Javascript jqueryif语句返回相同的数字，而不考虑输入

我有一个非常简单的HTML表单和一些用于计算数字的JQuery

我遇到的问题是if语句返回与输入无关的相同数字：

    $(document).ready(function()
    {
        $('input').change(function()
        {
            var behind = escape($('#behind').val());
            var speed = parseInt($('#speed').val());
            var diff;

if (behind = "nse") { diff = '0.83'; }
if (behind = "0.5") { diff = '1.66'; }
if (behind = "0.75") { diff = '2.49'; }

var hsr = speed-diff;
$('#hsr').html(hsr);

        });
    });

如果我在框中输入nse，计算将使用diff='2.49'，而不是我编程的diff='0.83'

包含完整代码的JSFIDLE页面位于

有人能解释一下代码有什么问题吗

提前感谢。

在if语句中使用比较运算符==而不是赋值运算符=

在if语句中使用比较运算符==而不是赋值运算符=

==比较

=分配

==比较

=分配

if语句中缺少a=

if (behind == "nse") { diff = '0.83'; }
if (behind == "0.5") { diff = '1.66'; }
if (behind == "0.75") { diff = '2.49'; }

使用one=赋值而不是比较。

在if语句中缺少a=

if (behind == "nse") { diff = '0.83'; }
if (behind == "0.5") { diff = '1.66'; }
if (behind == "0.75") { diff = '2.49'; }

使用一=指定一个值，而不是比较该值

if (behind == "nse") { diff = '0.83'; }
if (behind == "0.5") { diff = '1.66'; }
if (behind == "0.75") { diff = '2.49'; }