根据codeigniter中的第一个选择框选择从db获取第二个选择值

时间:2013-06-07 14:08:57

标签: php codeigniter codeigniter-2

我需要帮助,了解如何根据第一个选择框获取第二个选择框的值

这是观点:

$(document).ready(function() {

$('#state').change(function() {
// Get an instance of the select, and it's value.
    var state = $(this),
    stateID = state.val();
// Add if statement to check if the new state id
// is different to the last to prevent loading the same
// data again.

// Do the Ajax request.
$.ajax({
    url : 'http://localhost/ci_ajax/select/get_cities/'+stateID, // Where to.
    dataType : 'json', // Return type.
    success : function(data) { // Success :)
        // Ensure we have data first.
        if(data && data.Cities) {
            // Remove all existing options first.
            state.find('option').remove();
            // Loop through each city in the returned array.
            for(var i = 0; i <= data.Cities.length; i++) {
                // Add the city option.
                state.append($('option').attr({
                    value : data.Cities[i].value
                }).text(data.Cities[i].city));
            }
        }
    },
    error : function() {
        // Do something when an error happens?
    }
});

}); });

<form action="" method="post">
<select name="state" id="state">
    <option>Select</option>
    <?php foreach($states as $row):?>
        <option value="<?php echo $row->id?>"><?php echo $row->states;?></option>
    <?php endforeach;?>
</select>
<select id="cities" name="cities">
    <option>Select</option>
</select>

这是控制器:

类选择扩展CI_Controller {

function index(){

    $data['states']=$this->load_state();
    $this->load->view('form',$data);
}

function load_state(){

    $this->load->model('data_model');

    $data['states']=$this->data_model->getall_states();

    return $data['states'];
}

function get_cities() {
     // Load your model.
    $this->load->model('data_model');
    // Get the data.
    $cities = $this->data_model->get_cities();
    // Specify that we're returning JSON.
    header('content-type: application/json');
    // Return a JSON string with the cities in.
    return json_encode(array('Cities' => $cities));
}

}

这是型号:

类Data_model扩展了CI_Model {

function getall_states(){

    $query=$this->db->get('states');
    if($query->num_rows()>0){
        foreach($query->result() as $row){
            $data[]=$row;
        }
        return $data;
    }

}

function get_cities(){

    $this->db->select('id,cities');
    $this->db->from('cities');
    $this->db->where('s_id',$this->uri->segment(3));
    $query=$this->db->get();

    if($query->num_rows()>0){
        foreach($query->result() as $row){
            $data[]=$row;
        }
        return $data;
    }

}

}

请提供帮助,希望能提供正确的代码。

1 个答案:

答案 0 :(得分:0)

因为您直接访问get_cities()函数而不是控制器中的其他函数,所以您的return语句实际上不会将json数组打印到页面。

return json_encode(array('Cities' => $cities));

有三种打印方式:第一种是printecho json数组(不良做法),第二种是使用打印发送给它的原始文本的视图。即。

$this->load->view('raw', array('data' => json_encode(array('Cities' => $cities)));

raw.php只是:

<?php print $data; ?>

或者最后你可以使用set_output()类中的output函数,如下所示:

$this->output->set_output(array('data' => json_encode(array('Cities' => $cities)));

如果只能通过function load_state()功能访问index(),您可能还想要{{1}}私有。

您的代码可能还有其他问题,但这是最明显的问题。