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how can i put this code into php echo?

   <label for="upload-file"> </label>
</center>
<a href="/rotkanye/"  onclick="window.open('/upload/', 'newwindow', 'width=300, height=250'); return false;"> 
<div class="upload-button">

<div class="upload-cover">muck
    </div>

</div> <!-- .upload-button -->
</a>

i want to show this button after upload is done, so i used if what i should put after if?

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  • If I understand you correctly, this is what you want: You have a link/button. If that button is pressed, another window opens - this window contains an upload form. Once this upload has been started and completed, you want another button to appear on the first html page? If so, that's not possible using PHP only. You can save the upload status on your server once it is completed and check this status from the first page via AJAX. Then you can insert the button via JavaScript. Commented Apr 22, 2017 at 12:55
  • I want upload button appear on the page, where is the file uploaded. Commented Apr 22, 2017 at 14:16
  • Does the upload need to be completed for the button to appear or is it enough that the upload window has been opened? Commented Apr 22, 2017 at 17:15
  • new window isnt needed, but it was easier to code this way, purpose of the button is upload file and redirect user to anither page, where is this image used Commented Apr 24, 2017 at 22:00
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