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我有正常工作的情況如下:Jquery的Json的工作不正常
$('<li><a id=' + loc.locId + ' href="/DataEntry" rel="external">' + loc.locName + '</a></li>').appendTo("#btnList");
$("#btnList a").click(function() {
alert(siteName);
localStorage["dataEId"] = $(this).attr("id");
localStorage["dataESiteName"] = siteName;
localStorage["dataESysName"] = sysName;
localStorage["dataELocName"] = $(this).text();
}
當我有以下的,我甚至不能去點擊時顯示一個警告消息:
$.getJSON('/Home/GetLocType', { "locId": loc.locId }, function (result) {
var str = JSON.stringify(result);
if (str == '1') {
$('<li><a id=' + loc.locId + ' href="/DataEntry" rel="external">' + loc.locName + '</a></li>').appendTo("#btnList");
} else {
$('<li><a id=' + loc.locId + ' href="/DataEntry/PotableForm" rel="external">' + loc.locName + '</a></li>').appendTo("#btnList");
}
$("#btnList").listview('refresh');
});
$("#btnList a").click(function() {
alert(siteName);
localStorage["dataEId"] = $(this).attr("id");
localStorage["dataESiteName"] = siteName;
localStorage["dataESysName"] = sysName;
localStorage["dataELocName"] = $(this).text();
}
請注意有什麼區別。我需要使用基於價值的Json,我需要去兩個超鏈接中的任何一個。