在我的项目中，我使用ajax发送消息的问题是我无法在ajax函数中获得响应，该函数之前工作完美，无法找到问题的确切原因，帮助我解决它
var str = {
message:message, department_id:department, email:email, username:name
};
$.ajax( {
type: "POST",
url:'<?php echo base_url(); ?>home/savemessage',
dataType:"json",
data: str,
success: function(msg) {
$('#sentchat') . hide();
$('#chatmessage') . show();
$('#userchat') . html(msg . dataid);
$('.chat-box-content') . hide();
$('#adminname span').html('waiting for admins reply');
var elem = document . getElementById('userchat');
elem.scrollTop = elem . scrollHeight;
}
});
function savemessage() {
extract($this->input->post());
$data['message_id'] = $this->session->userdata('msgid');
$data['username'] = $username;
$data['email'] = $email;
$data['department_id'] = $department_id;
$data['message'] = $message;
$data['datetime'] = date('Y-m-d H:i:s');
$data['status'] = 'new';
$data['message_by'] = '1';
$this->db->insert('message', $data);
echo json_encode($username);
exit;
}
我无法得到响应，帮助我解决它
您使用ajax与PHP脚本通信，在PHP脚本中您可以拥有想要执行的函数的内容。例如在你的代码中:
$.ajax( {
type: "POST",
url:'<?php echo base_url(); ?>home/savemessage.php',
dataType:"json",
data: {myData:str},
success: function(msg) {
$('#sentchat') . hide();
$('#chatmessage') . show();
$('#userchat') . html(msg . dataid);
$('.chat-box-content') . hide();
$('#adminname span').html('waiting for admins reply');
var elem = document . getElementById('userchat');
elem.scrollTop = elem . scrollHeight;
}
});
然后，在服务器端php脚本"savemessage.php"将接收POST操作，因此您可以:
if(isset($_POST['myData']) && !empty($_POST['myData'])) {
$obj = $_POST['myData'];
//rest of your code
echo json_encode($username);
exit;
}
然而，从你的代码我不能看到$username定义，所以可能会返回一个错误。