[英]Ignore words less than or equal 3 javascript array
我正在建立我自己的boorkmarklet来分析当前页面中的单词,目前它工作正常,但我想过滤单词并只显示超过3个字母的单词,我是javascript的新手,但这里是我的代码:
var sWords = document.body.innerText.toLowerCase().trim().replace(/[,;.]/g,'').split(/[\s\/]+/g).sort();
// count duplicates
var iWordsCount = sWords.length;
// array of words to ignore
var ignore = ['and','the','to','a','of','for','as','i','with','it','is','on','that','this','can','in','be','has','if'];
ignore = (function(){
var o = {};
var iCount = ignore.length;
for (var i=0;i<iCount;i++){
o[ignore[i]] = true;
}
return o;
}());
谢谢你的时间!
您可以使用过滤功能:
function greaterThanThree(element){
return element.length > 3;
}
var longer_words = ['f','as','i','with','on','that','this','can','has','if'].filter(greaterThanThree);
//Will return ["with", "that", "this"]
希望这可以帮助。
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