Question

在我的表单中有多个文件上传，使用FormData只上传了一个文件，但我选择了多个要上传的文件，以下是代码

HTML

<form name="uploadImages" method="post" enctype="multipart/form-data">
<input type="file" name="photo[]" value="">
<input type="file" name="photo[]" value="">
<input type="file" name="photo[]" value="">
</form>

JS

     var ajaxData = new FormData();
     ajaxData.append( 'action','uploadImages');
     jQuery.each($("input[name^='photo']")[0].files, function(i, file) {
        ajaxData.append('photo['+i+']', file);
      });
     $.ajax({
        url: URL,
        data: ajaxData,
        cache: false,
        contentType: false,
        processData: false,
        type: 'POST',
        dataType:'json',
        success: function(data) {
            if (data.status == 'success') {
                location.reload();
            }
        }
       });

我在服务器上使用PHP，使用HTML attribute name i，e photo我只能保存文件，动态文件名对我不起作用。< / p>

Answer 1

你在javascript中有一个错误：你只在一个输入中迭代文件，请看看这个

var ajaxData = new FormData();
ajaxData.append( 'action','uploadImages');
$.each($("input[type=file]"), function(i, obj) {
        $.each(obj.files,function(j, file){
            ajaxData.append('photo['+j+']', file);
        })
});

示例jsfiddle

Answer 2

在前端

<form name="uploadImages" method="post" enctype="multipart/form-data">
    <input type="file" name="photo[]" value=""/>
    <input type="file" name="photo[]" value=""/>
    <input type="file" name="photo[]" value=""/>
    <button id="btn">btn</button>
</form>
        <script>
            $(function(){
     var ajaxData = new FormData();
     ajaxData.append( 'action','uploadImages');
     $.each($("input[type=file]"), function(i, obj) {
        $.each(obj.files,function(j, file){
            ajaxData.append('photo['+j+']', file);
          $('#btn').on('click',function(){
        $.ajax({
        url:'https://stores-govan.c9users.io/test',
          type:"POST",
          data:ajaxData,
          processData:false,
          contentType:false,
          success:function(){
            },
          crossDomain:true
        })
        })

        })
     });

})
</script>

在后端（nodejs）添加（使用multer）

var multer=require('multer')
app.post('/test',upload.array('photo[]',6),function(req ,res,next){
            var images=[]
               if(req.files){
               for(var i=0;i<req.files.length;i++){
               images[i]=req.files[i].filename }
               }
               console.log(images)
        })

Answer 3

xquery version "1.0-ml";

import module namespace admin = "http://marklogic.com/xdmp/admin" at "/MarkLogic/admin.xqy";

declare namespace db = "http://marklogic.com/xdmp/database";

let $config := admin:get-configuration()
let $database-id := xdmp:database()

let $existing-index := admin:database-get-range-element-indexes($config, $database-id)[
  db:namespace-uri = "http://www.dummy.com/namespaces/dummy"
    and db:localname = "dummyValue"
]
let $config := admin:database-delete-range-element-index($config, $database-id, $existing-index)

let $config :=  admin:database-add-range-element-index($config, $database-id, admin:database-range-element-index(
  "gYear",
  "http://www.dummy.com/namespaces/dummy",
  "dummyValue",
  (), 
  fn:false(),
  "reject"
))

return
  admin:save-configuration($config)

上一个答案有一个小错误，修复了下一个代码，并且可以通过ajax发送多个文件：

<input type="file" name="Attachment[]" class="form-control TheFiles" />

可选择只为你可以看到我的ajax配置

var formData = new FormData();
        $.each($(".TheFiles"), function (i, obj) {                
            $.each(obj.files, function (j, file) {                    
                formData.append('Attachment[' + i + ']', file); // is the var i against the var j, because the i is incremental the j is ever 0
            });
        });
        formData.append('Destination', Destination); //add more variables that you need
        formData.append('ReplyTo', ReplyTo);//add more variables that you need
        formData.append('Body', Body);//add more variables that you need

Answer 4

这些答案无效。

var ajaxData = new FormData();
ajaxData.append( 'action','uploadImages');
$.each($("input[type=file]"), function(i, obj) {
    $.each(obj.files,function(j,file){
        ajaxData.append('photo['+j+']', file);//i had to change "i" by "j"
    })
});

Answer 5

这很好：

var data = new FormData();
for( var i = 0, len = document.getElementById('attachment').files.length; i < len; i++ ){
    data.append( "files" + i, document.getElementById('attachment').files[i] );
}

JS：如何使用FormData发送多个文件（jQuery Ajax）

5 个答案: