PHP，帮助显示相关表的查询结果

Question

I'm new to working with db's and for practice I created a simple database called "Company", with three tables: employee, department, position. The fields in the employee table are: id, name, dept_id, pos_id and am using foreign keys to relate department and position to employee.

I'd like to display the information like this:

Department 1
Directors: Kevin, Jane, Tom
Managers: Joe, Fred, Mary

Deparment 2
Directors: Bill, Elizabeth, Frank
Managers: Jennifer, Brian, Nicole

I used a JOIN statement to return all employees with their related positions and departments, but I can't figure out the correct PHP to display them like the example above. Should I have not used JOINS?

SQL:

$sql = "SELECT employee.name, position.pos_name, department.dept_name 
        FROM employee
        INNER JOIN position
        ON employee.pos_id = position.id
        INNER JOIN department
        ON employee.dept_id = department.id";

    $result = $dbconn->query($sql);

Any help would be great,

Thanks

Answer 1

First of all: Adapt your SQL to sort as needed

    $sql = "SELECT employee.name, position.pos_name, department.dept_name 
            FROM employee
            INNER JOIN position
            ON employee.pos_id = position.id
            INNER JOIN department
            ON employee.dept_id = department.id
            ORDER BY dept_name,pos_name";

    $result = $dbconn->query($sql);

now do a double group change loop:

$dname=false;

while (true) {
  $row=mysql_fetch_row($result);

  //done!
  if (!$row) break;

  if (!($row[2]===$dname)) {
    //Group change #1: We have a new department
    $dname=$row[2];
    echo "<br>\nDepartment: $dname";
    $pname=false;
  }

  if (!($row[1]===$pname)) {
    //Group change #2: We have a new Position
    $pname=$row[1];
    echo "<br>\n$dname: ";
    $separator='';
  }

  echo $separator.$row[0];
  $separator=', ';
}

Answer 2

What is $result? Is it a db resource, or does the query() method return an array?

Assuming $result is an associative array returned by the query() method, you should be able to iterate through it and use conditionals to grab the information you need.

foreach($result as $row)
{
    switch($row['department.dept_name']) {
        case "department 1"
           if($row['position.pos_name']) == 'Director') {
               $department1['directors'][] = $row['employee.name'];
           } else if($row['position.pos_name'] == 'Manager') {
               $department1['managers'][] = $row['employee.name'];
           }
        break;

        case "department 2"
          //repeat for each department

    }
}

This will give you an associative array for each department, then you can iterate over those and echo out HTML in the format you described.

If $result is a DB resource and not an array, then you would need to do something like:

while($row = mysql_fetch_assoc($result))
{
    $data[] = $row;
}

and then use $data instead of $result in the foreach statement.

Answer 3

I can't test it and I'm not too sure if the condition should be in the JOIN or in a HAVING after, but try to test this:

    $sql = "
            SELECT department.dept_name,department.id,
            GROUP_CONCAT(DISTINCT directors.name SEPARATOR ', '),
            GROUP_CONCAT(DISTINCT managers.name SEPARATOR ', ')
            FROM department
                JOIN employee AS directors
                    ON directors.pos_id = x //director position
                    AND directors.dept_id = department.id
                JOIN employee AS managers 
                    ON managers.pos_id = x //managers position
                    AND managers.dept_id = department.id
            GROUP BY department.id"
$result = $dbconn->query($sql);
print_r($result->fetchAll());

EDIT: Sorry, my request was (very) wrong, should be fine now

Answer 4

while ($row = mysql_fetch_array($result)){
echo $row['name of column here']; // repeat this for all columns.
}