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Php 无法将值传递给控制器onclick复选框codeigniter_Php_Jquery_Ajax_Codeigniter_Codeigniter 2 - Fatal编程技术网

Php 无法将值传递给控制器onclick复选框codeigniter

Php 无法将值传递给控制器onclick复选框codeigniter,php,jquery,ajax,codeigniter,codeigniter-2,Php,Jquery,Ajax,Codeigniter,Codeigniter 2,我是CI新手,我在以下代码中遇到问题:我的控制器从DB动态显示国家的所有值,我需要在单击特定复选框时将值传递给我的控制器 我的观点的代码是: <?php foreach ( $countryDetails as $row ) { echo '<input id="country" type="checkbox" name="country" class="unique" value="'.$row->country_id.'">'.$row->countr

我是CI新手,我在以下代码中遇到问题:我的控制器从DB动态显示国家的所有值,我需要在单击特定复选框时将值传递给我的控制器

我的观点的代码是:

<?php  foreach ( $countryDetails as $row ) {
    echo '<input id="country" type="checkbox" name="country" class="unique" value="'.$row->country_id.'">'.$row->country_name.'<br/>';
}?>
<script>
    $('input.unique').click(function() {
        $('input.unique:checked').not(this).removeAttr('checked');
    });
    $(document).ready(function(){
        $(document).on('click','.unique',function(){
            var id=this.value;
            $.ajax({
                type: "POST",
                context: "application/json",
                data: {id:id},
                url: "http://localhost/index.php/fetch_country/index",
                success: function(msg){
                    alert('$id ');
                }
            });
        });
    });
</script>

从相同的源代码中编辑了以下代码!请帮助我更正错误。您现在面临的错误是什么?单击复选框后,我不会收到任何警报!未传递任何值。请尝试
alert(msg)
public function index() {
    $this->view_data['countryDetails'] = $this->get_county_model->getCountryDetails();
    $this->load->view('get_country' , $this->view_data );
}
public function fetch() {
    $id=$this->input->post('id');
    echo ('$id');
}
    $(document).on('click','.unique',function(){
        var id=$(this).val();
        $.ajax({
            type: "POST",
            context: "application/json",
            data: {id:id},
            url: "http://localhost/index.php/fetch_country/index",
            success: function(msg){
                alert(msg);
            }
        });
    });

public function fetch() {
    $id=$this->input->post('id');
    echo $id;
}