使用jquery从url获取url变量

Question

我有一个只显示价格的html表单,但不向服务器提交任何内容.这现在工作正常.

如何仅从url变量添加运行此表单的功能,因此如果您使用url变量来到此页面,则不必单击"提交"按钮？

这样的事情:www.my-domain.com/formurl/?smoker=yes&amount=500000&age=20应该在页面加载完成后显示价格,并根据url变量更改表单值.

这是我的HTML:

<form class="form-horizontal">
        <label class="control-label" for="smoker">Do you smoke?</label>
        <select id="smoker" name="smoker"><option value="No">No</option><option value="Yes">Yes</option></select>

        <label class="control-label" for="amount">Amount</label>
        <select id="amount" name="amount"><option value="500000">500 000</option><option value="1000000">1 000 000</option><option value="1500000">1 500 000</option><option value="2000000">2 000 000</option></select>

        <label class="control-label" for="age">Age</label>
        <input id="age" name="age" type="text" />


        <button class="btn" id="calc" type="submit">Show price</button>
</form>


<div class="alert alert-info hide" id="price-result"></div>

---

$(document).ready(function () {

    //JSON object
    var obj = {"data":
    [
        {"age": "18","500000": "645","1000000": "1018","1500000": "1391","2000000": "1764","smoker": "No"},
        {"age": "19","500000": "645","1000000": "1018","1500000": "1391","2000000": "1764","smoker": "No"},
        {"age": "20","500000": "645","1000000": "1018","1500000": "1391","2000000": "1764","smoker": "No"},
        {"age": "18","500000": "964","1000000": "1544","1500000": "2125","2000000": "2705","smoker": "Yes"},
        {"age": "19","500000": "964","1000000": "1544","1500000": "2125","2000000": "2705","smoker": "Yes"},
        {"age": "20","500000": "964","1000000": "1544","1500000": "2125","2000000": "2705","smoker": "Yes"}
    ]

    };        

    //Find the value when form is submitted
    $('#calc').click(function(e) {
        e.preventDefault();

        var data = jQuery.grep(obj.data, function(element, index){
          return element.age && element.smoker; // retain appropriate elements

        });    

        var selectedSmoke = $('#smoker').val().toString().toLowerCase();
        var selectedAmount= $('#amount').val().toString().toLowerCase();
        var selectedage = $('#age').val().toString().toLowerCase();

        var amount = "";

        $.each(data,function(k, v){

            if( data[k].age.toString().toLowerCase() == selectedage &&  
                data[k].smoker.toString().toLowerCase() == selectedSmoke &&
                data[k][selectedAmount] != undefined){

                amount = data[k][selectedAmount];


            }

        });


        //Empty the div
        $('#price-result').empty();

        //Show the result in div
        var displayText = "<h3>Price:" + amount + "</h3>";

        $("#price-result").append(amount == "" ? "Not a valid age" : displayText).removeClass("hide");

        return false;//Stop page from reloading

    });


});

Answer 1

填写input表单URL

function getURLParameter(name) {
  return decodeURI(
   (RegExp(name + '=' + '(.+?)(&|$)').exec(location.search)||[,null])[1]
   );
}
var smoker = getURLParameter('smoker');
var amount = getURLParameter('amount');
var age = getURLParameter('age');

$('#smoker').val( smoker );
$('#amount').val( amount );
if(age != 'null'){
    $('#age').val( age );
    calculAmount(obj);
}

//Find the value when form is submitted
$('form').submit(function(e) {
    e.preventDefault();
    calculAmount(obj);
});

为了将数据发送到服务器,您可以使用get,post或ajaxjQuery的

替换 $('#calc').click(function(e) 为 $('form').submit(function(e)

使用get:

 //Empty the div
 $('#price-result').empty();

$.get("www.my-domain.com/formurl/",
 { smoke: selectedSmoke , amount: amount ,age: selectedage} );

要么

$.post('www.my-domain.com/formurl/', $("form").serialize(), function(data) {
  console.log("data sent");
});

要么

$.ajax({
         type : 'POST',
         url : 'www.my-domain.com/formurl/',
         data: $("form").serialize(),
         success : function(data){                                               
                   console.log("data sent");
          },
          error: function (request, status, error) {
                 console.log(request.responseText);
          } 
      });