[英]Retrieve and display json data from codeigniter function
如何使用ajax(json)和codeigniter或php顯示數據庫(SQL)中的數據? 我試圖這樣做,但結果是“未定義”,這是我的代碼:
jQuery代碼:
$('[id^="menuitem_"]').on('click',function(){
var menuID = $(this).attr('id').split("_")[1];
$.ajax({
url:"<?php echo base_url();?>Vendors_search/Get_Business_By_Type",
method:"POST",
dataType: "text",
data: {menuID:menuID},
success:function(resp){
var obj = $.parseJSON(resp);
var values = Object.values(obj);
$.each(obj,function(key,val){
$('.business_id').append('<strong>' + val.business_name + '</strong>');
});
}
});
Codeigniter代碼:
function Get_Business_By_Type(){
$obj = $this->input->post('menuID');
if($this->Search_obj_model->getBusinessType($obj)){
$bus_type = $this->Search_obj_model->getBusinessType($obj);
$result['business_details'] = $this->Search_obj_model->getBusinessType($obj);
}
else{
$result['message'] = 'Oooops! We could not find your request. Try again later';
}
$this->output->set_content_type('application/json');
$this->output->set_output(json_encode($result));
$string = $this->output->get_output();
echo $string;
exit();
}
執行以下操作以顯示JSON數據
在Ajax調用中將dataType
更改為json
無需添加output->set_content_type('application/json')
$result['business_details'] = $this->Search_obj_model->getBusinessType($obj);
}
else{
$result['message'] = 'Oooops! We could not find your request. Try again later';
}
echo $result;
從Ajax獲取數據
$.each(resp, function (key, data) {
$.each(data, function (index, data) {
data.business_name // Your Value
});
});
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