我有一個問題與在控制器代碼點火器中傳遞兩個Ajax形式有關。我的第一種形式是一個文件var formData = new FormData($('#form-upload')[0]);
ajax通過codeigniter傳遞兩種形式
和我的第二種形式是由配置文件數據的$('#frm_patientreg').serialize()
現在我的問題是我怎麼能傳遞阿賈克斯這兩種形式? 我已經嘗試過這樣的代碼:
var fileToUpload = inputFile[0].files[0];
if(fileToUpload != 'undefine') {
var formData = new FormData($('#form-upload')[0]);
$.ajax({
type: "POST",
url: siteurl+"sec_myclinic/addpatient",
data: $('#frm_patientreg').serialize()+formData,
processData: false,
contentType: false,
success: function(msg) {
alert("Successfully Added");
$('#frm_patientreg')[0].reset();
}
});
}
else {
alert("No File Selected");
}
但它返回我一個錯誤。 當我試圖只通過data:formData,
,我的圖像文件已成功上傳,但是當我添加$('#frm_patientreg').serialize()
時,它會輸出一個錯誤。我怎樣才能通過這兩種形式?
這裏是我的控制器:
public function addpatient() {
$config['upload_path'] = './asset/uploaded_images/';
$config['allowed_types'] = 'gif|jpg|jpeg|png';
$config['max_size'] = 1024 * 8;
$this->load->library('upload', $config);
if($this->upload->do_upload("file")) {
$upload_data = $this->upload->data();
$file_name = base_url().'asset/uploaded_images/'.$upload_data['file_name'];
$mypatiendid = $this->genpatient_id();
$patient_bday = $this->input->post('pabdate');
$DB_date = date('Y-m-d', strtotime($patient_bday));
$patient_height = $this->input->post('paheight');
$DB_height = $patient_height . " cm";
$patient_weight = $this->input->post('paweight');
$DB_weight = $patient_weight . " kg";
$data = array (
'patient_id' => $mypatiendid,
'patient_fname' => $this->input->post('pafname'),
'patient_mname' => $this->input->post('pamname'),
'patient_lname' => $this->input->post('palname'),
'patient_address' => $this->input->post('paaddress'),
'patient_contact_info' => $this->input->post('pacontact'),
'patient_bday' => $DB_date,
'patient_age' => $this->input->post('paage'),
'patient_height' => $DB_height,
'patient_weight' => $DB_weight,
'patient_sex' => $this->input->post('psex'),
'patient_civil_status' => $this->input->post('pmartialstat'),
'patient_photo' => $file_name,
);
var_dump($data);
}
else {
echo "File cannot be uploaded";
$error = array('error' => $this->upload->display_errors()); var_dump($error);
}
}
爲什麼不使用單個表單並單獨發送所有值。你使用html和css單獨顯示它們。 –
如果您使用單獨的表單,那麼我認爲對於第二種形式,您必須爲第二種形式的每個元素添加「$ .each」形式的數據.. –
我該怎麼做@RahulSharma –