[英]MySQL CodeIgniter Select or not from second table based on column of first Select
[英]getting values of second select from db based of first select box selection in codeigniter
我需要有關如何基於第一個選擇框獲取第二個選擇框的值的幫助
這是視圖:
$(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>
這是控制器:
類Select擴展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));
}
}
這是模型:
Data_model類擴展了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;
}
}
}
請對此提供幫助,以提供正確的代碼。
因為您是直接訪問get_cities()
函數,而不是從控制器中的另一個函數訪問,所以return語句實際上不會將json數組打印到頁面上。
return json_encode(array('Cities' => $cities));
有3種打印方法:第一種是print
或echo
顯json數組(不好的做法),第二種是使用視圖打印發送給它的原始文本。 IE瀏覽器
$this->load->view('raw', array('data' => json_encode(array('Cities' => $cities)));
與raw.php只是:
<?php print $data; ?>
或者最后,您可以在output
類中使用set_output()
函數,如下所示:
$this->output->set_output(array('data' => json_encode(array('Cities' => $cities)));
如果只想從index()
函數訪問它,您可能還想使它的function load_state()
私有。
您的代碼可能還有其他問題,但這是最明顯的問題。
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