Question

如何返回内容部分与模式匹配的u_c_pages中第一位的url部分（不区分大小写）？如果找不到页面，则应返回空字符串。

e.g。以下示例

var pg = [ "|www.cam.ac.uk|Cambridge University offers degree programmes and world class research." , "!www.xyz.ac.uk!An great University" , "%www%Yet another University" ]
var pt = "alt";

url1(pages,"ALT") returns "www.xyz.ac.uk" 
url1(pages,"xyz") returns ""

这是我到目前为止所做的。目前它只过滤掉分隔符“|”但我希望它检查任何符号（不使用正则表达式）...

function url1_m1(u_c_pages, pattern) {

    // we create an array to store all the arrays that have the seperator "|"
    // this has been done for situations in which there are invalid arrays lacking the seperator "|"
    // if we find the seperator "|" within an array we also need to know about the position of "|" in the array element
    var seperator = [];
    var seperatorPos = [];
    if (pattern) {

        // looping through the u_c_pages to find occurences of seperator "|"
        // the found variable is initialised with a value of true
        // if we do not find the seperator "|" in an array element then we set the found variable to false
        for (var i = 0; i < u_c_pages.length; i++) {
            var found = true;
            if ((u_c_pages[i].indexOf("|")) < 0) {
                found = false;
            }

            // whereas if we do find a seperator "|" in an array element
            // we create a new array element in seperator containing the page index
            // we create a new array element containing the position of the seperator "|"
            else {
                seperator[seperator.length] = i;
                seperatorPos[seperatorPos.length] = (u_c_pages[i].indexOf("|"));
            }
        }

        // if no arrays with the seperator "|" have been found then we end the program
        if (seperator.length == 0) {
            return ("");
        }

        // otherwise we initialise variable found2
        // we loop through the pages with the seperator "|" as we have stored the index of all the pages in array seperator
        // using the stored index of in array seperator we check the content (after the seperator) for the pattern 
        // we also jump the seperator "|" so it is not returned as one of the results
        else var found2 = ""; {
            for (var j = 0; j < seperator.length; j++)

            // if the the pattern has been found then we extract the url (before the seperator)
            // break the loop once it has been found
            // return URL to user
            {
                if (u_c_pages[seperator[j]].substring(seperatorPos[j] + 1, u_c_pages[j].length).toLowerCase().indexOf(pattern.toLowerCase()) >= 0) {
                    found2 = (u_c_pages[j].substring(0, seperatorPos[j]));
                    break;
                }
            }
            return (found2)
        }
    }
    else {
        return ("");
    }
}

alert(url1_m1(pg, pt));

Answer 1

有一种非常简单的方法可以实现它：

function returnData(arr, splitPattern) {

  var result = []
  for (x in arr) {
    current = arr[x].split(splitPattern);
    var url = current[1]; // www.cam.ac.uk
    var alt = current[2]; // Cambridge University offers degree programmes and world class research.
    result.push({
        url: url,
        alt: alt
    })
  }
  return result
}

var arr = ["|www.cam.ac.uk|Cambridge University offers degree programmes and world class research.",
"|www.xyz.ac.uk|An great University"];

returnData(arr, "|")[0].url // www.cam.ac.uk

returnData(arr, "|")[0].alt // Cambridge University offers degree programmes and world class research.

returnData(arr, "|")[1].url // www.xyz.ac.uk!An great University

returnData(arr, "|")[1].alt // An great University

var anotherArr = ["!%www.cam.ac.uk!%Cambridge University offers degree programmes and world class research.",
"!%www.xyz.ac.uk!%An great University"];

returnData(anotherArr, "!%")[0].url // www.cam.ac.uk

returnData(anotherArr, "!%")[0].alt // Cambridge University offers degree programmes and world class research.

returnData(anotherArr, "!%")[1].url // www.xyz.ac.uk!An great University

returnData(anotherArr, "!%")[1].alt // An great University

http://jsfiddle.net/QsBxD/5/

返回网址部分

1 个答案: