Question

我想创建一个json架构对象以在json架构表单中使用。这个对象看起来像这个

 this.FormValueData = {

        'discountauthorizationelementsname': "userroleid,metalid,purityid,discountmodeid,categoryid,discounttypeid,discountvalue,creditpercent,creditvalue",
        'discountauthorizationelementstable': "discountauthorization,discountauthorization,discountauthorization,discountauthorization,discountauthorization,discountauthorization,discountauthorization,discountauthorization,discountauthorization",
        'discountauthorizationelementscolmn':"userroleid,metalid,purityid,discountmodeid,categoryid,discounttypeid,discountvalue,creditpercent,creditvalue",
        'discountauthorizationelementspartabname': "discountauthorization,discountauthorization,discountauthorization,discountauthorization,discountauthorization,discountauthorization,discountauthorization,discountauthorization,discountauthorization",
        'discountauthorizationviewfieldids':"110",
        'viewfieldids':"110",
        'moduleid':"110",
        'discauthpassword':"21212121212"

    };

现在我写了以下几行

                let hidden_qp = {}

                let elementsname = this.lowermodname.concat("elementsname");
                let elementsnamevalue = "userroleid,metalid,purityid,discountmodeid,categoryid,discounttypeid,discountvalue,creditpercent,creditvalue";

                let elementstable = this.lowermodname.concat("elementstable");
                let elementstablevalue = "discountauthorization,discountauthorization,discountauthorization,discountauthorization,discountauthorization,discountauthorization,discountauthorization,discountauthorization,discountauthorization";

                let elementscolmn = this.lowermodname.concat("elementscolmn");
                let elementscolmnvalue = "userroleid,metalid,purityid,discountmodeid,categoryid,discounttypeid,discountvalue,creditpercent,creditvalue";

                let elementspartabname = this.lowermodname.concat("elementspartabname");
                let elementspartabnamevalue = "discountauthorization,discountauthorization,discountauthorization,discountauthorization,discountauthorization,discountauthorization,discountauthorization,discountauthorization,discountauthorization";

                let viewfieldids = this.lowermodname.concat("viewfieldids");
                let viewfieldidsvalue = "110";

                let moduleid = "moduleid";
                let moduleidvalue = "110"

                let discauthpassword = "discauthpassword";
                let discauthpasswordvalue = "215521512521"

                hidden_qp[elementsname] = elementsnamevalue;
                hidden_qp[elementstable] = elementstablevalue;
                hidden_qp[elementscolmn] = elementscolmnvalue;
                hidden_qp[elementspartabname] = elementspartabnamevalue;
                hidden_qp[viewfieldids] = viewfieldidsvalue;
                hidden_qp['viewfieldids'] = viewfieldidsvalue;
                hidden_qp[moduleid] = moduleidvalue;
                hidden_qp[discauthpassword] = discauthpasswordvalue;

这给了我相同的输出，但这里的键名是双引号而不是单引号，就像第一个那样，这就是在我的架构表单构建器中创建问题。如何获取像“discountauthorizationelementsname”而不是“discountauthorizationelementsname”这样的关键值。下面是我在控制台中看到的两个变量。！（https://imgur.com/a/ICmi1）

Answer 1

我知道在做json.stringify后我再次需要做json.parse。所以使用这一行

                  this.FormValueData = JSON.parse(JSON.stringify(hidden_qp));

给了我正确的输出。

如何在这样的打字稿中动态创建json架构对象？

1 个答案: