Question

我有以下javascript来遍历记录数组，并为每个字段提醒数组中找到的匹配数：

    mymusic=[{title:"a",artist:"b",artwork:"c",tracks:[{tracktitle:"d",trackmp3:"e"}]}];
tracksArray=[];
trackTitles=[];
var albumScore=0;
var artistScore=0;
var tracksScore=0;
stringToSearchFor="d";
for(i=0;i<mymusic.length;i++){
    if((mymusic[i].title).match(stringToSearchFor))
        albumScore+=1;
    }
if(albumScore!=0)
    alert(albumScore+" match(es) found in Albums");
else
    alert("No matches found in Albums");
for(d=0;d<mymusic.length;d++){
    if((mymusic[d].artist).match(stringToSearchFor))
        artistScore+=1;
    }
if(artistScore!=0)
    alert(artistScore+" match(es) found in Artists");
else
    alert("No matches found in Artists");
for(f=0;f<mymusic.length;f++){
    tracksArray[f]=mymusic[f].tracks;
    for(g=0;g<tracksArray;g++){
        trackTitles[g]=tracksArray[g].tracktitle;
        }
    for(h=0;h<trackTitles.length;h++){
        if(trackTitles(h).match(stringToSearchFor))
            {
            tracksScore+=1;
            }
        }
    }
if(tracksScore!=0)
    alert(tracksScore+" match(es) found in Tracks");
else
    alert("No matches found in Tracks");

适用于“标题”和“艺术家”记录，但始终警告“轨迹”记录“找不到匹配”，即使有匹配也是如此。我想问题是通过trackTitles数组的嵌套for循环，但我看不到我可以改变它使它工作。有任何想法吗？感谢

Answer 1

if(trackTitles(h)

你正在调用一个数组。应该是方括号。

你可以将数组处理内容分解为可重用的函数，以提高可读性并减少这些杂散变量的数量。

由于已经有程序方法的答案，这里有一个基于功能类数组处理的额外乐趣（*）：

function countItemsContaining(seq, prop, str) {
    return seq.map(itemGetter(prop)).filter(function(s) {
        return s.indexOf(str)!==-1;
    }).length;
}

function itemGetter(prop) {
    return function(o) {
        return o[prop];
    };
}


mymusic= [{title:"a",artist:"b",artwork:"c",tracks:[{tracktitle:"d",trackmp3:"e"}]}];
needle= 'd';

var titleScore= countItemsContaining(mymusic, 'title', needle);
var artistScore= countItemsContaining(mymusic, 'artist', needle);

// Calling concat is a JavaScript idiom to combine a load of lists into one
//
var mytracks= [].concat.apply([], mymusic.map(itemGetter('tracks')));
var tracksScore= countItemsContaining(mytracks, 'tracktitle', needle);

array.map和array.filter在ECMAScript第五版中已标准化，但尚未在IE中提供，因此为了兼容性，您可以像这样定义它们：

if (!('map' in Array.prototype)) {
    Array.prototype.map= function(f, that) {
        var a= new Array(this.length);
        for (var i= 0; i<this.length; i++) if (i in this)
            a[i]= f.call(that, this[i], i, this);
        return a;
    };
}

if (!('filter' in Array.prototype)) {
    Array.prototype.filter= function(f, that) {
        var a= [];
        for (var i= 0; i<this.length; i++) if (i in this)
            if (f.call(that, this[i], i, this))
                a.push(this[i]);
        return a;
    };
}

（*：答案中包含的实际乐趣数量可能有限）

Answer 2

查看名为underscore.js的库。它是为这种东西而制作的。这些任务通常可以归结为一行或两行易于阅读的代码。

它在可用时使用本机方法，填充缺少的位（取决于浏览器）并且是可链接的。它甚至使内置数组方法可链接。

Answer 3

请改为尝试：

var tracksScore=0;
stringToSearchFor="d";
for(var f=0;f<mymusic.length;f++){
    var tracksArray=mymusic[f].tracks;
    for(var g=0;g<tracksArray.length;g++) {
        var tracktitle=tracksArray[g].tracktitle;
        if(tracktitle.match(stringToSearchFor))
        {
                tracksScore+=1;
        }
    }
}
if(tracksScore!=0)
    alert(tracksScore+" match(es) found in Tracks");
else
    alert("No matches found in Tracks");

Answer 4

您有许多基本错误，这些错误最终源于拥有太多变量。这是您重构的代码： -

mymusic=[{title:"a",artist:"b",artwork:"c",tracks:[{tracktitle:"d",trackmp3:"e"}]}];
var albumScore=0;
var artistScore=0;
var tracksScore=0;
stringToSearchFor="d";

for (var i=0; i < mymusic.length; i++)
{
    if( mymusic[i].title.match(stringToSearchFor))
        albumScore += 1;

    if( mymusic[i].artist.match(stringToSearchFor))
        artistScore += 1;

    for (var j = 0; j < mymusic[i].tracks.length; j++)
    {
        if (mymusic[i].tracks[j].tracktitle.match(stringToSearchFor))
            tracksScore += 1
    }
}

if (albumScore != 0)
    alert(albumScore + " match(es) found in Albums");
else
    alert("No matches found in Albums");

if (artistScore != 0)
    alert(artistScore + " match(es) found in Artists");
else
    alert("No matches found in Artists");

if (tracksScore != 0)
    alert(tracksScore+" match(es) found in Tracks");
else
    alert("No matches found in Tracks");

Answer 5

AnthonyWJones和bobince所说的话（虽然我需要花一些时间阅读bobince的回答）。

另一种解决方案：当我看到数据结构时，我想到了“递归！”并且认为看看我是否能够提出适用于任何（未知）深度级别的任何大小数据结构的解决方案会很有趣。

我不经常编码，所以下面可能有不好的做法，但它的工作原理:)。让我知道你的想法。

myMusic=[{title:"a",artist:"b",artwork:"c",year:"d",tracks:[{tracktitle:"d",trackmp3:"e"}]}];

function find_match(dataObj,stringToSearchFor,resultObj){

  resultObj = (resultObj)?resultObj:{}; //init resultObj

  for (dataKey in dataObj){ //loop through dataObj
    if (typeof(dataObj[dataKey]) == "object"){ //if property is array/object, call find_match on property
     resultObj = find_match(dataObj[dataKey],stringToSearchFor,resultObj); 
    }else if (dataObj[dataKey].match(stringToSearchFor)){ //else see if search term matches    
      resultObj[dataKey] = (resultObj[dataKey] )?resultObj[dataKey] +=1:1; //add result to resultObj, init key if not yet found, use dataObj key as resultObj key 
    }
  }

  return resultObj; //return resultObj up the chain

}

results = find_match(myMusic,"d");

alertString = "";

for (resultKey in results){ //loop  through results and construct alert msg.
  alertString += results[resultKey] + " match(es) found in " + resultKey + "\n";
}

alert(alertString );

获取javascript来搜索数组中的数组

5 个答案: