我需要你的帮助,在下面的代码,花了寻找解决方案,没有一个是有用的,当我试图保存不保存在数据库中的任何东西,你能帮我吗。
这是我的观点
<div class="modal fade" id="add_modal_para" tabindex="-1" role="dialog" aria-labelledby="title" aria-hidden="true">
<div class="modal-dialog" role="document">
<div class="modal-content">
<div class="modal-header">
<h5 class="modal-title" id="title" style="font-weight: 700;">Agregar</h5>
<button type="button" class="close" data-dismiss="modal" aria-label="Close">
<span aria-hidden="true">×</span>
</button>
</div>
<div class="modal-body">
<form method="post" id="form_add_para" enctype="multipart/form-data">
<div class="input-group mb-3">
<div class="input-group-prepend">
<span class="input-group-text" id="basic-addon1"><i class="fas fa-poop"></i></span>
</div>
<input type="text" name="txt_firtsname_para" id="txt_firtsname_para" class="form-control" placeholder="Nombres" maxlength="50" aria-label="Nombres" aria-describedby="basic-addon1" data-validation-error-msg-container="#txt_firtsname-error-dialog" data-validation="required custom length" data-validation-regexp="^[a-zA-Z\s]+$" data-validation-length="min3" data-validation-length="max50">
</div>
<div id="txt_firtsname-error-dialog"></div>
<div class="input-group mb-3">
<div class="input-group-prepend">
<label for="txt_avatar">Imagen</label>
</div>
<input type="file" class="form-control-file" id="txt_avatar_para" name="txt_avatar_para">
</div>
</div>
<div class="modal-footer">
<button type="submit" id="btn_add_para" class="btn btn-success"><i class="fas fa-save"></i> Guardar</button>
<button type="button" class="btn btn-secondary" data-dismiss="modal"><i class="fas fa-sign-in-alt"></i> Cerrar</button>
</div>
</form>
</div>
</div>
</div>
这是我的ajax代码
$('#submit').submit(function(e){
e.preventDefault();
$.ajax({
url: baseurl+'app/Paramilitar/add_Para',
type: "POST",
data: new FormData(this),
dataType: "JSON",
processData:false,
contentType:false,
cache:false,
async:false,
success: function(data){
$("#form_add_para")[0].reset();
$('#add_modal_para').modal('hide');
show_para_table();
}
});
});
这是我的控制器
public function add_Para(){
$config['upload_path']='./assets/uploads/';
$config['allowed_types']='jpeg|jpg|png';
$config['encrypt_name'] = TRUE;
$this->load->library('upload', $config);
if($this->upload->do_upload('txt_avatar_para'))
{
$data_ima = array('upload_data' => $this->upload->data());
$data_person = array(
'id_person_type' => '3',
'firtsname' => $this->input->post('txt_firtsname_para'),
'avatar' => $data_ima['upload_data']['file_name'],
'active' => '1'
);
$data_result= $this->Person_model->Insert_Person($data_person);
echo json_encode($data_result);
}
这是我的模型示例:
公共函数insert\u person($data\u person){$data\u result=$this->db->insert('tbl\u person',$data\u person);返回$data_result;}
暂无答案!
目前还没有任何答案,快来回答吧!