以下是javascript函数：

var baseurl = 'http://fadli.com/';
            function upload_item(kode,field,table)
        {
            $("#upload_item_"+table+"_"+kode).hide(); //hide submit button
            $("#loading_image_upload_"+table+"_"+kode).show(); //show loading image
            console.log('masuk ke fungsi upload');


            $.ajaxFileUpload({
                url             : baseurl+"admin/level/upload_item/"+kode+"/"+field+"/"+table, 
                //url           :'./admin/level/upload_file/',
                secureuri       : false,
                fileElementId   :'userfile_'+table+'_'+field+'_'+kode,
                dataType        : 'json',
                data            : {
                    'title'             : $('#sub_modul_title_'+kode).val()
                },
                success : function (data, status)
                {
                    if(data.status != 'error')
                    {
                        // $('#files').html('<p>Reloading files...</p>');
                        // refresh_files();
                        // $('#title').val('');
                        alert('sukses brow');
                        $("#upload_item_"+table+"_"+kode).show(); //show submit button
                        $("#loading_image_upload_"+table+"_"+kode).hide(); //hide loading image
                    }
                    console.log(data.msg);
                },
                error:function (xhr, ajaxOptions, thrownError){
                    console.log(thrownError);
                    $("#upload_item_"+table+"_"+kode).show(); //show submit button
                    $("#loading_image_upload_"+table+"_"+kode).hide(); //hide loading image
                }
            });
            return false;
        }

这是HTML代码：

<input type="file" name="userfile" id="userfile_sub_modul_sub_modul_id_<?php echo $sub->sub_modul_id?>" class="styled-finputs">
<button class="btn btn-labeled btn-success btn-xs" type="button" id="upload_item_sub_modul_<?php echo $sub->sub_modul_id; ?>" onclick="upload_item('<?php echo $sub->sub_modul_id?>','sub_modul_id','sub_modul')"><span class="btn-label icon fa fa-picture-o"></span> Upload gambar</button>                                                           
<img src="http://fadli.com/assets/img/input-spinner.gif" id="loading_image_upload_sub_modul_<?php echo $sub->sub_modul_id?>" style="display:none" />

问题出在第一个控制台日志之后，它并没有全部停止运行。

任何帮助都会受到赞赏。

由于

Answer 1

对于ajaxFileUpload，它以大写字母A开头，尝试将ajaxFileUpload更改为AjaxFileUpload

Answer 2

您只需使用ajax（无页面重新加载）提交表单即可只需使用此代码。它也会提交您的文件数据。

$(document).on("submit", "form", function(event)//I wrote it for any form.You can change it for specific form
{
    event.preventDefault();
    $.ajax({
        url: $(this).attr("action"),//make sure your form has action url
        type: $(this).attr("method"),//Your form method type GET OR POST
        dataType: "JSON",
        data: new FormData(this),
        processData: false,
        contentType: false,
        success: function (data, status)
        {

        },
        error: function (xhr, desc, err)
        {


        }
    });
});

您还可以使用FormData添加附加数据。 https://developer.mozilla.org/en-US/docs/Web/Guide/Using_FormData_Objects 您可以像这样修改上述功能

$(document).on("submit", "form", function(event)
{
    var formData = new FormData(this);
    formData.append("MORE_INPUT_KEY", "MORE_DATA");
    event.preventDefault();
    $.ajax({
        url: $(this).attr("action"),//set any link
        type: $(this).attr("method"),//set any metho
        dataType: "JSON",
        data: formData ,
        processData: false,
        contentType: false,
        success: function (data, status)
        {

        },
        error: function (xhr, desc, err)
        {


        }
    });
});

使用codeigniter的JQuery ajaxfileupload

以下是javascript函数：

这是HTML代码：

2 个答案: