javascript - 如何删除查询字符串的某些元素？

Question

我正在编写一个脚本，您可以向它传递一个像 /search?filter1=question-1&filter2=question2 这样的网址，当问题 1 或问题 2 发生更改时，它将获取 url，并将 question-x 替换为问题值。

我想要构建的一件事是，如果该值为空，我希望它删除查询字符串部分。例如，如果 Question-1 的值为 something，但 2 还没有值，则 url 将为 /search?filter1=something。

我认为可行的方法是这样的

$url.match('/([^?&]*)' + name.toSearch + '/g') // toSearch is a value like question-1

但是返回 null。谁能帮我弄清楚我需要改变什么才能获得我想要的输出？

Answer 1

Given the url /search?filter=question-1, I need to see if the element with the name question[1] has a value, if it does, replace question-1 with the value, and if it doesn't have one, remove the total filter=question-1 string.

从评论中更好地了解您的要求，我使用部分原始答案和您的一些代码完全重写了我的答案:

// Given url
var url = '/search?filter1=question-1&filter2=question-2';

// Specify the filters so no extra query string noise enters the final url
var filters = ["filter1", "filter2", "filter3"];

// Split the query string parts
var urlParts = url.split(/([&?])/);
var reassembled = [];

// Break the url parts into key:value pairs
var qs = (function(a) {
  if (a === "") return {};
  var b = {};
  for (var i = 0; i < a.length; ++i)
  {
    var p=a[i].split('=', 2);
    if (p.length == 1)
      b[p[0]] = "";
    else
      b[p[0]] = decodeURIComponent(p[1].replace(/\+/g, " "));
  }
  return b;
})(urlParts);

// This include a param:value in the reassembled array
function includeQSParam(param, value) {
  if(qs[param]) {
    reassembled.push(param + "=" + value);
  }
}

// Run through the filters
for(var ind in filters) {
  var filter = filters[ind];

  // Check that the filter exists and the supplied value is of type question-
  if(qs[filter] && qs[filter].indexOf("question-") >= 0) {
    // Turns question-number into question[number] so it's a valid selector.
    var inputSelector = "question["+(qs[filter]).replace(/\D/g, "")+"]";  

    // Get the input, and the value (author-supplied code)
    var $input = $('[name="' + inputSelector + '"]');

    // TODO: confirm this is how you get the value
    var value = $input.closest('.question').val();

    if($input.length > 0 && (value !== '' && value !== undefined)) {

      // Replace the parameter's original value with the question value
      includeQSParam(filter, value);
    } else {
      // Nothing to do. This filter will be removed automatically
    }
  }
}

// Reassemble the URL
var fixedUrl = urlParts[0] + (reassembled.length > 0 ? "?"+reassembled.join("&") : "");

同样，这是根据我原来的答案进行修改的，因此会有一些臃肿，但我不想放弃对你的问题。