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摇曳的蔷薇
var obj= {dEabc: 1,dEasdf: 2,dEjfhs: 5,bhsn: 7,oki: 9}let aa = Object.keys(obj).reduce((acc,ele)=>{return (ele.indexOf('dE') != -1) ? {...acc,...{[ele]:obj[ele]}} : acc},{});console.log(aa);
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天涯尽头无女友
使用 ,可以按以下方式完成:reduceobject1 = {dEabc: 1,dEasdf: 2,dEjfhs: 5,bhsn: 7,oki: 9}let finalObj = Object.keys(object1).filter(i=>i.includes('dE')).reduce((obj, curr)=>{ return {...obj, ...{[curr]: object1[curr]}};},{});console.log(finalObj);
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叮当猫咪
您可以使用它来解构对象,它,并使用:Object.entries()filter()Object.fromEntries()const o = { dEabc: 1, dEasdf: 2, dEjfhs: 5, bhsn: 7, oki: 9};const result = Object.fromEntries(Object.entries(o).filter(([k]) => k.includes('dE')));console.log(result);不需要正则表达式的表达式如此简单。