我在类构造函数中声明了一个属性,并通过使用“ this”声明为“ static”的方法进行访问,并且该方法不可访问。如何获取静态方法中的构造函数(类)变量?

export class Reporter {
    constructor() {
        this.jsonReports = path.join(process.cwd(), "/reports/json")

        this.cucumberReporterOptions = {
            jsonFile: targetJson,
            output: htmlReports + "/cucumber_reporter.html",
            reportSuiteAsScenarios: true,
            theme: "bootstrap",
        }
    }

    static createHTMLReport() {
        try {
            reporter.generate(this.cucumberReporterOptions);
        } catch (err) {

        }
    }
}


更新:

按照“ @CodingIntrigue”,我已经在'reporter.js'文件中做到了这一点,并在我的配置文件中将该方法称为Reporter.createHTMLReport()并按预期工作。但不确定这是否是最佳做法。

const jsonReports = path.join(process.cwd(), "/reports/json")

const cucumberReporterOptions = {
    jsonFile: targetJson,
    output: htmlReports + "/cucumber_reporter.html",
    reportSuiteAsScenarios: true,
    theme: "bootstrap",
}

export class Reporter {
    static createHTMLReport() {
        try {
            reporter.generate(cucumberReporterOptions);
        } catch (err) {

        }
    }
}

最佳答案

如果要继续使用class语法,也可以只设置jsonReportscucubmerReporterOptions静态属性:

export class Reporter {
    static createHTMLReport() {
        try {
            reporter.generate(Reporter.cucumberReporterOptions);
        } catch (err) {

        }
    }
}

Reporter.jsonReports = path.join(process.cwd(), "/reports/json")

Reporter.cucumberReporterOptions = {
    jsonFile: targetJson,
    output: htmlReports + "/cucumber_reporter.html",
    reportSuiteAsScenarios: true,
    theme: "bootstrap",
}

10-08 11:13