[英]Open a jQuery (modal form) dialog form in ASP.NET gridview
我真的需要你的幫助)) 我有一個綁定到 SQL 數據庫的 gridview。 如何通過單擊表中的圖像或 SQL ID 號來打開彈出窗口?
腳本
<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>
這是我的網格視圖:
<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>
好吧,如果我使用這個按鈕
<button id="create-user">Create new user</button>
然后它工作正常,我打開我的 EditTmplate,在那里我可以更改我的 SQL 數據庫中的所有數據,但我必須通過單擊表的每個元素來執行相同的彈出; 我知道它有幾行或只有 1 個命令可以更改,但是在 JS、jQuery 等中 id 非常糟糕=)我會很高興為您提供幫助。
我還是做了一些圖片來更好地解釋它=) http://ipic.su/img/img7/fs/dsadsadsada.1406036339.jpg
將打開彈出窗口的代碼放在下面的函數中 -
$(document).on("click", "[id*=idViewPopup]", function () {
});
這里的 [id*=idViewPopup] 只不過是 GridView 中對象的 ID。 因此,當我單擊 ID = idViewPopup 的圖像時,將顯示模型彈出窗口。
查看以下鏈接以獲取詳細示例 - http://www.aspsnippets.com/Articles/Display-GridView-Row-details-inside-jQuery-Dialog-Modal-Popup-in-ASPNet.aspx
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