如何在函数外返回响应?

问题描述 投票:0回答:0

我正在尝试返回正文作为响应。该值正确地显示在

console.log(result);
中,但在我尝试
return result
主体时不起作用,但它没有返回任何内容。这是我的代码

const getPhotosData = async (photosUrl,cks) => {
    const curl = new Curl()
    const urld = photosUrl
    curl.setOpt(Curl.option.URL, urld)
    curl.setOpt(Curl.option.SSL_VERIFYPEER, 0)
    curl.setOpt(Curl.option.COOKIEFILE, 'cookies/'+cks)
    curl.setOpt(Curl.option.FOLLOWLOCATION, true)
    curl.on('end', function(statusCode, body, headers) {
       const result = body;
       
        // var bodii = console.info(this.getInfo(Curl.info.TOTAL_TIME));        
        this.close(); 
        console.log(result);
        return result;
    });   
    curl.perform();       
}
node.js libcurl
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