如何根据特定的键值通过比较两个表（在数据库中）获得多个计数

Question

this is my 1st post so excuse me if my question is a bit weird. I work on a test project for an android application that send some data to a server. The java part is already done but I don't have much knowledge on serverside things and databases.So I need to find a way to compare two tables. A small one and a much bigger. For example,the small one is a list of a company's employees and the bigger is a database of companies and their employees. I need to make an array with the counts of the similar rows from the comparison of the two tables. I know my English are a bit rusty so I will try to be more clear:

company x:

|id  |emp_name|age|
|2314|bob     |33 |
|2314|sam     |45 |
|2314|paul    |28 |

companies database:

|id  |emp_name|age|
|1342|nick    |33 |
|1342|helen   |38 |
|1342|john    |52 |
|...              |
|6742|bob     |33 |
|6742|julia   |36 |
|6742|paul    |28 |
|...              |
|7654|sam     |45 |
|6742|bob     |33 |
|6742|paul    |28 |
|...              |

so for the above example I need to make an array like this:

|0|2|3|.....| 

I don't care about the company id, just for names and age.

Answer 1

May be this will help

$arr=array(0=> array("id"=>'1', "emp_name"=>"bob", "age"=>33),1=> array("id"=>'2', "emp_name"=>"sam", "age"=>45),2=> array("id"=>'3', "emp_name"=>"paul", "age"=>   28));
$company=array(0=> array("id"=>'1', "emp_name"=>"bob", "age"=>33),1=> array("id"=>'2', "emp_name"=>"samd", "age"=>45),2=> array("id"=>'3', "emp_name"=>"paul", "age"=>  28),3=> array("id"=>'1', "emp_name"=>"bobe", "age"=>33),4=> array("id"=>'2', "emp_name"=>"same", "age"=>45),5=> array("id"=>'3', "emp_name"=>"paul", "age"=>    28),6=> array("id"=>'1', "emp_name"=>"bob", "age"=>33),7=> array("id"=>'2', "emp_name"=>"sam", "age"=>45),8=> array("id"=>'3', "emp_name"=>"paul", "age"=>  28));


$result=array();


foreach($arr as $ar1){
  $i=0;
  foreach($ar1 as $ar2key=>$ar2val){

    foreach($company as $comp1){
      foreach($comp1 as $comp2key=>$comp2val){
        if($ar1["emp_name"]==$comp1[$comp2key]){
           $i+=1;   
           $result[$comp2val]=$i;

         }
      }    
    } 
  }
}

print_r($result);
//$result=array_values($result);
//print_r($result);

Answer 2

Simply use an aggregate SQL query to count the matching companies between both tables. Then, have PHP run the query using its MySQL PDO or other database API into an associative array and/or designate a separate array and append query counts into it:

SQL Query

  SELECT s.emp_name, s.age
         COUNT(*) As emp_count
    FROM smalltable s
   INNER JOIN largetable l
      ON s.emp_name = l.emp_name
     AND s.age = l.age
GROUP BY s.emp_name, s.age

PHP Script

try {
    // OPEN CONNECTION
    $dbh = new PDO("mysql:host=$hostname;dbname=$database",$username,$password);    
    $dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

    $sql = "SELECT s.emp_name, s.age
                   COUNT(*) As emp_count
            FROM smalltable s
            INNER JOIN largetable l
            ON s.emp_name = l.emp_name
            AND s.age = l.age
            GROUP BY s.emp_name, s.age";

    // QUERY AND FETCH RESULTS
    $STH = $dbh->query($sql);
    $STH->setFetchMode(PDO::FETCH_ASSOC);
}

catch(PDOException $e) {  
    echo $e->getMessage();
    exit;
}

$values = [];
// ITERATE THROUGH ASSOCIATIVE ARRAY, $row
while($row = $STH->fetch()) {  
      $values[] = $row['emp_count'];  // APPEND emp_count TO $values ARRAY
      echo $row['emp_count'];         // ECHO emp_count
}

// CLOSE CONNECTION
$dbh = null;