Question

我试图用检索到的所有值填充一个数组（我处于循环中），然后删除YearsJustSelected数组中已经存在的值：

if Text == i:

例如：
循环1-> yearsJustSelected = [1、2、3]-yearsJustSelectedAll = [1、2、3]-yearsJustSelectedFiltered = []
循环2-> yearsJustSelected = [2，3，4，5]-yearsJustSelectedAll = [1，2，3，4，5]-yearsJustSelectedFiltered = [4，5]->因为[2，3]已经存在在YearsJustSelectedAll在第一个循环中

此代码：

yearsJustSelected.forEach((el:any)=>{
    yearsJustSelectedAll.push(el);
  });


  yearsJustSelectedFiltered = yearsJustSelected.filter((el: any) => !yearsJustSelectedAll.includes(el));

总是返回一个空数组。

Answer 1

var con = mysql.createConnection({
    host: "localhost",
    user: "root",
    password: "",
    database: "node"
  });
                                        ↓
con.query(`DELETE FROM NodeSQL WHERE id='${id}';`, function (err, result) {     ?????
    if (err) throw err;
});

Answer 2

您可以检查目标数组是否没有该项。然后将项目推到目标数组。

function select(source, target) {
    source.forEach(v => target.includes(v) || target.push(v));
}

var yearsJustSelectedAll = [];

select([1, 2, 3], yearsJustSelectedAll);
console.log(...yearsJustSelectedAll);

select([2, 3, 4, 5], yearsJustSelectedAll);
console.log(...yearsJustSelectedAll);

select([4, 5], yearsJustSelectedAll);
console.log(...yearsJustSelectedAll);

过滤数组返回空

2 个答案: