inserting status for four inputs from select php - php

Question

I have four HTML inputs and a select option. i want to insert the status for the three inputs = "0" and for the selected one = "1". kindly help me and tell some thing about mysql query for
<input placeholder="Choice A:" name="a">
<input placeholder="Choice B:" name="b">
<input placeholder="Choice C:" name="c">
<input placeholder="Choice D:" name="d">
<select class="form-control" name="select">
<option>A</option>
<option>B</option>
<option>C</option>
<option>D</option>
</select>
i want the status for a is 1 and for the remaing 0 in Database :

Answer 1

I don't know what you want to achieve but as per your comments you are looking for this:
HTML:
<form method="post" action="">
<input type="text" placeholder="Choice A:" name="a">
<input type="text" placeholder="Choice B:" name="b">
<input type="text" placeholder="Choice C:" name="c">
<input type="text" placeholder="Choice D:" name="d">
<select class="form-control" name="select">
<option value="A">A</option>
<option value="B">B</option>
<option value="C">C</option>
<option value="D">D</option>
</select>
<input type="submit" name="submit" value="Submit Now">
</form>
PHP:
if(count($_POST) > 0){
$updateZero = array();
$updateOne = array();
$selectArr = array('A','B','C','D'); // default array
foreach ($selectArr as $key => $value) {
if($value == $_POST['select']){
$updateOne[] = $value; // store those selected
}
else{
$updateZero[] = $value; // store those not selected
}
}
echo "<pre>";
print_r($updateOne); // selected array
print_r($updateZero); // remaining array
}
Result:
Array
(
[0] => A
)
Array
(
[0] => B
[1] => C
[2] => D
)
Now you can use these array in MYSQL Statement as you need.