如何从MySQL中的索引字段输出另一个表的值？

Question

I have two MySQL tables

1. users with id, user_formOfAdress and serveral additional fields. Field user_formOfAdress contains the id of table formofadress

2. formofadress with id, formOfAdress_german and serveral additional fields, for example id 1 = Mr., id 2 = Mrs.

The record of the users table is identified by a Session variable.

To output not the id of the field user_formOfAdress but the value of the table formofadress.formOfAdress_german (for example Mr. or Mrs.) I have written this:

if(array_key_exists("id", $_SESSION) && $_SESSION['id']){
        $uid = $_SESSION['id'];
        $link = mysqli_connect("localhost:3307", "root", "Dinah123", "proficrm");
        $query = "
        SELECT formofadress.ID AS formofadress_ID, formofadress.formOfAdress_german, users.ID, users.user_formOfAdress 
        FROM `formofadress` 
        LEFT JOIN users 
        ON formofadress.formofadress_ID = users.user_formOfAdress
        WHERE `users.ID` = '".$uid."' 
        LIMIT 1
        ";
        $result = mysqli_query($link, $query);
        $record = mysqli_fetch_assoc($result); 
    $user_formOfAdress = $record['formOfAdress_german'];
        }

"FROM formofadress" because I want to output the Mr. or Mrs. of this table, but I have to use also the users table because of the Session ID which is also the id of the users record ...

Not every record in the users table has a value in user_formOfdAdress (value 1, 2 or NULL) but every record in the formofadress table has a fixed value.

Error is: Undefined variable: user_formOfAdress located in the last row

It's my first time to use JOINs and I'm unfortunately not able to solve this issue even after a long time of searching.

Correct code:

if(array_key_exists("id", $_SESSION) && $_SESSION['id']){
        $uid = $_SESSION['id'];
        $link = mysqli_connect("localhost:3307", "root", "Dinah123", "proficrm");
        $query = "
            SELECT formofadress.ID, formofadress.formOfAdress_german, users.ID, users.user_formOfAdress 
            FROM formofadress 
            LEFT JOIN users 
            ON formofadress.ID = users.user_formOfAdress 
            WHERE users.ID = '".$uid."' 
        ";
        $result = mysqli_query($link, $query);
        $record = mysqli_fetch_assoc($result); 
        $user_formOfAdress = $record['formOfAdress_german'];
    }

Answer 1

There were problems with your SQL syntax. You used LEFT JOIN to join users to formofadress while users sometimes can't follow and you will possibly get a NULL value. I also cleaned up your other query syntax so it is more readable.

Also, check to see if the database is expecting an INT for $uid . If not then return the apostrophes '' .

if(array_key_exists("id", $_SESSION) && $_SESSION['id'])
{
    $uid = $_SESSION['id'];
    $link = mysqli_connect("localhost:3307", "root", "Dinah123", "proficrm");
    $query = "SELECT u.ID, u.user_formOfAdress, f.ID, f.formOfAdress_german,  
    FROM users u LEFT JOIN formofadress f ON f.formofadress_ID = u.user_formOfAdress 
    WHERE u.ID = ".$uid." LIMIT 1";
    // if database is not expecting `INT` for `u.ID` then return the apostrophes ' '

    $result = mysqli_query($link, $query);

    // mysqli_fetch_assoc returns case sensitive keys
    $record = mysqli_fetch_assoc($result);

    $user_formOfAdress = $record['formOfAdress_german'];
 }