Ajax 업로드 이미지

Ajax 업로드 이미지

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Q.1이 양식을 ajax로 변환하고 싶지만 내 ajax 코드에 뭔가 부족한 것 같습니다. 제출시에는 아무것도하지 않습니다.

Q2. 또한 제출을 기다리지 않도록 파일이 선택되었을 때 변경시 함수가 실행되기를 원합니다.

여기 JS입니다.

$('#imageUploadForm').on('submit',(function(e) {
    e.preventDefault()
    $.ajax({
        type:'POST',
        url: $(this).attr('action'),
        data:$(this).serialize(),
        cache:false
    });
}));

PHP와 HTMl.

<form name="photo" id="imageUploadForm" enctype="multipart/form-data" action="<?php echo $_SERVER["PHP_SELF"];?>" method="post">
    <input type="file" style="widows:0; height:0" id="ImageBrowse" hidden="hidden" name="image" size="30"/>
    <input type="submit" name="upload" value="Upload" />
    <img width="100" style="border:#000; z-index:1;position: relative; border-width:2px; float:left" height="100px" src="<?php echo $upload_path.$large_image_name.$_SESSION['user_file_ext'];?>" id="thumbnail"/>
</form>

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먼저 ajax 호출에서 성공 및 오류 기능을 포함하고 오류가 발생하는지 확인하십시오.

코드는 다음과 같아야합니다.

$(document).ready(function (e) {
    $('#imageUploadForm').on('submit',(function(e) {
        e.preventDefault();
        var formData = new FormData(this);

        $.ajax({
            type:'POST',
            url: $(this).attr('action'),
            data:formData,
            cache:false,
            contentType: false,
            processData: false,
            success:function(data){
                console.log("success");
                console.log(data);
            },
            error: function(data){
                console.log("error");
                console.log(data);
            }
        });
    }));

    $("#ImageBrowse").on("change", function() {
        $("#imageUploadForm").submit();
    });
});

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HTML 코드

<div class="rCol"> 
     <div id ="prv" style="height:auto; width:auto; float:left; margin-bottom: 28px; margin-left: 200px;"></div>
       </div>
    <div class="rCol" style="clear:both;">

    <label > Upload Photo : </label> 
    <input type="file" id="file" name='file' onChange=" return submitForm();">
    <input type="hidden" id="filecount" value='0'>

다음은 Ajax 코드입니다.

function submitForm() {

    var fcnt = $('#filecount').val();
    var fname = $('#filename').val();
    var imgclean = $('#file');
    if(fcnt<=5)
    {
    data = new FormData();
    data.append('file', $('#file')[0].files[0]);

    var imgname  =  $('input[type=file]').val();
     var size  =  $('#file')[0].files[0].size;

    var ext =  imgname.substr( (imgname.lastIndexOf('.') +1) );
    if(ext=='jpg' || ext=='jpeg' || ext=='png' || ext=='gif' || ext=='PNG' || ext=='JPG' || ext=='JPEG')
    {
     if(size<=1000000)
     {
    $.ajax({
      url: "<?php echo base_url() ?>/upload.php",
      type: "POST",
      data: data,
      enctype: 'multipart/form-data',
      processData: false,  // tell jQuery not to process the data
      contentType: false   // tell jQuery not to set contentType
    }).done(function(data) {
       if(data!='FILE_SIZE_ERROR' || data!='FILE_TYPE_ERROR' )
       {
        fcnt = parseInt(fcnt)+1;
        $('#filecount').val(fcnt);
        var img = '<div class="dialog" id ="img_'+fcnt+'" ><img src="<?php echo base_url() ?>/local_cdn/'+data+'"><a href="#" id="rmv_'+fcnt+'" onclick="return removeit('+fcnt+')" class="close-classic"></a></div><input type="hidden" id="name_'+fcnt+'" value="'+data+'">';
        $('#prv').append(img);
        if(fname!=='')
        {
          fname = fname+','+data;
        }else
        {
          fname = data;
        }
         $('#filename').val(fname);
          imgclean.replaceWith( imgclean = imgclean.clone( true ) );
       }
       else
       {
         imgclean.replaceWith( imgclean = imgclean.clone( true ) );
         alert('SORRY SIZE AND TYPE ISSUE');
       }

