Question

我知道在某个地方必须有一个库，但我要做的是传入一个字符串数组和一个搜索字符串，让它按照它与搜索字符串的相关性对数组进行排序。

我一直很难搞清楚要解决这个问题，有人指出我的方向正确吗？

最好是在php或其他服务器端语言

Answer 1

我真的不知道'revelant'是什么意思..

但是，如果要根据搜索字符串找到最佳字符串，可以使用Levenshtein算法。它计算两个字符串之间的“距离”。

此处有更多信息：http://php.net/manual/fr/function.levenshtein.php

Answer 2

这有点棘手，我无法用英语解释，所以我只需要向您展示一个有效的代码。如果您有疑问，可以提问。

<!DOCTYPE html>
   <head>
      <title>Search Array</title>
      <script>
      //so let's set an array of values we will be searching

    function searchArray(dis) {

        var bigContent = new Array('Jesus loves you','Angle','God in Heaven','Bala','Overcomer','Be born again','Adesuwa','Heaven is real','James','Apple','John the baptist');//this is the array that serves as your database

        var searchi = dis.value, result = Array();
        searchi = searchi.toLowerCase();

        //order the array alphabetically
        bigContent = bigContent.sort();

        var content;

        if(searchi !='') {
            //Define the arrays for initial pre-storage
            var keys = new Array(), contentArray = new Array();

            //Loop through the content array to search for all occurence of string
            for(var i=0;i<bigContent.length;i++) {
                content = bigContent[i].toLowerCase();
                if(content.indexOf(searchi) > -1) {//found the search in this value
                    //get the position of the search string in content
                    var pos = content.indexOf(searchi);

                    //make position the key for your content array
                    if(contentArray[pos]) {//check if this position has already been assigned to a content
                        //if yes, append this content.
                        contentArray[pos] += '[]'+bigContent[i];
                    } else {//else, set the content
                        contentArray[pos] = bigContent[i];
                        //only save the position the first time you find it, to avoid duplication
                        keys[keys.length] = pos;
                    }
                }
            }
            //sort the key so it can order the values in ascending order(relevance)
            keys = keys.sort();

            //loop thru the key
            for(var i=0;i<keys.length;i++) {
                var key = keys[i];//remember the value of "var key" is the key for contentArray value
                if(contentArray[key]) {//just to be sure it's ther
                    var thisContent = contentArray[key];
                    //check if the content has more than 1 value
                    if(thisContent.indexOf('[]') < 0) {//if it doesn't
                        result[result.length] = contentArray[key];
                    } else {//if it does
                        //convert content into array
                        var thisContentAr = thisContent.split('[]');
                        //Loop thru the array
                        for(var j=0;j<thisContentAr.length;j++) {
                            result[result.length] = thisContentAr[j];
                        }
                    }
                }
            }
        }
        document.getElementById('ov').innerHTML = '';
        for(var i = 0; i<result.length;i++) {
            document.getElementById('ov').innerHTML += '<div>'+result[i]+'</div>';
        }
    }

</script>
</head>
<body>
    <div><input type="text" onkeyup="searchArray(this);" autofocus="autofocus" /></div>
    <div id="ov"></div>
 </body>
</html>

按相关搜索结果排序数组

2 个答案: