How to show div depending on the value in a PHP variable

I am trying to show a div which contains an input for file upload, the rule that I need to comply with is to show div it only if the status is equal to 1, otherwise it will hide it.

I have the status declared in a PHP variable called $status, to hide the div I use the display='none'.

The following is the validation code that I use in a Javascript function.

function showFileInput (){
    var element = document.getElementById('content');
    if ($status == '1'){
        element.style.display='block';
    }else{
        element.style.display='none';
    }
}

In my HTML code I have the following div that I want to display when the status is equal to 1.

<div id="content" class="form-group row">
                <label for="fileToUpload" class="col-sm-3 col-form-label">File Upload:</label>
              <div class="col-sm-8">
                <input type="file" name="fileToUpload" id="XmlToUpload" class="btn" accept=".xml" onchange="ValidateFile()" required="">
              </div>
          </div>

It’s important to note that I make use of the event onchange for other additional validations that I use when uploading the file.

With what I have previously, the div is still not hidden when the status is different from 1, it continues to be shown. Is there something that I am doing wrong in my validation or do I need additional?

>Solution :

You can directly put a check on $status variable in css like this.

<div style=<?php echo $status == 1 ? "display:block;" : "display:none;" ?>>My content</div>