    });
    return false;
  }//end size
  else
  {
      imgclean.replaceWith( imgclean = imgclean.clone( true ) );//Its for reset the value of file type
    alert('Sorry File size exceeding from 1 Mb');
  }
  }//end FILETYPE
  else
  {
     imgclean.replaceWith( imgclean = imgclean.clone( true ) );
    alert('Sorry Only you can uplaod JPEG|JPG|PNG|GIF file type ');
  }
  }//end filecount
  else
  {    imgclean.replaceWith( imgclean = imgclean.clone( true ) );
     alert('You Can not Upload more than 6 Photos');
  }
}

다음은 PHP 코드입니다.

$filetype = array('jpeg','jpg','png','gif','PNG','JPEG','JPG');
   foreach ($_FILES as $key )
    {

          $name =time().$key['name'];

          $path='local_cdn/'.$name;
          $file_ext =  pathinfo($name, PATHINFO_EXTENSION);
          if(in_array(strtolower($file_ext), $filetype))
          {
            if($key['name']<1000000)
            {

             @move_uploaded_file($key['tmp_name'],$path);
             echo $name;

            }
           else
           {
               echo "FILE_SIZE_ERROR";
           }
        }
        else
        {
            echo "FILE_TYPE_ERROR";
        }// Its simple code.Its not with proper validation.

여기 업로드 및 미리보기 부분이 완료되었습니다. 이제 페이지와 폴더에서 이미지를 삭제하고 제거하려면 삭제를위한 코드가 여기에 있습니다. Ajax 부분 :

function removeit (arg) {
       var id  = arg;
       // GET FILE VALUE
       var fname = $('#filename').val();
       var fcnt = $('#filecount').val();
        // GET FILE VALUE

       $('#img_'+id).remove();
       $('#rmv_'+id).remove();
       $('#img_'+id).css('display','none');

        var dname  =  $('#name_'+id).val();
        fcnt = parseInt(fcnt)-1;
        $('#filecount').val(fcnt);
        var fname = fname.replace(dname, "");
        var fname = fname.replace(",,", "");
        $('#filename').val(fname);
        $.ajax({
          url: 'delete.php',
          type: 'POST',
          data:{'name':dname},
          success:function(a){
            console.log(a);
            }
        });
    } 

다음은 PHP 부분 (delete.php)입니다.

$path='local_cdn/'.$_POST['name'];

   if(@unlink($path))
   {
     echo "Success";
   }
   else
   {
     echo "Failed";
   }

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jquery.form.js 플러그인을 사용하여 ajax를 통해 서버에 이미지를 업로드 할 수 있습니다.

http://malsup.com/jquery/form/

다음은 샘플 jQuery ajax 이미지 업로드 스크립트입니다.

(function() {
$('form').ajaxForm({
    beforeSubmit: function() {  
        //do validation here


    },

    beforeSend:function(){
       $('#loader').show();
       $('#image_upload').hide();
    },
    success: function(msg) {

        ///on success do some here
    }
}); })();  

If you have any doubt, please refer following ajax image upload tutorial here

http://www.smarttutorials.net/ajax-image-upload-using-jquery-php-mysql/

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Here is simple way using HTML5 and jQuery:

1) include two JS file

<script src="jslibs/jquery.js" type="text/javascript"></script>
<script src="jslibs/ajaxupload-min.js" type="text/javascript"></script>

2) include CSS to have cool buttons

<link rel="stylesheet" href="css/baseTheme/style.css" type="text/css" media="all" />

3) create DIV or SPAN

<div class="demo" > </div>

4) write this code in your HTML page

$('.demo').ajaxupload({
    url:'upload.php'
});

5) create you upload.php file to have PHP code to upload data.

You can download required JS file from here Here is Example

Its too cool and too fast And easy too! :)

참고URL : https://stackoverflow.com/questions/19447435/ajax-upload-image