list = list.filter(({ field1, field2, field3, field4, field5, field6, field7}) => {
let found = false;
if (field1.includes(this.state.filterString) ||
field2.includes(this.state.filterString) ||
field3.includes(this.state.filterString) ||
field4.includes(this.state.filterString) ||
field5.includes(this.state.filterString) ||
field6.includes(this.state.filterString) ||
field7.includes(this.state.filterString) ||
) {
found = true;
}
return found
});
上面的代码看起来很重复,我想知道是否可以使其更简洁或者不违反 DRY 原则?
最佳答案
使用属性数组后跟一个 .some
测试:
const itemProperties = ['field1', 'field2', 'field3', 'field4', 'field5', 'field6', 'field7'];
const { filterString } = this.state;
const filteredList = list.filter(
item => itemProperties.some(prop => item[prop].includes(filterString))
);
关于javascript - 干燥过滤?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/53908758/