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如何在没有任何提交按钮的情况下动态显示基于下拉菜单中所选年份的记录

[英]How to display records based on selected year from dropdown dynamically without any submit button

I am working on PHP code-igniter. 我正在研究PHP代码点火器。

My view code goes like below: 我的查看代码如下所示:

<form method="post" action="">  
    <div class="col-md-4 col-sm-4 ">
        <div class="form-group">                            
            <label class="h5" for="Mode"><b>See year wise commitments</b></label>
            <select class="form-control" name="Year" id="Year"  onchange="this.form.submit()">
                <option value="">Select Year</option>
                <option value="2015">2015</option>
                <option value="2016">2016</option>
                <option value="2017">2017</option>
            </select>
        </div>
    </div>
</form>

Just by selecting any year from dropdown, I have to pass the selected year to my controller and inturn to model and get the query results based on the selected year. 只需从下拉列表中选择任何年份,我就必须将所选年份传递给我的控制器,然后进行建模并根据所选年份获得查询结果。

In my controller I am receiving the posted value like: 在我的控制器中,我收到以下发布的值:

$selectedYear = $this->input->post('Year');

And before selecting any dropdown value the records in my view has to be of Current year. 在选择任何下拉值之前,我认为记录必须是当年的。

My ajax code is 我的ajax代码是

$( "#year" ).change(function() {
    year = $(this).val();
    $.ajax({
          url : "controllerfilename/controllername",
          data : {"year": year},
          type : "post",
          success: function(data){
             console.log(data);
          }
   });

}); });

You can make an ajax call to your controller on change function of dropdown 您可以在下拉菜单的更改功能上对控制器进行ajax调用

    $( "#year" ).change(function() {
        year = $(this).val();
        $.ajax({
              url : "controller url",
              data : {"year": year},
              type : "post",
              success: function(data){
                 console.log(data);
              }
       });
   });

You defined id as id="Year" but in the ajax you call it like $( "#year" ).change(function() { . Means In ajax its year and in view its Year (uppercase). 您将id定义为id="Year"但在ajax中将其命名为$( "#year" ).change(function() { 。意味着在ajax中,其year而在Year大写。

Try this code: 试试这个代码:

$("#Year").change(function() {
year = $(this).val();
$.ajax({
      url : "<?php echo base_url('controllerfilename/controllername') ?>",
      data : {"year": year},
      type : "post",
      success: function(data){
         console.log(data);
      }
});
});

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