Jquery获取iD和Value，显示URL参数

Jquery Get iD and Value and display URL parameters

本文关键字：显示 URL 参数 Value 获取 iD Jquery      更新时间：2023-09-26

我用在下面的HTML &JS脚本获取表单中每个元素的ID和值。

但是，我需要得到所有这些值和下面一样，并添加到顶部的URL当用户提交表单重定向预订引擎。

低于console.log的输出:

someaction ? fromLocation = undefined& departureDate = undefined& tripType = undefined& returnDate =定义

HTML: 

  <form method="POST" id="bookingForm" action="someaction">
<div class="col-lg-6">
  <div class="form-group">
    <label for="fromLocation">From:</label>
    <select class="select form-control input-lg" id="fromLocation" form="fromLocation" name="fromLocation" required>
      <option value="">From</option>
      <option value="ADL">Adelaide</option>
      <option value="DRW">Darwin</option>
      <option value="MEL">Melbourne</option>
      <option value="PER">Perth</option>
    </select>
  </div>
  <div class="form-group">
    <label for="departure">Departure Date:</label>
    <input class="departureDate form-control" id="departureDate" type="text" placeholder="Departure Date" name="Departure Date" required>
  </div>
  <div class="form-group" id="tripType">
    <label for="tripType">Trip Type:</label>
    <div class="col-lg-12">
      <div class="col-lg-2">
        <input class="radio tripType" id="ReturnTrip" type="radio" name="tripType" value="Return" checked="">Return
      </div>
      <div class="col-lg-2">
        <input class="radio tripType" id="SingleTrip" type="radio" name="tripType" value="One-way">One-way
      </div>
      <div class="col-lg-8"></div>
    </div>
  </div>
</div>
<div class="col-lg-6">
  <div class="form-group">
    <label for="return">Return Date:</label>
    <input class="returnDate form-control" type="text" id="returnDate" placeholder="Return Date" name="Return Date" required>
  </div>
  <div class="form-group" id="booking-btn--container">
    <button class="btn btn-primary" type="submit">Submit</button>
  </div>
</div>
</form>

JS: 

var url = $("#bookingForm").attr("action") + "?";
var urlElements = [];
$("#bookingForm").find('.form-group').each(function(){
urlElements.push($(this).find('.form-control').attr("id") + "=" + $(this).find('.form-control').attr("value") || $(this).find('.form-control').attr("id") + "=" + $(this).find('.form-control').attr("value"));
});
urlElements = urlElements.join("&");
url += urlElements;
console.log(url);

只需修改您的名称属性以匹配您的id，并使用.serialize():

$("form").on("submit", function(event) {
  event.preventDefault();
  var form = $(this);
  var url = form.attr('action') + '?' + form.serialize();
  
  console.log(url);
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<form method="POST" id="bookingForm" action="someaction">
  <div class="col-lg-6">
    <div class="form-group">
      <label for="fromLocation">From:</label>
      <select class="select form-control input-lg" id="fromLocation" name="fromLocation" required>
        <option value="">From</option>
        <option value="ADL">Adelaide</option>
        <option value="DRW">Darwin</option>
        <option value="MEL">Melbourne</option>
        <option value="PER">Perth</option>
      </select>
    </div>
    <div class="form-group">
      <label for="departure">Departure Date:</label>
      <input class="departureDate form-control" id="departureDate" type="text" placeholder="Departure Date" name="departureDate" required>
    </div>
    <div class="form-group" id="tripType">
      <label for="tripType">Trip Type:</label>
      <div class="col-lg-12">
        <div class="col-lg-2">
          <input class="radio tripType" id="ReturnTrip" type="radio" name="tripType" value="Return" checked="">Return
        </div>
        <div class="col-lg-2">
          <input class="radio tripType" id="SingleTrip" type="radio" name="tripType" value="One-way">One-way
        </div>
        <div class="col-lg-8"></div>
      </div>
    </div>
  </div>
  <div class="col-lg-6">
    <div class="form-group">
      <label for="return">Return Date:</label>
      <input class="returnDate form-control" type="text" id="returnDate" placeholder="Return Date" name="returnDate" required>
    </div>
    <div class="form-group" id="booking-btn--container">
      <button class="btn btn-primary" type="submit">Submit</button>
    </div>
  </div>
</form>

作为奖励，它将正确地url编码您的值

试一下

var $input = $(this).find('.form-control');
urlElements.push($input.attr("id") + "=" + $input.val());
相关文章：