.splice（）從數組中刪除2個對象而不是1

Question

例如，當輸入'John Smith'時，slice會刪除前兩個員工姓名，而不是僅刪除John的員工姓名。 知道為什么會這樣嗎？

 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.'); } 

Answer 1

您可以使用filter而不是forEach使事情變得更簡單：

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

如果您仍想使用splice ，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); 

Answer 2

傑基傑克遜仍在名單中

你像這樣遍歷列表：

1
2
3
4
5

對於第一次iterration你在索引0 。 然后刪除索引0 （John Smith）。 此時 Jackie Jackson是新的指數0 ，但iterration跳轉到下一個元素（索引1 ），什么是Chris Jones 。

新索引0永遠不會登出到控制台！ 但他仍然在名單中！

Answer 3

只需使用Array＃filter功能即可刪除這些項目。 您不需要先檢查（ 迭代 ）然后使用forEach循環（ 迭代 ）。 你有2次迭代。 您只能在一次迭代中執行此操作。

 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); 

Answer 4

您可以使用findIndex查找對象的索引，其中name與提示的輸入相同。 使用該索引，您可以使用splice從employees數組中刪除項目

 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) 

Answer 5

你不需要forEach第三個第三個參數。 只需簡單地splice employees數組，如下所示。

 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)