如何将 Controller 的响应返回给 Jquery Javascript?
Javascript
$('.signinform').submit(function() {
$(this).ajaxSubmit({
type : "POST",
url: 'index.php/user/signin', // target element(s) to be updated with server response
cache : false,
success : onSuccessRegistered,
error: onFailRegistered
});
return false;
});
数据返回 null(空白)!
function onSuccessRegistered(data){
alert(data);
};
Controller -
public function signin() {
$arr = array('a' => 1, 'b' => 2, 'c' => 3, 'd' => 4, 'e' => 5);
echo json_encode( $arr );
}
最佳答案
return $this->output
->set_content_type('application/json')
->set_status_header(500)
->set_output(json_encode(array(
'text' => 'Error 500',
'type' => 'danger'
)));
关于javascript - CodeIgniter - 如何从 Controller 返回 Json 响应,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/18821492/