如何通过选择dropdownlist用json数据填充文本框
本文关键字:数据 填充 文本 json 何通过 选择 dropdownlist | 更新日期: 2023-09-27 17:58:56
`public分部类DropDownApi:System。网状物UI。页{string _URL="/api/保险";
protected void Page_Load(object sender, EventArgs e)
{
WebRequest req = WebRequest.Create(_URL);
req.Method = "GET";
// req.Headers.Add("key");
req.ContentType = "application/json; charset=utf-8";
WebResponse resp = req.GetResponse();
Stream stream = resp.GetResponseStream();
StreamReader re = new StreamReader(stream);
String json = re.ReadToEnd();
json = "{'"InsuranceDetails'":" + json + "}";
wrapper w = (wrapper)new JavaScriptSerializer().Deserialize(json, typeof(wrapper));
ddlinsurance.DataSource = w.InsuranceDetails;
ddlinsurance.DataBind();
}
public class wrapper
{
public List<insurance_master> InsuranceDetails { get; set; }
}`<asp:DropDownList ID="ddlinsurance" runat="server" AutoPostBack="false" DataTextField="policy_number"></asp:DropDownList>`
`我有一个充满json数据(policy_number)的下拉列表,现在当我从下拉列表中选择policy_number时,它必须通过匹配所选的policy_number来在文本框中显示剩余的详细信息。我的项目使用了MVC WebApi服务。有人能帮帮我吗!!!
提前感谢。
这不是答案,但我认为这会对你有所帮助。。
Html
<select id="list">
<option value="">Choose</option>
</select>
<input type="text" id="inp" />
JavaScript
// sample json data
var sampleJson = [
{ 'policy_number': 1, 'data': 'policy 1', 'display': 'one' },
{ 'policy_number': 2, 'data': 'policy 2', 'display': 'two' },
{ 'policy_number': 3, 'data': 'policy 3', 'display': 'three' }
];
//get select element using id
var select = document.getElementById('list');
//append option to select list with values from sampleJson
for (var i = 0; i < sampleJson.length; i++) {
var option = document.createElement("option");
option.value = sampleJson[i]['policy_number'];
option.text = sampleJson[i]['display'];
select.appendChild(option);
}
var input = document.getElementById('inp');
//onchange event handler for select
select.onchange = function()
{
for (var i = 0; i < sampleJson.length; i++) {
//selecetd value present in sampleJson set it into textbox
if(sampleJson[i]['policy_number'] == this.value)
{
input.value = sampleJson[i]['data'];
}
// check the value is not empty(for default Choose option)
if(this.value == '')
{
input.value = this.value;
}
}
};
这是Demo