementById(divname).appendChild(newDiv);

},
错误: function (XMLHttpRequest,textStatus,errorThrown){
alert(' 得到错误');
}
});

}
.ajax({ url: "webmethod.aspx/dropdownbine", data: JSON.stringify(obj), dataType: "json", type: "POST", contentType: "application/json; charset=utf-8", success: function (r) { var result = r.d; for (i = 0; i < result.length; i++) html += "<option value='" + result[i] + "'>" + result[i] + "</option>"; html += '</select>'; newDiv.innerHTML = html; document.getElementById(divname).appendChild(newDiv); }, error: function (XMLHttpRequest, textStatus, errorThrown) { alert('Got an Error '); } }); }


这篇关于创建不适合我的动态下拉列表的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