javascript 如果文件名相同，则更改事件功能不适用于输入类型文件
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Change event function not working on input type file if file name is same
提问by user1881845
This is my html code to upload excel file
这是我上传excel文件的html代码
<span class="control-fileupload" >
<label for="file1" class="text-left">{{filePlaceHolder}}</label>
<input type="file" id="file1" value="" (change)="openFile($event)" >
</span>
But the problem is if i'm uploading same file twice the change function is not executing again because there is no change in input field.
但问题是，如果我上传相同的文件两次，更改功能不会再次执行，因为输入字段没有更改。
Suppose i have uploaded abc.xls file once and there are some validation on this file and if i change the content of abc.xls and re upload it then change function is not re validating it again.
假设我已经上传了一次 abc.xls 文件并且对该文件进行了一些验证，如果我更改了 abc.xls 的内容并重新上传它，那么更改功能不会再次重新验证它。
What changes i should make to work change function every time i upload a file whether file name is same or not.
每次上传文件时，无论文件名是否相同，我应该对工作更改功能进行哪些更改。
I want to know how to write this click function in type script as i'm new to this.
我想知道如何在类型脚本中编写这个点击函数，因为我是新手。
回答by guramidev
In angular 2 you can do it like this:
在 angular 2 中，您可以这样做：
<span class="control-fileupload" >
<label for="file1" class="text-left">{{filePlaceHolder}}</label>
<input #fileInput type="file" id="file1" (click)="fileInput.value = null" value="" (change)="openFile($event)" >
</span>
This way every time you click on file input it will clear it's value so even if you select the same file change will fire.
这样每次单击文件输入时，它都会清除它的值，因此即使您选择相同的文件更改也会触发。
