Question

我试图在JSON字符串中转义换行符并失败。变量为newissuedetailsvalue。

 $(newissueshtml).filter('.newissue').each(function () {
            var newissuehtml = $(this).html();
            var issueguid = $(newissuehtml).find(".issueguid").val();
            var newissuetypevalue = $(newissuehtml).find('.newissuetypeinput').val();
            newissuesjsonstring += '{"newissuetype":"' + newissuetypevalue + '",';
            newissuesjsonstring += '"newissueguid":"' + issueguid + '",';
            var newissuedetailsvalue = $(newissuehtml).find('.newissuedetailsinput').val();
            newissuedetailsvalue = newissuedetailsvalue.replace(/"/g, '\'');
            newissuedetailsvalue = newissuedetailsvalue.replace(/\\/g, ' ');
            newissuedetailsvalue = newissuedetailsvalue.replace(/[\\n]/g, 'x')
                                                       .replace(/[\\r]/g, 'x')
                                                       .replace(/\t/g, 'x');
            newissuesjsonstring += '"newissuedetails":"' + newissuedetailsvalue + '",';
            var newissueid = $(newissuehtml).find(".newissueid").val();
            newissuesjsonstring += '"newissueid":"' + newissueid + '",';
            var selectedresolutiondetail = $(newissuehtml).find(".resdetails").val();
            newissuesjsonstring += '"selectedresolutiondetail":"' + selectedresolutiondetail + '",';
            var resdetailslong = $(newissuehtml).find(".resdetlong").val();
            newissuesjsonstring += '"resolutiondetailslong":"' + resdetailslong + '",';
            var rescost = $(newissuehtml).find(".rescost").val();
            newissuesjsonstring += '"resolutioncost":"' + rescost + '",';
            var newissuesourceinput = $(newissuehtml).find(".newissuesourceinput").val();
            newissuesjsonstring += '"newissuesource":"' + newissuesourceinput + '",';
            var resolutiondate = $(newissuehtml).find(".resdate").val();
            newissuesjsonstring +=  '"resolutiondate":"' + resolutiondate + '"},';
        });
        newissuesjsonstring = newissuesjsonstring.slice(0, -1);
        newissuesjsonstring += ']';

Answer 1

为什么要手动构建JSON字符串？为什么不使用JSON.stringify并让库句柄为你逃避？

newissuesjsonstring = JSON.stringify({
    newissuetype: $(newissuehtml).find('.newissuetypeinput').val(),
    newissueguid: $(newissuehtml).find(".issueguid").val(),
    newissuedetails: $(newissuehtml).find('.newissuedetailsinput').val(), //no need for manual escaping!
    /*...*/
});

JS在JSON中转义换行符

1 个答案: