json_encode 响应警报成功 bootstrap

Question

我有 2 个条件 return 像这样 json

public funtion do_upload{
//1st
return json_encode(['upload_success' => 'Successfully import ' . $count_validEntries . ' document]);

//2nd
return json_encode(['upload_error' => 'Your document is not valid, at row '. $implode_err. 'Please use given data']);
}

我的Ajax打电话

  $("#upload_btn").on('click', function(){
    var mydata = new FormData(document.getElementById("form_upload"));
    $.ajax({
      type : "POST",
      data: mydata,
      contentType: false,
      dataType : "json",
      url : "<?php echo base_url();?>/C_MRCR_A/do_upload",
      cache: false,
      processData: false,
      beforeSend:function(){
      },
      success:function(response){
        console.log('succ ',response);
      },
      error:function(xhr, status, error){
      },
      complete:function(){
        // do nothing for now
      }
    });
  });

我想 alert 如果 'upload_success' 成功，如果 'upload_error' 使用上面的响应则警告错误？或者还有其他方法吗？

Answer 1

您需要在成功函数中应用条件。如下所示 -

success:function(response){
    //console.log('succ ',response);
    if(response.upload_success){
      $('#response').html('<div class="alert alert-success" role="alert"> T'+response.upload_success+' </div>');
    } else if(response.upload_error){
      $('#response').html('<div class="alert alert-success" role="alert"> '+response.upload_error+' </div>');
    }
}

在您的 HTML 中创建一个 div 以显示响应。

<div id="response"></div>

你可以append/show任何你想要的回复。

json_encode 响应警报成功 bootstrap

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