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Javascript 节点js变量在外部函数中不起作用_Javascript_Node.js_Axios - Fatal编程技术网

Javascript 节点js变量在外部函数中不起作用

Javascript 节点js变量在外部函数中不起作用,javascript,node.js,axios,Javascript,Node.js,Axios,有人能告诉我如何在axios之外使用var dataTitle吗 express = require('express') / const bodyParser = require('body-parser') const axios = require('axios'); var randomInt = require('random-int'); const URL = 'url'; var randomNumber = randomInt(11) axios.get(URL + rand

有人能告诉我如何在axios之外使用var dataTitle吗

express = require('express') / 
const bodyParser = require('body-parser')
const axios = require('axios');
var randomInt = require('random-int');
const URL = 'url';
var randomNumber = randomInt(11)
axios.get(URL + randomNumber)
.then(function (response) {
var dataTitle = response.data.question;
console.log(dataTitle)
})
.catch(function (error) {
console.log(error);
});

您的问题似乎是关于JavaScript中的变量范围。 不能在函数外部使用在函数内部声明的变量。因此,您必须在axios之前声明它,并在axios中使用它,如下所示:

express = require('express');
const bodyParser = require('body-parser');
const axios = require('axios');
var randomInt = require('random-int');
const URL = 'url';
var randomNumber = randomInt(11);

var dataTitle;

axios.get(URL + randomNumber)
.then(function (response) {
     dataTitle = response.data.question;
     console.log(dataTitle)
})
.catch(function (error) {
     console.log(error);
});

请参阅,以及其他几十篇类似的文章。当我在axios之外尝试console.log(dataTitle)时,可能会重复我未定义的内容