优雅高效的方式插入很多表MySQL

I am making a survey and currently the database has 19 tables and roughly 100 columns. All of them will be inserted to as they are required fields. I am looking for an Elegant and Efficient way of inserting into that many tables and columns. So far what I have come up with is to create a multidimensional array that contains the first key as the table name and the second key as the column name with the field. Something like below:

    $tableArray =  array( 
        'ownerTable' => array(
            'firstNameRow' => $firstName,
            'lastNameRow' => $lastName
        ),
        'dealerTable' => array(
            'dealerNameRow' => $dealerName,
            'dealerCityRow' => $dealerCity
            )
        );

foreach($tableArray as $row => $key) {
    foreach($tableArray[$row] as $row1) {
        $sql = "(INSERT INTO $tableArray[$key] ($tableArray[$row]) VALUES ($row1)"; 
    }   
}

I didn't test this code but I am thinking something along those lines would work. I think one problem I see with this is a separate INSERT for each column instead of one INSERT for each table. I can always work on writing the code to just load all the values from the array at once to solve this problem but before I start getting to carried away I want to make sure I am not making a big mistake and waste time if there is a better way to do it.

$tableArray =  array( 
        'ownerTable' => array(
            'firstNameRow' => $firstName,
            'lastNameRow' => $lastName
        ),
        'dealerTable' => array(
            'dealerNameRow' => $dealerName,
            'dealerCityRow' => $dealerCity
            )
        );

//it's for bulj query execution in single statment;
$i=1;
$sql="";
foreach($tableArray as $row => $key) {
    foreach($tableArray[$row] as $row1) {
        //here you can update i value as per row for execution
        if($i<=25)
        {
         $sql = $sql  + "(INSERT INTO $tableArray[$key] ($tableArray[$row]) VALUES ($row1);"; 
        }
        else
        {
            //execute sql statement here;
            $i=0;
            $sql="";
        }
    }   
}