通过连接两个表获取数据：PHP

Question

I am trying t fetch the user_email based on the foreign Key I have set in the other table.

Suppose in one Table 'Users'I Have:

Users: id, user_email, user_phone etc.

and I Have other Table named patients as

patient_id, patient_name, doctor_id.

Where doctor_id is the foreign key to the Users Table.

I have the following code:

$query = "SELECT 'users.user_email' as 'name'
FROM users,patients 
WHERE 'users.id'='patients.doctor_id' AND 'users.id'='1';
" ;
 $result = mysqli_fetch_assoc($conn,$query) ;

while ($row = mysqli_fetch_assoc($result)) {
    echo $row['name'] ;
}

THe problem is its not working. its giving an error:

Warning: mysqli_fetch_assoc() expects exactly 1 parameter, 2 given in 
C:\xampp\htdocs\joli\data\foreigndata.php on line 10

Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, null 
given in C:\xampp\htdocs\joli\data\foreigndata.php on line 12

PS: The connection to the database is already established.

Answer 1

You should use mysqli_query() to run the query and then mysqli_fetch_assoc() to get the result as used below:

$query = "SELECT users.user_email as name
FROM users,patients 
WHERE users.id=patients.doctor_id AND users.id=1;
" ;

$result=mysqli_query($conn,$query);
while ($row = mysqli_fetch_assoc($result)) {
    echo $row['name'] ;
}

Answer 2

Try this

select users.user_email as name from users inner join patients on users.id=patients.doctor_id where users.id='1'

Answer 3

Try this

$query = "SELECT 'users.user_email' as 'name' FROM users
inner join patients on 'users.id'='patients.doctor_id'
WHERE 'users.id'='1'";

$result = mysqli_query($query);

while ($row = myslqi_fetch_array($result)) {
    echo $row['name'] ;
}

Answer 4

Like the error says, " mysqli_fetch_assoc() expects exactly 1 parameter, 2 given in "

mysqli_fetch_assoc() only expects 1 parameter, but you're passing 2.

Try this.

$query = "SELECT 'users.user_email' as 'name'
FROM users,patients 
WHERE 'users.id'='patients.doctor_id' AND 'users.id'='1';
" ;

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

while ($row = mysqli_fetch_assoc($result)) {
    echo $row['name'] ;
}

Answer 5

try

 $query = "SELECT 'users.user_email' as 'name' FROM users
    inner join patients on 'users.id'='patients.doctor_id'
    WHERE 'users.id'='1'";

    $result = mysqli_query($query);

    while ($row = myslqi_fetch_array($result)) {
        echo $row['name'] ;
    }