Javascript 从serevr侧显示弹出窗口
单击链接按钮时显示弹出错误我不知道如何解决此问题 弹出窗口 按钮点击按钮打开弹出窗口Javascript 从serevr侧显示弹出窗口,javascript,jquery,asp.net,Javascript,Jquery,Asp.net,单击链接按钮时显示弹出错误我不知道如何解决此问题 弹出窗口 按钮点击按钮打开弹出窗口 <asp:Button OnClick="Unnamed_Click" ID="popup" runat="server" /> 服务器端的按钮事件 protected void Unnamed_Click(object sender, EventArgs e) { string dQuery = ""; string DCode = Convert.ToString(Sessi
<asp:Button OnClick="Unnamed_Click" ID="popup" runat="server" />
服务器端的按钮事件
protected void Unnamed_Click(object sender, EventArgs e)
{
string dQuery = "";
string DCode = Convert.ToString(Session["DCode"]);
dQuery = @"SELECT * FROM TABLE";
DataTable dtButton = obj.GetDataTable(dQuery);
rptButtonMaster.DataSource = dtButton;
rptButtonMaster.DataBind();
// Page.ClientScript.RegisterStartupScript(this.GetType(), "myScript", "OpenPopUp();", true);
ScriptManager.RegisterStartupScript(this, Page.GetType(), "success", "<script> OpenPopUp(); </script>", false);
}
单击按钮时调用Javascript函数
<script>
function OpenPopUp() {
debugger;
$('#myModal').modal('show');
}
</script>
在加载时获取所有时间相同的错误
错误:模型不是函数
我遗漏了什么???将您的呼叫包装成一份准备好文档的声明
ScriptManager.RegisterStartupScript(this, Page.GetType(), "success", "<script>$(function(){OpenPopUp();})</script>", false);
确保在引导和jquery库之后注册/加载脚本在哪里调用OpenPopUp?从服务器端--ScriptManager.RegisterStartupScriptthis,Page.GetType,success,OpenPopUp,错误的见下面我的答案
ScriptManager.RegisterStartupScript(this, Page.GetType(), "success", "<script>$(function(){OpenPopUp();})</script>", false);