我有匹配的动态数组,如下所示:
[
"Rivers United - Kano Pillars",
"MFM FC - Abia Warriors",
"Lobi Stars - Enugu Rangers",
"Sunshine Stars - Dakkada",
"Ifeanyi Ubah - Kwara United",
"Enyimba - Adamawa United",
"Empoli - Torino U19",
"Bologna U19 - Ascoli",
"Lazio - Inter U19",
"Juventus U19 - Spal",
"Milan - Fiorentina U19",
"Sassuolo U19 - AS Roma U19",
"Genclerbirligi - Denizlispor",
"Goztepe - Fenerbahce",
"Kayserispor - Demir Grup Sivasspor",
"Trabzonspor U19 - Istanbul Basaksehir AS",
"Fatih Karagumruk - KASIMPASA AS",
"Portugal - Israel",
"Wydad AC Casablanca - DHJ ",
"Rsb Berkane - Athletic Youssoufia"
]当我使用基于上述数组的supertest库进行API调用时,我能够获取单个对手并保存到单个变量中,我的代码是:
const response = await api.getRequestAsync("https://sports/api/feeds/prematch/mostpopularsports/en/1/5/6/", `2021-06-06`, "", "");
const otp = api.jsonPath(response.body, `$[0].AreaMatches[0].Items..ItemName`);
let opponent1 = otp[0].split(" - ")[0];
let opponent2 = otp[0].split(" - ")[1];
let opponent20 = otp[5].split(" - ")[1];
console.log(opponent1 + " | " + opponent2 + " | " + opponent20);它将变量生成为:opponent1, opponent2...opponent20
控制台输出为:[0-0] Rivers United | Kano Pillars | Adamawa United
但这会创建手动工作来为每个单独的变量赋值。
如何将此代码传输到loop logic,以便动态地为每个对手提供变量:Rivers United, Kano Pillars...Athletic Youssoufia
和变量为:opponent1, opponent2...opponent20
发布于 2021-06-09 21:03:09
例如,你可以使用数组来存储多个值,就像你已经在使用的那样。您只需自己创建这些变量即可。
请参阅Mozilla Web文档中的JavaScript - Array
在您的示例代码中,您将访问otp,它将对手作为字符串。我们可以在otp上创建一个循环,用split()调用的结果填充数组。
下面是一个简单的示例代码,您可以根据自己的喜好进行调整:
let opponentsLeft = [];
let opponentsRight = [];
for(var i=0; i < otp.length; i++) {
var split = otp[i].split(" - ");
opponentsLeft.push(split[0]);
opponentsRight.push(split[1]);
}在更现代的JavaScript中,您可以使用forEach()
let opponentsLeft = [];
let opponentsRight = [];
otp.forEach(element => {
var split = element.split(" - ");
opponentsLeft.push(split[0]);
opponentsRight.push(split[1]);
});您可以随心所欲地处理生成的数组。同样,只需根据您的喜好进行修改。
然后,您可以在这些数组上编写一个循环,以执行您希望执行的操作:
for(var o = 0; o < opponentsLeft.length; o++) {
console.log('left', opponentsLeft[o]);
console.log('right', opponentsRight[o]);
}发布于 2021-06-09 21:58:58
嘿@Peter Krebs感谢你的努力,但我需要的解决方案是:
let opponentsA = [];
let opponentsB = [];
for (var i = 0; i < otp.length; i++) {
opponentsA[i] = otp[i].split(" - ")[0];
opponentsB[i] = otp[i].split(" - ")[1];
}
console.log(opponentsA[0]); //access the 1st opponent dynamically
console.log(opponentsA[20]); //access the 20h opponent dynamicallyhttps://stackoverflow.com/questions/67904388
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