js原生

1.数组  shift unshift pop push

           头删增         尾删增

//  数组  shift unshift pop push
    var str="a,b,c,d,e,f";
    var arr=str.split(",");
    console.log("arr: "+arr);
    var shift1=arr.shift();
    console.log("shift1: "+shift1);   //返回删除的数组第一个元素 a
    console.log("arr: "+arr);         // b c d e f
    var unshift1=arr.unshift('aa');
    console.log("unshift1: "+unshift1);  // 返回长度 6
    console.log("arr: "+arr);             // 数组头部添加  aa b c d e f

    var pop1=arr.pop();
    console.log("pop1: "+pop1);     //  返回删除的数组最后一个元素 f
    console.log("arr: "+arr);       //  aa b c d e

    var push1=arr.push("ff");
    console.log("push1: "+push1);  // 返回长度 6
    console.log("arr: "+arr);      // aa b c d e ff
View Code

 2.string chatAt() indexOf() split() substr()

array join() reverse()

<!DOCTYPE html>
<html>
<head lang="en">
    <meta charset="UTF-8">
    <title></title>
    <style>
        p{line-height: 40px;
            border-bottom: 1px solid green;}
    </style>
</head>

<body>

string   chatAt() indexOf() split() substr()
<script>
  var str='hello';
  console.log("chartAt(): "+str.charAt(0));  // e    // 根据索引返回 char 字符
  console.log("indexOf(): "+str.indexOf('e')); //1    //  根据字符返回 index 索引
  var arr = str.split('');
  for(let i of arr){
    console.log("split(): "+i);  // split 转成数组
  }
  var str2=str.substr(0,3);
  console.log("substr(): "+str2);  // hel   截取字符串
</script>
array shift() unshift() pop() push()   join() reverse()
<script>
  var arr2=['a','b','c','d'];
  var str3=arr2.join(''); // 用空格连接
  console.log("join(): "+str3);
  var str4=arr2.reverse().join('');
  console.log("reverse(): "+str4);
</script>

<script>
  function unique(arr) {
    let obj = {};
    let data =[];
    for(let i =0,l=arr.length;i<l;i++){
      if(!obj[arr[i]]){
        obj[arr[i]]=true;
        data.push(arr[i]);
      }
    }
    return data;
  }
  var arr4=[11,12,13,13,14,15,16,16,17];
  console.log("arr4: "+unique(arr4));
</script>
</body>
</html>
View Code

 

posted @ 2016-09-28 14:27  gyz418  阅读(165)  评论(0)    收藏  举报