联接mySQL表时如何从主键和外键获取值？

Question

I have two tables, employee as the parent and license as the child. They both have a Lic_ID column for reference, this column is the PK in license and the FK in employee . The license table also has a column Lic_Type which holds the name of the license.

I am trying to create a table with list boxes so the employee table can be updated. The list box value needs to be populated with the license.Lic_ID and the license.Lic_Type is to be displayed in the option . Here is what I have:

(Employee name, Id, etc. called out up here)

<?php
echo "<select name=\"Lic\">";
echo "<option value=\"\">Select...</option>";

$sql = $mysqli->query("SELECT Lic_ID, Lic_Type FROM license");

while($row = $result->fetch_assoc())
     {
     echo "<option value=\"" . $row['Lic_ID'] . "\">" . $row['Lic_Type'] . "</option>";
     }

echo "</select>";
?>

So that works good, it shows the license type and has the value set to the license id. What I want to do is have <option selected="selected"> if the license id is set for an employee. This code doesn't work, but I think it illustrates what I'm trying to do:

<?php
echo "<select name=\"Lic\">";
echo "<option value=\"\">Select...</option>";

$sql = $mysqli->query("SELECT license.Lic_ID, license.Lic_Type, employee.Lic_ID FROM employee INNER JOIN license ON employee.Lic_ID = license.Lic_ID");

while($row = $result->fetch_assoc())
     {
     echo "<option value=\"" . $row['license.Lic_ID'] . "\"";
         if($row['employee.Lic_ID'] = $row['license.Lic_ID']){echo "selected=\"selected\";}
     echo ">" . $row['license.Lic_Type'] . "</option>";
     }

echo "</select>";
?>

Is there a way to accomplish what I'm trying to do?

Answer 1

I think there may have been some confusion on what exactly I was trying to accomplish, I apologize for not being very clear. Anyways, I stumbled over the answer today, so I thought I should post it.

$sql1 = ("SELECT Emp_Name, Lic_MAT_ID FROM employee");

if(!$result_employee_query = $mysqli->query($sql1))
    {
    die ("There was an error getting the records from the employee table");
    }

while($employee = $result_employee_query->fetch_assoc())
    {
    echo "Employee Name: " . $employee['Emp_Name'] . "<br>";

    echo "License: ";
    echo "<select>";

    $sql2 = ("SELECT Lic_MAT_ID, Lic_MAT_Type FROM license_mat");

    if(!$result_license_query = $mysqli->query($sql2))
        {
        die ("There was an error getting the records from the license table");
        }

    while($license = $result_license_query->fetch_assoc())
        {
        echo "<option value=\"" . $license ['Lic_MAT_ID'] . "\"";
            if($license['Lic_MAT_ID'] == $employee['Lic_MAT_ID'])
                {
                echo " selected=\"selected\"";
                }
        echo ">" . $license ['Lic_MAT_Type'] . "</option>";
        }

    echo "</select><br>";
    }

Answer 2

As I understand your problem: You want to see if the License has been added to ANY users or has it been unassigned. If any of the users have the license set then it's "selected", othervize not.

First you have to assign the keyword "multiple" to your select object to make it a listbox:

echo "<select name=\"Lic\" multiple=\"multiple\">";

Second: I would write this kind of query:

$sql = $mysqli->query("SELECT l.Lic_ID, l.Lic_Type, e.cnt FROM licence l left outer join (select Lic_id, count(*) cnt from employee group by Lic_id) e on l.Lic_ID=e.Lic_id");

It selects the Lic_ID , Lic_Type and the count of how many employees have the respective Lic_ID set to it ( left outer join )

and in the code just check, if the count is higher then 0

if($row['cnt'] > 0){ 
  echo "selected=\"selected\";
}