嗨我正在使用下面的代码来构建一个字符串并复制它,但在输出时,当我粘贴它时,换行符不适用
function copyToClipboardShipto() {
var $temp = $("<input>");
$("body").append($temp);
var str1 = "@(Model.firstName)"; var str2 = " "; var str3 = "@(Model.lastName)"; var str4 = "\n";
var str5 = "@(Model.shiptoes[0].address.address1)";
var str6 = ",";
var str7 = "@(Model.shiptoes[0].address.address2)";
var str8 = "\n";
var str9 = "@(Model.shiptoes[0].address.city)"; var str10 = ","; var str11 = "@(Model.shiptoes[0].address.state)"; var str12 = "\n";
var str13 = "@(Model.shiptoes[0].address.zip)";
var str = str1 + str2 + str3 + str4 + str5 + str6 + str7 + str8 + str9 + str10 + str11 + str12 + str13;
$temp.val(str).select();
document.execCommand("copy");
$temp.remove();
}
}@H_403_3@
名字lastname223 E JACKSON AVE,city,statezip
有任何帮助
