MySQL syntax to select table name from more than one choice that matches criteria

Question

Question is how to dynamically build one statement to return just one table name from array $tables that has the most records that match criteria $table.status = 'ready' .

I already understand I could create a combined subquery and count each, then sort using small php function, but part of my mission is to use MySQL statement for the heavy lifting because I am assuming it will be faster.

For example,

$tables = array('foo','bar','beyond','repair') ;

// start building the statement
   $query = "SELECT TABLE_NAME" ;
          // ? What else add here $query .= '????' ;


// loop to build part of the statement
   foreach($tables as $table){
         // create this statement dynamically
         // somehow I need to incorporate count(*)
         // and $table.status = 'ready'
         // note: all tables in $tables have a column named 'status'
         // ? What else add here $query .= "????" ;
    }

// add any remaining syntax 
// ? What else add here $query .= "????" ;

And as stated in the question, I want to select based on the highest count(*) of rows where $table.status = 'ready'

(My server environment is PHP 5.3.29 and MySql 5.1.73-cll. )

I can already solve this with one query and some php but I am asking if there is an answer to just return the TABLE_NAME with the one statement.

For example, I can solve like this:

$choices = array() ;

$query = "SELECT";
    foreach($tables as $table) {
           // loop and create subqueries, append table name with _count and we will strip it later 
           $query .= "(SELECT COUNT(*) FROM $table WHERE $table.status = 'ready') as ".$table."_count," ;
    }

$query = rtrim($query, ',');
$result = mysqli_query($db_connection,$query) or die("Sql error: " . mysqli_error($db_connection));
    while($row =  mysqli_fetch_assoc($result)) {
            while (list($key, $val) = each($row)) {
                    $choices[ str_replace('_count', '', $key) ] = $val;
            }
    }
arsort($choices ); // sort by value  reverse
$table_with_highest_count = key($choices) ; // the key will be table name now

Again, it solves the question with one query but forces me into php to complete, thus the question I wrote above.

Answer 1

If you look here you can see how to pass in variables and how to do a prepared statement:

http://php.net/manual/en/pdo.prepared-statements.php

It is important to use prepared statements to avoid SQL injection.

$tables = array('foo','bar','beyond','repair') ;

$maxcount = 0;
$maxTableName = '';

foreach ($tables as $table) {
    $stmt = $dbh->prepare("SELECT count(*) FROM :name WHERE status = 'ready'");
    $stmt->bindParam(':name', $table);

    $count = $stmt->fetch();

    if ($count > $maxCount) {
        $maxCount = $count;
        $maxTableName = $table;
    }
}

// Now $maxTableName is the one you are looking for.