我有一个清单
let list = [
{
id: "247",
order_number: "21251",
tel: 13911111,
weight: "10kg"
},
{
id: "245",
order_number: "223",
tel: 31,
weight: "10kg"
},
{
id: "123",
order_number: "312312321",
tel: 3213123,
weight: "10kg"
}
];
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现在我只想删除特定的列,例如“电话”,以获取新列表。有什么优雅的方法可以做到吗?或者我必须循环整个数据才能使用 splice 方法?
我已经使用 Element-ui 编写了表格代码。我需要可排序的列,但它不起作用。我不知道是什么原因。是不是数据格式不对?
太棒了
<el-table v-loading="listLoading" :data="list">
<el-table-column type="selection" > </el-table-column>
<el-table-column label="NO" fixed width="100" sortable><template slot-scope="scope">{{ scope.row.number }}</template></el-table-column>
<el-table-column v-for="head in formThead" :key="head.value" :label="head.title" :min-width="head.celwidth ==null ? 100: head.celwidth" :sortable="head.sortable">
<template slot-scope="scope">
{{ scope.row[head.value] }}
</template>
</el-table-column>
</el-table>
formThead: [
{ value: 'dest', title: 'Dest', celwidth: 110, sortable: true }, { value: 'date', title: 'Date' }, { value: 'pcs', title: 'PCS', celwidth: 80 }],
list?[
{"number": "2", "dest": "AAA", "date": "2019-01-01", "pcs": "2"},
{"number": "3", "dest": "ABB", "date": "2019-01-02", "pcs": …Run Code Online (Sandbox Code Playgroud) 我有这个对象数组
[{
"A": "thisA",
"B": "thisB",
"C": "thisC"
}, {
"A": "thatA",
"B": "thatB",
"C": "thatC"
}]
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我试图将这种格式作为最终结果: [["thisA","thisB","thisC"], ["thatA","thisB","thatC"]]
我知道我们可以使用map()函数和特定键(A,B,C).
newarray = array.map(d => [d['A'], d['B'], d['C']])
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但我需要一个通用的功能来转移它而不使用密钥,因为数组的内容会有所不同,关键会有所不同.有什么好的解决方案吗?