我有一个问题,当我张贴表单使用jQuery和Codeigniter的响应显示错误501方法没有实现。我使用智能手机浏览器,如android.this时得到这个错误是我的代码:错误501方法没有在jQuery codeigniter上实现
$(document).ready(function(){
var obj=$('#recoveryForm');
obj.find(':submit').removeClass('disabled');
obj.validate({
submitHandler: function() {
$.ajax({
cache:false,
dataType:'json',
url: "<?=base_url();?>reseller/changePassword",
type: "POST",
data: obj.serialize(),
beforeSend:function(){
obj.find(':submit').button('loading');
},
success:function(data){
if(data.status=='false'){
$('#alert').html('<div class="alert alert-danger center">'+data.noty+'</div>');
obj.find(':submit').button('reset');
}else{
$('#alert').html('<div class="alert alert-success center">'+data.noty+'<br><br><a href="<?=base_url();?>" class="btn btn-pink">Back to Home</a></div>');
obj.fadeOut('slow',function(){obj.remove()});
}
},
error: function (xhr, ajaxOptions, thrownError) {
$('#alert').html('<div class="alert alert-danger center">'+xhr.status+' '+ thrownError+'</div>');
obj.find(':submit').button('reset');
}
});
}
});
});
这我的PHP代码
public function changePasswordR(){
if($this->check_login()){
$data=$this->input->post(null,true);
if(!empty($data['password']) || !empty($data['newpassword']) || !empty($data['konfirmpassword'])){
$user=$this->reseller_data();
$row=$this->madmin->read(__CLASS__,'where email= "'.$user['email'].'"');
if(!empty($row)){
if($this->salt($data['password'])==$row[0]->password){
if($this->madmin->update(__CLASS__,'email',$row[0]->email,array('password'=>$this->salt($data['newpassword'])))){
//noty
$this->madmin->create('notification',array('id'=>null,'time'=>date('Y-m-d H:i:s'),'msg'=>$row[0]->nama.' mengubah passwordnya'));
die(json_encode(array('status'=>'true','noty'=>'<script type="text/javascript">alert("Selamat '.$row[0]->nama.', password anda berhasil diubah. Silahkan melakukan login ulang menggunakan password baru anda.");window.location.href = "'.base_url().'logout";</script>')));
}else{
die(json_encode(array('status'=>'false','noty'=>'Gagal mengubah password, silahkan coba beberapa saat lagi.')));
}
}else{
die(json_encode(array('status'=>'false','noty'=>'Password lama salah.')));
}
}else{
die(json_encode(array('status'=>'false','noty'=>'Akun tidak ditemukan.')));
}
}else{
die(json_encode(array('status'=>'false','noty'=>'Data tidak lengkap.')));
}
}else{
die(json_encode(array('status'=>'false','noty'=>'Login kembali untuk melanjutkan.')));
}
}
enter code here
请帮帮我!
您应该添加一些文字解释一下这个答案确实为什么它为特定问题提供解决方案。代码转储很少用于其他人。 – Paritosh 2016-02-10 03:37:48