let a = false;
let b = [{ judge: 1 }, { judge: 3 }, { judge: 0 }]
for (let i = 0; i < b.length; i++)
{
a = b[i].judge === 0;
}
if (a) return;
请教各位大佬还有跟好的方法吗
let b = [{ judge: 1 }, { judge: 3 }, { judge: 0 }]
if(b.findIndex((v, i)=>v.judge==0)) return;
那你可以这样写,通过判断数组里是否有等于0的,来return
已经很简洁了,能优化的也就循环部分了,使用es6新特性 for...of
let b = [{ judge: 1 }, { judge: 3 }, { judge: 0 }]
if (b.filter(item => item.judge === 0).length > 0) return;
这样写算更好吗