我有一个函数，它根据 3 个输入变量格式化字符串。我已经使用 if / else if 语句完成了它，但我相信应该有一种更简单的方法来做到这一点。我有 3 个变量，它们都可以是字符串或 null。我正在使用 javascript / 角度。我可以想象有一个类似的情况，有 5 个变量，这会增加显着的 if 数量，如何简化它？formatDistance(){let distance;let swim;let bike;let run;swim = this.sport.swim ? this.sport.swim : null;bike = this.sport.bike ? this.sport.bike : null;run = this.sport.run ? this.sport.run : null;if(swim && bike && run) { distance = swim + ' / ' + bike + ' / ' + run; }else if(swim && bike && !run) { distance = swim + ' / ' + bike; }else if(swim && !bike && run) { distance = swim + ' / ' + run; }else if(!swim && bike && run) { distance = bike + ' / ' + run; }else if(!swim && !bike && run) { distance = run; }else if(!swim && bike && !run) { distance = bike; }else if(swim && !bike && !run) { distance = swim; }else { distance = '';} return distance;}
2 回答
慕盖茨4494581
TA贡献1850条经验 获得超11个赞
您可以将过滤器与连接一起使用。序列由数组中的序列定义。将any 替换为null，可以看到它会与/ 连接在一起。
下面的片段中的示例。
[swim, bike, run].filter(item => item !== null).join('/')
swim = null
bike = null
run = 'c'
console.log('Should return c =>', [swim, bike, run].filter(item => item !== null).join('/'))
swim = 'a'
bike = null
run = 'c'
console.log('Should return a/c =>', [swim, bike, run].filter(item => item !== null).join('/'))
swim = null
bike = 'b'
run = 'c'
console.log('Should return b/c =>', [swim, bike, run].filter(item => item !== null).join('/'))
swim = 'a'
bike = 'b'
run = 'c'
console.log('Should return a/b/c =>', [swim, bike, run].filter(item => item !== null).join('/'))
杨__羊羊
TA贡献1943条经验 获得超7个赞
尝试使用这个
formatDistance(){
let distance = "";
let swim;
let bike;
let run;
swim = this.sport.swim ? this.sport.swim : null;
bike = this.sport.bike ? this.sport.bike : null;
run = this.sport.run ? this.sport.run : null;
if(swim != null)
distance = distance + swim + ' / ';
if(bike != null)
distance = distance + bike + ' / ';
if(run != null)
distance = distance + run + ' / ';
distance = distance.substring(0, distance.length - 1);
return distance;
}
我所做的是根据可用性附加值并删除最终的值/
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