javascript - 如何添加ajax参数？

Question

面临这样一个问题，就是无法给ajax请求添加参数，因为有一个“Form Data”，添加更多选项是不行的。当您在参数“data”中添加另一个变量时，就会发生错误。如何做到这一点是在一个请求中发布一个文件和参数？

请原谅这个错误，无论错误与否，php 文件根本不输出任何内容，而是空值

//Here the file is sent without problems, but the parameters no more don't accept php file, nothing displays in the result
    var fd = new FormData();
    fd.append('file', input[0].files[0]);
    $.ajax({
      url: "/controllers/createNewsController.php",
      data: fd,
      type: "POST",
      processData: false,
      contentType: false,
      success: function(data) {
        $(".news .containers").append(data);
      }
    });

//When this code runs the PHP file won't take anything, although I was expecting the output variables and the file. Using var_dump, $_POST and $_FILES displays array(0){}
var fd = new FormData();
                fd.append('file', input[0].files[0]);
                $.ajax({
                    url: "/controllers/createNewsController.php",
                    data: {fd, title:newsHeader, description:description, hashTag:hashTag, themeHashTag:themeHashTag, viewNews:viewNews},
                    type: "POST",
                    processData: false, 
                    contentType: false,
                    success: function(data){
                        $(".news .containers").append(data);
                    } 
                });

//Similarly, nothing appears in the php file
var fd = new FormData();
                fd.append('file', input[0].files[0]);
                $.ajax({
                    url: "/controllers/createNewsController.php",
                    data: {fd:fd, title:newsHeader, description:description, hashTag:hashTag, themeHashTag:themeHashTag, viewNews:viewNews},
                    type: "POST",
                    processData: false, 
                    contentType: false,
                    success: function(data){
                        $(".news .containers").append(data);
                    } 
                });

Answer 1

就像附加文件一样，您可以向其中附加更多数据，如下所示:

fd.append('file', input[0].files[0]);

fd.append('var1', val1);
fd.append('var2', val2);