下划線刪除值而不替換原始變量

Question

我有一個對象數組，我試圖在對象中查找值，然后刪除已存在對象中找到的值。 例如，

當前的JSON對象：

exist=[{"x":6811,"y":15551,"a":["aa","ab","ac"]},{"x":6811,"y":15552,"a":["bb"]},{"x":6812,"y":15551,"a":["aa","cc"]}]

我想找到值為aa的“ a”鍵

最后的結果是

exist= [{"x":6811,"y":15552,"a":["bb"]}]
found= [{"x":6811,"y":15551,"a":["aa","ab","ac"]},{"x":6812,"y":15551,"a":["aa","cc"]}]

Answer 1

您可以在不下划線的情況下進行操作：

 let exist= [ { x: 6811, y: 15551, a: ['aa', 'ab', 'ac'] }, { x: 6811, y: 15552, a: ['aa', 'bb'] }, { x: 6812, y: 15551, a: ['cc'] }, ]; const found = exist.filter(({ a }) => a.includes('aa')); exist = exist.filter(({ a }) => !a.includes('aa')); console.log('found:', found); console.log('exist:', exist); 

Answer 2

您可以將.reject 與 .contains和_.result組合

 exist=[ {"x":6811,"y":15551,"a":["aa","ab","ac"]},{"x":6811,"y":15552,"a":["bb"]},{"x":6812,"y":15551,"a":["aa","cc"]}, {"x":6812,"y":15551} ] exist = _.reject(exist, function(item){ return _.contains(_.result(item, "a"), "aa"); }) console.log(exist); 

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