Question

我想将下拉列表中所选元素的值保存到JSON文件中。

<td contenteditable=true bgcolor="#F4F2F2">
    <select>
       <option value="default">${list.days}</option>
       <option value="1">1</option>
       <option value="2">2</option>
       <option value="3">3</option>
       <option value="4">4</option>
    </select>
</td>

function(i) {
    var $tds = $(this).find('td'), status = $tds.eq(0).text();
    console.log(status)
    json = json + " {\"deviceType\": \"" + $tds.eq(1).text()+ "\"," + "\"days\": \"" + $tds.eq(2).text()+ "\"},";});

Answer 1

我建议不要尝试从头开始创建自己的JSON字符串。而是使用JSON.stringify（）方法：

$("#prw").on('click', function(e){

var url = window.location.origin + "/newsletter/preview";

var title = $('#title').val();
var intro = $('#intro').val();
// That table reference is a bit unclear to me, is this a 3rd party library?
var checkbox = false; // Grab your checkbox value here
var inputVal = null; // Grab your input value here

var data = { 
title: title,
intro: intro,
checkbox: checkbox,
inputVal: inputVal 
};

$.ajax({
    type: "POST",
    url: url,
    data: JSON.stringify(data),
    contentType: 'application/json'
}).done(function(data){
    console.log("Response", data);
});

e.preventDefault();

此代码示例假定$ tds.eq（1）和$ tds.eq（2）均包含一个select元素。

Answer 2

也许您可以将值发送给PHP并保存

$('.input').change(function(){
  //Save value of this dropdown menu
  $value = $(this).val();

  //AJAX the value to the PHP

  $.post('balbal.php',{
    value: $value
  });


});

Answer 3

知道使用@ ohio818输入后对JSON对象进行了更改

json = json + " {\"deviceType\": \"" + $tds.eq(1).text()+ "\"," +"\"days\":\""+ $tds.eq(2).find("select").val()+ "\"},";

将所选下拉值保存到JSON

3 个答案: