我无法将我的数组从我的控制器获取到 codeigniter 中的 ajax

Posted 2023-05-07

【中文标题】我无法将我的数组从我的控制器获取到 codeigniter 中的 ajax

【英文标题】：I can't get my array from my controller to ajax in codeigniter

【发布时间】：2021-04-01 17:08:06

【问题描述】：

我正在尝试从我的控制器中获取一个数组，但我的响应总是这样 我做错了什么？

这是 ajax 中的 console.log，它返回这些我无法理解的脚本

if (!window.console) console = ;console.log = console.log || function();console.warn = console.warn || function();console.error = console.error || function();console.info = console.info || function();console.debug = console.debug || function();</script>["matricula":"1","fullname":"aaa,  bbbbbb","dni":"123","matricula":"2","fullname":"dddd, ddddd","dni":"1234"]




//CONTROLLER FUNCTIONS

      public function uploadExcel()
            $this->loadUploadConfiguration();            
            if(!$this->upload->do_upload('file'))
               echo $this->upload->display_errors();
            else 
                $path = $this->upload->data()['file_name'];
                $this->getExcelData($path); // I call the next function
            
           

        private function getExcelData($savedPath)
            $profesionals = array();

            $path = Homec::PATH_TO_SAVE.$savedPath;
            $object = phpExcel_IOFactory::load($path);
            $rowNumber = 2;
            $worksheet = $object->setActiveSheetIndex(0);
                while($worksheet->getCellByColumnAndRow(0,$rowNumber)->getValue() != "")
                    if($this->isProfesionalAvailable($worksheet->getCellByColumnAndRow(0,$rowNumber)->getValue()))
                          $profesional = array('matricula' => $worksheet->getCellByColumnAndRow(0, $rowNumber)->getValue(), 
                         'fullname' => $worksheet->getCellByColumnAndRow(1, $rowNumber)->getValue(),
                          'dni' => $worksheet->getCellByColumnAndRow(2, $rowNumber)->getValue()
                          );
                          array_push($profesionals,$profesional);
                    
                    $rowNumber++;
                
                  echo json_encode($profesionals);
            $this->deleteSavedFile($path);
        

AJAX 当我尝试循环我的数据时，它是一个字符串而不是 json，当我尝试将内容类型更改为 application/json 时，它返回空

$( '#formUpload' )
  .submit( function( e ) 
    $.ajax( 
      url: 'uploadExcel',
      type: 'POST',
      data: new FormData( this ),
      processData: false,
      contentType: false, //when i change my contentype to appalication/json, my data inside success is empty 
      success: function(data)
                        console.log(data); // console.log return this
<script type="text/javascript">
if (!window.console) console = ;console.log = console.log || function();console.warn = console.warn || function();console.error = console.error || function();console.info = console.info || function();console.debug = console.debug || function();</script>["matricula":"1","fullname":"aaa,  bbbbbb","dni":"123","matricula":"2","fullname":"dddd, ddddd","dni":"1234"]
                 //operations = JSON.parse(data) // if i do this, it return an error

              
     );

    e.preventDefault();
   );

查看

<div class="upload ">
   <form class="form-inline" id="formUpload" method='post' enctype="multipart/form-data" >
      <div class="form-group ">
         <label class="" for="file">Subir excel</label>
         <input class="form-control-file" type='file' name='file'>
      </div>
      <button class="btn  btn-danger" id="btnUpload" >Subir</button>

   </form>

</div> ```

【问题讨论】：

对不起，我是新来的，我也做错了

为什么你的JS中间有<script...>标签？ if (!window.console) ... 是什么东西？删除所有这些 - 只留下 console.log(data)，编辑您的问题并粘贴结果。

【参考方案1】：

  $this->deleteSavedFile($path);
  echo json_encode($profesionals);die;

按此顺序使用。一旦你回显 json 数据需要退出，否则它会继续另一行。