Show button whose value is present in database else hide the button

I have 5 columns in database id, a1, b1, c1 and d1. I want to show button a if the value in database for a1 is not there and if two values eg. c1 and b1 is present in database i want to show buttons for a and d.

Simple thing I want to show the button for that if there is no value in database.

I have tried the it by entering 16 conditions but it take a lot of time and space.

if(($row['a1'] == '') && ($row['b1'] != '') && ($row['c1'] != '') && ($row['d1'] != '')) {
?>
    <form action = "a1.php" method="POST">
        <input type="submit" name="a1" value="a1">
    </form>
<?php
} elseif (($row['b1'] == '') && ($row['a1'] != '') && ($row['c1'] != '') && ($row['d1'] != '')) {
?>
    <form action = "b1.php" method="POST">
        <input type="submit" name="b1" value="b1">
    </form>
<?php       
} elseif (($row['c1'] == '') && ($row['a1'] != '') && ($row['d1'] != '') && ($row['b1'] != '')) {
?>
    <form action = "c1.php" method="POST">
        <input type="submit" name="c1" value="c1">
    </form>
<?php
} elseif (($row['d1'] == '') && ($row['a1'] != '') && ($row['b1'] != '') && ($row['c1'] != '')) {
?>
    <form action = "d1.php" method="POST">
        <input type="submit" name="d1" value="d1">
    </form>
<?php   
}
?>

Is there any other way to do this.

Thanks in advance

Answer

here is the logic you can try.if you are going for specific row

step1: select the row according to id

$sql =  SELECT * FROM Customers WHERE id=xyx(your preferred id);

step2: fetch a assosiative array.

  $result=mysqli_query($con,$sql);

  // Associative array
 $row=mysqli_fetch_assoc($result);

step2: check for the condition.

 if(isset($row['a1'])){
       echo "<button></button>"
   }

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Answer