我在这里错过了什么？ （Javascript screen.width）

Question

我不知道我在这里缺少什么。我想制作一个响应屏幕分辨率的javascript，但我无法让它工作。有什么输入吗？

<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")}

问题是如果

，你在第二个和第三个条件下有错误的条件

screen.width ＆gt; = 480＆amp;＆amp; screen.width ＆lt; = 767

screen.width ＆gt; = 768＆amp;＆amp; screen.width ＆lt; = 989

Answer 2

您对ands的简写不起作用，您必须明确地写出每个条件。你也得到了第一个条件

来自：

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

到此：

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