.splice() 正在从数组中删除 2 个对象而不是 1 个答案

【问题标题】：.splice() is removing 2 objects from array instead of 1.splice() 正在从数组中删除 2 个对象而不是 1 个
【发布时间】：2018-03-12 01:08:17
【问题描述】：

例如，当输入“John Smith”时，slice 会删除前两个员工姓名，而不是只删除 John 的姓名。知道为什么会这样吗？

let removeEmployee = '';
let employees = [
  {
    name: 'John Smith'
  }, {
    name: 'Jackie Jackson'
  }, {
    name: 'Chris Jones'
  }, {
    name: 'Amanda Cullen'
  }, {
    name: 'Jeremy Goodwin'
  }, ]

removeEmployee = prompt('Enter the name of the employee to be removed:');

function employeeExists(employee) {
  return employees.some(function(el) {
    return el.name === employee;
  });
}

if (employeeExists(removeEmployee)) {
  employees.forEach(function(employee, index, object) {
    if (employee.name === removeEmployee) {
      object.splice(index, 1);
    } else {
      console.log(employee.name);
    }
  });
} else {
  console.log('That employee does not exist, please try again.');
}

【问题讨论】：

标签： javascript arrays loops foreach slice

【解决方案1】：

您可以使用filter 而不是forEach 使事情变得更简单：

if (employeeExists(removeEmployee)) {   
    employees = employees.filter(e => e.name !== removeEmployee);
}

如果你仍然想使用splice，你可以使用findIndex：

let employees = [ {name: 'John Smith'}, {name: 'Jackie Jackson'}, {name: 'Chris Jones'}, {name: 'Amanda Cullen'}, {name: 'Jeremy Goodwin'} ];
var removeEmployee = 'Chris Jones';
var index = employees.findIndex(e => e.name === removeEmployee);
employees.splice(index, 1);
console.log(employees);

【讨论】：

【解决方案2】：

杰基杰克逊仍在名单中

您可以像这样遍历列表：

对于第一次迭代，您位于索引 0。然后删除索引0 (John Smith)。此时 Jackie Jackson 是新的索引0 但迭代会跳转到下一个元素（索引1），Chris Jones 是什么。

新的索引 0 永远不会注销到控制台！但他还在名单上！

【讨论】：

【解决方案3】：

您可以使用findIndex 查找名称与提示输入相同的对象的索引。使用该索引，您可以使用 splice 从员工数组中删除项目

let removeEmployee = '';
let employees = [{
  name: 'John Smith'
}, {
  name: 'Jackie Jackson'
}, {
  name: 'Chris Jones'
}, {
  name: 'Amanda Cullen'
}, {
  name: 'Jeremy Goodwin'
}, ]


removeEmployee = prompt('Enter the name of the employee to be removed:');

function employeeExists(employee) {
  let ifEmployee = employees.findIndex(function(el) {
    return el.name === employee.trim();
  })
  return ifEmployee;
}
var employeIndex = employeeExists(removeEmployee);
if (employeIndex !== -1) {
  employees.splice(employeIndex, 1)
} else {
  console.log('That employee does not exist, please try again.');
}
console.log(employees)

【讨论】：

【解决方案4】：

forEach 中不需要第三个参数。只需简单的splice employees 数组如下。

let removeEmployee = '';
let employees = [{
  name: 'John Smith'
}, {
  name: 'Jackie Jackson'
}, {
  name: 'Chris Jones'
}, {
  name: 'Amanda Cullen'
}, {
  name: 'Jeremy Goodwin'
}, ]

// let letters = ['a', 'd', 'c']
removeEmployee = prompt('Enter the name of the employee to be removed:');

function employeeExists(employee) {
  return employees.some(function(el) {
    return el.name === employee;
  });
}

if (employeeExists(removeEmployee)) {
  employees.forEach(function(employee, index) {

    if (employee.name === removeEmployee) {
      employees.splice(index, 1);
    } else {
      console.log(employee.name);
    }
  });
} else {
  console.log('That employee does not exist, please try again.');
}

console.log(employees)

【讨论】：

【解决方案5】：

只需使用 Array#filter 函数即可删除项目。您无需先检查 (iteration)，然后使用 forEach(iteration) 循环。您有 2 次迭代。您只能在一次迭代中执行此操作。

let employees = [
  { name: 'John Smith', }, 
  { name: 'Jackie Jackson' }, 
  { name: 'Chris Jones' }, 
  { name: 'Amanda Cullen' }, 
  { name: 'Jeremy Goodwin'} 
];

let name = prompt('Enter the name of the employee to be removed:');

employees = employees.filter(emp => emp.name.localeCompare(name));

console.log(employees);

【讨论】：