convert string to variable within prepared statement - php

Question

I am creating a php script that imports a csv file into an existing mySQL database.
I am selecting the matching column headings by checking if the heading in in an array and trying to use the column number to create my input into my prepared statement.
while (($data = fgetcsv($handle, 1000, ",")) !== FALSE) {
$numCols = count($data);
$row = array();
// process the first row to select columns for extraction
if($isFirstRow) {
$num = count($data);
for($col=0; $col<count($columns); $col++){
for($c=0; $c<$numCols; $c++)
if(!in_array($data[$c], $columns[$col])){
if($c == ($numCols-1))
{
$matchingCol[$col] = '""';
}
}
else{
$matchingCol[$col] = 'data['.$c.']';
apc_store("foo$c", $matchingCol[$col]);
}
}
$isFirstRow = false;
}
$data = array(
'contractorName' => (apc_fetch('foo1')) ,
'contractorType' => $matchingCol[3]);
$query = "INSERT INTO uploadSQL SET" . bindFields($data);
$result = $pdo->prepare($query);
$result->execute($data);
The data posted into the database is '$data[3]', '$data[5]' etc.
How can I get the INSERT to input the data stored at $data[3] and not the string '$data[3]'?

Answer 1

Use $data[$c] instead of 'data['.$c.']'...!?
Single quoted strings will display things almost completely "as is." Variables and most escape sequences will not be interpreted.

Answer 2

Not an exact solution to the problem but a work around. I'm sure there's a more efficient way.
if($isFirstRow) {
$csvRow1 = $data;
}
$num = count($data);
for($col=0; $col<count($columns); $col++){
for($c=0; $c<$numCols; $c++){
if(!in_array($csvRow1[$c], $columns[$col])){
if($c == ($numCols-1))
{
$matchingCol[$col] = '""';
}
}
else{
$matchingCol[$col] = "$data[$c]";
$c = $numCols;
}
}
}