我需要你的帮助))我有一个与sql数据库绑定的gridview。如何通过单击表格中的immage或SQL-ID号码来打开弹出窗口。 脚本v
<script>
$(function () {
var dialog, form,
// From http://www.whatwg.org/specs/web-apps/current-work/multipage/states-of-the-type-attribute.html#e-mail-state-%28type=email%29
emailRegex = /^[a-zA-Z0-9.!#$%&'*+\/=?^_`{|}~-]+@[a-zA-Z0-9](?:[a-zA-Z0-9-]{0,61}[a-zA-Z0-9])?(?:\.[a-zA-Z0-9](?:[a-zA-Z0-9-]{0,61}[a-zA-Z0-9])?)*$/,
name = $("#name"),
email = $("#email"),
password = $("#password"),
allFields = $([]).add(name).add(email).add(password),
tips = $(".validateTips");
function updateTips(t) {
tips
.text(t)
.addClass("ui-state-highlight");
setTimeout(function () {
tips.removeClass("ui-state-highlight", 1500);
}, 500);
}
function checkLength(o, n, min, max) {
if (o.val().length > max || o.val().length < min) {
o.addClass("ui-state-error");
updateTips("Length of " + n + " must be between " +
min + " and " + max + ".");
return false;
} else {
return true;
}
}
function checkRegexp(o, regexp, n) {
if (!(regexp.test(o.val()))) {
o.addClass("ui-state-error");
updateTips(n);
return false;
} else {
return true;
}
}
function addUser() {
var valid = true;
allFields.removeClass("ui-state-error");
valid = valid && checkLength(name, "username", 3, 16);
valid = valid && checkLength(email, "email", 6, 80);
valid = valid && checkLength(password, "password", 5, 16);
valid = valid && checkRegexp(name, /^[a-z]([0-9a-z_\s])+$/i, "Username may consist of a-z, 0-9, underscores, spaces and must begin with a letter.");
valid = valid && checkRegexp(email, emailRegex, "eg. ui@jquery.com");
valid = valid && checkRegexp(password, /^([0-9a-zA-Z])+$/, "Password field only allow : a-z 0-9");
if (valid) {
$("#users tbody").append("<tr>" +
"<td>" + name.val() + "</td>" +
"<td>" + email.val() + "</td>" +
"<td>" + password.val() + "</td>" +
"</tr>");
dialog.dialog("close");
}
return valid;
}
dialog = $("#dialog-form").dialog({
autoOpen: false,
height: 400,
width: 350,
modal: true,
buttons: {
"Create an account": addUser,
Cancel: function () {
dialog.dialog("close");
}
},
close: function () {
form[0].reset();
allFields.removeClass("ui-state-error");
}
});
form = dialog.find("form").on("submit", function (event) {
event.preventDefault();
addUser();
});
$("#create-user").button().on("click", function () {
dialog.dialog("open");
});
});
</script>
和我的gridview
<asp:GridView id="Objektkatalog" runat="server" OnClick = "Page_Changed" AllowPaging="true" PageSize="4" DataSourceID="SqlDataSource2" DataKeyNames="ObjektID" AutoGenerateColumns="false" OnRowDataBound="GridView1_RowDataBound">
<columns>
<asp:TemplateField>
<ItemTemplate>
Nr.
<%#Eval("ObjektID")%>
</ItemTemplate>
</asp:TemplateField>
<asp:TemplateField>
<ItemTemplate>
<asp:ImageButton ID="ImageButton1" runat="server" Height="55px" Width="55px" ImageUrl='<%# Eval("Picture","~/Objekte/{0}") %>' /> //it comes to a second itemTemplate, dono why i cant add this in code section...
</ItemTemplate>
</asp:TemplateField>
</columns>
</asp:GridView>
并且,如果我使用这个按钮<button id="create-user">Create new user</button>
然后它工作正常我打开我的EditTmplate,我可以在我的SQL数据库中更改所有数据,但我必须通过点击每个弹出相同的弹出窗口我的桌子的元素,我知道它的几行或只有1个命令要改变,但在JS,Qjerry等身份非常糟糕=)我将非常高兴你的帮助。
我还是做了一些图片来更好地解释它=)http://ipic.su/img/img7/fs/dsadsadsada.1406036339.jpg
答案 0 :(得分:0)
将打开弹出窗口的代码放在下面的函数中 -
$(document).on("click", "[id*=idViewPopup]", function () {
});
这里[id * = idViewPopup]只是GridView中对象的ID。因此,当我点击ID = idViewPopup的图像时,将显示模型弹出窗口。
检查以下链接以获取详细示例 - http://www.aspsnippets.com/Articles/Display-GridView-Row-details-inside-jQuery-Dialog-Modal-Popup-in-ASPNet.aspx