我正在尝试通过ajax上传多个文件,但我不知道如何在PHP中获取上传的文件。我发送给他们
var attachments = $('.attachment-file');
var post_data = new FormData();
if (attachments.length > 0) {
attachments.each(function(i, v) {
post_data.append('my_file', $(v)[0].files);
});
}
$.ajax({
cache: false,
contentType: false,
processData: false,
method: 'post',
type: 'post',
data: post_data,
url: form.attr('action'),
})
.done(function(response) {
...
但是当我尝试像这样获取 PHP 中的文件时:
$data = $this->request->request->all();
$files = $data['my_file']; \Log::Info(var_dump_safe($files));
foreach ($files as $file) {
它抱怨“为 foreach() 提供的参数无效”。转储显示字符串(17)“[object FileList]”。如果该文件在那里,我该如何获取它?
文件的 JS 日志 console.log($(v)[0].files);看起来像这样:
FileList(1)
0: File
lastModified: 1604656019000
name: "Screenshot_20201106_204451.png"
size: 66866
type: "image/png"
webkitRelativePath: ""
<prototype>: FilePrototype { name: Getter, lastModified: Getter, webkitRelativePath: Getter, … }
length: 1
<prototype>: FileListPrototype { item: item(), length: Getter, … }
\Log::Info(var_dump_safe($data)); 的整个表单数据转储;看起来像这样:
array(4) {
["topic"]=>
string(2) "24"
["title"]=>
string(5) "Title"
["content"]=>
string(4) "Test"
["my_file"]=>
string(8) "Array(1)"
}
以及 \Log::Info(var_dump_safe($data['my_file'])); 数组的转储:
array(1) {
[0]=>
string(17) "[object FileList]"
}
如何获取文件数据?