Tinymce 使用 file_picker_callback 和图片上传器获取图片 _description

Question

TL:DR 我正在尝试使用 javascript 获取 image_description 字段的值以传递我的 post xhr 请求

原题如下

我正在使用 file_picker_callback 类型的图片 https://www.tinymce.com/docs/configure/file-image-upload/#file_picker_callback

我在

中启用了 image_description 输入字段

tinymce.init({
     ....,
     image_description: true,
     ...

一切都在上传，但我还想传递 image_description 字段以存储在数据库中。但是我抓不到数据

下面是我的两个函数，直接取自tinymce网站

  file_picker_callback: function(cb, value, meta) {

    var input = document.createElement('input');
    input.setAttribute('type', 'file');
    input.setAttribute('accept', 'image/*');

    input.onchange = function() {
      console.log(this.files);
      var file = this.files[0];

      console.log( meta ); //I thought it might be here in the meta bt it isn't

     console.log( $('#mceu_62').val() ); //I tried to get it from its id but it returns undefined i think that field is created after this function is created.



      var id = file.name;
      var blobCache = tinymce.activeEditor.editorUpload.blobCache;
      var blobInfo = blobCache.create(id, file);
      blobCache.add(blobInfo);


      // call the callback and populate the Title field with the file name
      cb(blobInfo.blobUri(), { title: file.name });
    };

    input.click();
  },
  images_upload_handler: function (blobInfo, success, failure) {
      var xhr, formData;

      xhr = new XMLHttpRequest();
      xhr.withCredentials = false;
      xhr.open('POST', '/articles/postAcceptor');

      xhr.onload = function() {
        var json;

        if (xhr.status != 200) {
          failure('HTTP Error: ' + xhr.status);
          return;
        }

        json = JSON.parse(xhr.responseText);

        if (!json || typeof json.location != 'string') {
          failure('Invalid JSON: ' + xhr.responseText);
          return;
        }

        success(json.location);
      };

      formData = new FormData();
      formData.append('file', blobInfo.blob(), blobInfo.filename());
      formData.append('description', /*but i can't get the value*/);

      xhr.send(formData);
    }

@卡尔·莫里森

Answer 1

试试这个以获得价值：

images_upload_handler: function (blobInfo, success, failure) {
      var xhr, formData;

      xhr = new XMLHttpRequest();
      xhr.withCredentials = false;
      xhr.open('POST', '/articles/postAcceptor');

      xhr.onload = function() {
        var json;

        if (xhr.status != 200) {
          failure('HTTP Error: ' + xhr.status);
          return;
        }

        json = JSON.parse(xhr.responseText);

        if (!json || typeof json.location != 'string') {
          failure('Invalid JSON: ' + xhr.responseText);
          return;
        }

        success(json.location);
      };


      var description = '';

      jQuery(tinymce.activeEditor.dom.getRoot()).find('img').not('.loaded-before').each(
    function() {
        description = $(this).attr("alt");
        $(this).addClass('loaded-before');
    });

      formData = new FormData();
      formData.append('file', blobInfo.blob(), blobInfo.filename());
      formData.append('description', description); //found now))

      xhr.send(formData);
    }

Answer 2

以下可能会对您有所帮助 https://www.tinymce.com

<!DOCTYPE html>
<html>
<head>
<script src="https://cloud.tinymce.com/stable/tinymce.min.js"></script>
<script>tinymce.init({ selector:'textarea' });</script>
</head>
<body>
<textarea>Next, get a free TinyMCE Cloud API key!</textarea>
</body>
</html>

Tinymce 使用 file_picker_callback 和图片上传器获取图片 _description

Tinymce get image _description with file_picker_callback and image uploader

javascript

tinymce-4