小编How*_*arn的帖子

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 …

我有一个带有column(bit)的表,我只想在单个查询中将列值从true更新为false,将false更新为true.我只是想交换价值观,任何帮助表示赞赏,谢谢