我在这里错过了什么?(Javascript screen.width)

Question

我不知道我在这里错过了什么。我想做一个响应屏幕分辨率的javascript，但我无法让它工作。任何输入？

<SCRIPT LANGUAGE="javascript">
if (screen.width <= 479) 
{document.write("Less tgab 479");}

if (screen.width <= 480 && <= 767)
{document.write("Between 480 and 767")}

if (screen.width <= 768 && <= 989)
{document.write("Between 768 and 989")}

if (screen.width >= 990)
{document.write("Above 990")}
</SCRIPT>

感谢所有帮助。

Answer 1

这应该有效

if (screen.width <= 479) {document.write("Less tgab 479");}
if (screen.width >= 480 && screen.width <= 767) {document.write("Between 480 and 767")}
if (screen.width >= 768 && screen.width <= 989) {document.write("Between 768 and 989")}
if (screen.width >= 990) {document.write("Above 990")}

问题是你在第二个和第三个有错误的条件，如果

screen.width >= 480 && screen.width<= 767

screen.width >= 768 && screen.width<= 989

Answer 2

你的 ands 简写不起作用，你必须明确地写出每个条件。你也得到了第一个条件倒过来

由此：

if (screen.width <= 480 && <= 767)
{document.write("Between 480 and 767")}

对此：

if (screen.width >= 480 && screen.width < 768)
{document.write("Between 480 and 767")}