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  • 根据字数截取字符串,不能截断url

    今天收到个需求:

    1,给一个文字,对输出的文字进行截取,保留400个字符

    2,截取内容最后如果是url,保留完整url地址

    3,添加省略号......

    ----

    其中对url的保留比较麻烦,尤其是有两个相同url时不能采用indexOf获取其字符位置。

    处理结果:

    相关代码:

    String.prototype.sizeAt = function(){
    var nLen = 0;
    for(var i = 0, end = this.length; i<end; i++){
    nLen += this.charCodeAt(i)>128?2:1;
    }
    return nLen;
    };
    String.prototype.cutStr = function(n, sCut){
    if(this.sizeAt() <= n){
    return this;
    }
    sCut = sCut || "";
    var max = n-sCut.sizeAt();
    var nLen = 0;
    var s = this;
    for(var i =0,end = this.length;i<end;i++){
    nLen += this.charCodeAt(i)>128?2:1;
    if(nLen>max){
    s = this.slice(0,i);
    s += sCut;
    break;
    }
    }
    return s.toString();
    };
    String.prototype.cutStrButUrl = function(n, sCut){
    if(this.sizeAt() <=n){
    return this.toString();
    }
    sCut = sCut || "";
    var max = n-sCut.sizeAt();
    var s = this;
    //查找所有包含的url
    var aUrl = s.match(/https?:\/\/[a-zA-Z0-9]+(\.[a-zA-Z0-9]+)+([-_A-Z0-9a-z\$\.\+\!\*\/,:;@&=\?\~\#\%]*)*/gi);
    //当第max个字符刚好在url之间时,bCut会被设置为flase;
    var bCut = true;
    if(aUrl){
    //对每个url进行判断
    for(var i=0, endI = aUrl.length;i<endI;i++){
    var sUrl = aUrl[i];
    //可能出现两个相同url的情况
    var aP = s.split(sUrl);
    var nCurr = 0;
    var nLenURL = sUrl.sizeAt();
    var sResult = "";
    for(j = 0, endJ = aP.length; j<endJ; j++){
    nCurr +=aP[j].sizeAt();
    sResult +=aP[j];
    sResult += sUrl;
    //当前字数相加少于max但添加url超过max:即会截到url
    if(nCurr < max && nCurr + nLenURL>max){
    s = sResult + sCut;
    bCut = false;
    break;
    }
    nCurr += nLenURL;
    }
    if(bCut === false){
    break;
    }
    };
    }
    if(bCut){
    s = s.cutStr(n, sCut);
    }
    return s.toString();
    };
    console.log('正常截取20个字符'.cutStrButUrl(20,'......'));
    console.log('正常截取20个字符,但我超了'.cutStrButUrl(20,'......'));
    console.log('有url的字符串http://www.baidu.com你能截取到吗?'.cutStrButUrl(20,'......'));
    console.log('http://www.baidu.com有两个相同url的字符串http://www.baidu.com好吗?'.cutStrButUrl(51, '......'));



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  • 原文地址:https://www.cnblogs.com/Gukw/p/2319774.html
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