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getElementsByTagName排除脚本标签

[英]getElementsByTagName exclude script tag

As the title suggests I want to exclude the script tag. 如标题所示,我想排除脚本标签。

Cause while using regex (at least I think that's the right name :P) 原因在使用正则表达式时(至少我认为这是正确的名称:P)

I get to a point where something 我到了某个地方

var wdc = /something/g;

is included inside the 包含在

var foundwdc = words.match(wdc).length;

So when I alert foundwdc it gives 3 "somethings" instead of the desired two inside the body 因此,当我提醒foundwdc时,它会给出3个“东西”,而不是体内所需的两个东西

var words = document.getElementsByTagName('body')[0].innerHTML;

Hope this is clear enough :D and hope the title is right :P 希望这是足够清晰的:D并希望标题是正确的:P

Use replace() to remove the script tag from string 使用replace()从字符串中删除脚本标签

 var words = document.getElementsByTagName('body')[0].innerHTML.replace(/<script[^>]*>[\\s\\S]*?<\\/script>/gi, ''); console.log(words); 
 hi hello <script> </script> 


Regex explanation here . 正则表达式的解释在这里

正则表达式可视化

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