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在下面的代码中有两个函数: OppSelect() 函数正在调用另一个函数,即 exOpp()。现在,如何将 OppSelect() 的“结果”值放入“exOpp()”函数并附加到 url?请帮忙。

JS:

function exOpp() {

    alert("inside");

    document.forms[0].action = "/Mgmt/ExportOpp?loadData=xls&customerprofile=" + $customerIDH + "&role=" + $roleIDH + "&targetDate=" + $targetDateH + "&actualDate=" + $actualDateH;
    document.forms[0].submit();
}

function OppSelect() {

    var menu = document.getElementById('selectedServersID').length;
    alert(menu);
    var result = "";
    for (var i = 0; i < menu; i++) {

        var selectedcol = document.getElementById('selectedServersID')[i].value;
        //var selectedcol = document.getElementById('selectedServersID')[i].innerHTML;
        // alert(selectedcol);

        if (i > 0) {
            result += ',';
        }
        result += selectedcol;
    }

    alert(result);
    exOpp();
}
4

1 回答 1

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嗨,你可以将它作为参数传递给exOpp(result);

    function exOpp(data) {

var urldata = data;

              document.forms[0].action = "/Mgmt/ExportOpp?loadData=xls&customerprofile="+$customerIDH+"&role="+$roleIDH+"&targetDate="+$targetDateH+"&actualDate="+$actualDateH;
              document.forms[0].submit();

        }

现在根据需要将 urldata 添加到 urstring..

于 2013-10-08T10:22:07.370 回答