If you are using MySql 8.0+ you can create a CTE that returns the rows of the ids that you want to update and their marks and join it to the table:

WITH cte(studentId, marks) AS (VALUES
  ROW(1, 25), ROW(2, 22), ROW(5, 27), ROW(7, 30), ROW(9, 24)
)
UPDATE tablename t
INNER JOIN cte c ON c.studentId = t.studentId
SET t.marks = c.marks

See the demo.

For previous versions, instead of the CTE you can join a query that uses UNION ALL:

UPDATE tablename t
INNER JOIN (
  SELECT 1 studentId, 25 marks UNION ALL
  SELECT 2, 22 UNION ALL
  SELECT 5, 27 UNION ALL
  SELECT 7, 30 UNION ALL
  SELECT 9, 24 
) c ON c.studentId = t.studentId
SET t.marks = c.marks

See the demo.

You can try building a single update with a CASE statement, like so:

UPDATE table1
    SET column1 = CASE
        WHEN cart_id = 1 THEN 11
        WHEN cart_id = 2 THEN 12
        ...
        ELSE cart_id = cart_id
    END

You can execute one query only by using this :

$sql = "update live_tracking set swim_rank = '".$rank1[$i]."' WHERE id IN (";
for(i<7;$i++){    
   $sql .= i].",";
}

query = mysql_query($sql) or die("error : ".mysql_error());

if($query){
   echo "success";
} 

UPDATE mytable SET
    fruit = CASE WHEN id=1 THEN 'orange' ELSE 'strawberry' END,
    drink = CASE WHEN id=1 THEN 'water'  ELSE 'wine'       END,
    food  = CASE WHEN id=1 THEN 'pizza'  ELSE 'fish'       END
WHERE id IN (1,2);

Personally, using CASE WHEN THEN END looks clumsy.

You could code this using the IF function.

UPDATE mytable SET
    fruit = IF(id=1,'orange','strawberry'),
    drink = IF(id=1,'water','wine'),
    food  = IF(id=1,'pizza','fish')
WHERE id IN (1,2);

Give it a Try !!!

CAVEAT : CASE WHEN THEN END is only handy when dealing with multiple values (more than 2)

Here's my experimental attempt, though it fails. I'm posting this because you asked to see what I had tried. I don't claim this is a full solution to your question.

drop table if exists mytable;

create table mytable (
  id serial primary key,
  meta json
);

insert into mytable set meta = '...your example data...';

SELECT mytable.id, j.*
FROM mytable CROSS JOIN JSON_TABLE(
  mytable.meta,
  '$.industries[*]' COLUMNS(
    industry_num FOR ORDINALITY,
    industry_name VARCHAR(20) PATH '$.name',
    NESTED PATH '$.interests[*]' COLUMNS(
      interest_num FOR ORDINALITY,
      interest_name VARCHAR(20) PATH '$.name'
    )
  )
) AS j;

Output:

+----+--------------+---------------+--------------+---------------+
| id | industry_num | industry_name | interest_num | interest_name |
+----+--------------+---------------+--------------+---------------+
|  1 |            1 | Science       |            1 | Visual Arts   |
|  1 |            1 | Science       |            2 | Data Science  |
|  1 |            2 | Arts          |            1 | Music         |
|  1 |            2 | Arts          |            2 | Visual Arts   |
+----+--------------+---------------+--------------+---------------+

Now that we have a method for identifying the array index at each level of the JSON nested objects, we can try to UPDATE the value:

WITH cte AS (
  SELECT mytable.id, j.*
  FROM mytable CROSS JOIN JSON_TABLE(
    mytable.meta,
    '$.industries[*]' COLUMNS(
      industry_num FOR ORDINALITY,
      industry_name VARCHAR(20) PATH '$.name',
      NESTED PATH '$.interests[*]' COLUMNS(
        interest_num FOR ORDINALITY,
        interest_name VARCHAR(20) PATH '$.name'
      )
    )
  ) AS j
)
UPDATE mytable JOIN cte USING (id)
SET meta = JSON_REPLACE(meta, CONCAT('$.industries[', cte.industry_num-1, '].interests[', cte.interest_num-1, '].name'), 'New Interest Name')
WHERE cte.interest_name = 'Visual Arts';

The problem that I have not solved is that this appears to only do one replacement.

JSON_REPLACE() takes variable arguments, so you can change the value at more than one path, but constructing that call would require dynamic SQL. I'm not interested in doing any more work on this, because in my opinion, the work above should demonstrate that managing this data in JSON is already more trouble than it's worth.

Finally i got it, after several tries i manage to get what i needed, so i'll share my answer.

First i made one procedure to update one single row in the column:

delimiter //
create procedure update_amount_products (in id int)
begin
update categories set products_amount= 
           (SELECT DISTINCT COUNT( products_to_categories.products_id ) 
FROM categories_description
INNER JOIN products_to_categories ON products_to_categories.categories_id = categories_description.categories_id
INNER JOIN products_description ON products_description.products_id = products_to_categories.products_id
WHERE categories_description.categories_id =id)
WHERE categories_id = id;
end
//

DELIMITER ;

Then i realized i could handle the whole column by calling the procedure with another procedure with a loop on it

DELIMITER $$
CREATE PROCEDURE update_products_amount()

   BEGIN
      DECLARE a INT Default 1 ;
      DECLARE category_length int;

      select max(categories_id) into category_length from categories_description;

      simple_loop: LOOP
         SET a=a+1;

         call update_amount_products(a);

         IF a >= category_length THEN
            LEAVE simple_loop;
         END IF;
   END LOOP simple_loop;
END $$
DELIMITER ;

commit;

Finally, all i have to do now, is to call my last procedure and voilá, all my column will be automatically updated.

call update_products_amount();

just remove this from school from your update query and give a name to your table .

like that

 $schoolinfo = mysqli_prepare($con, "UPDATE school SET firstname=?, lastname=? where   foreignkey='$id'");

you mixed between SELECT and UPDATE.

i dont know if im wrong or , you have binded firstname and lastname before the loop .

try this

$schoolinfo = mysqli_prepare($con, "UPDATE school SET firstname=?, lastname=? WHERE foreignkey='$id'");

for ($i=0;$i<count($_POST['row']);$i++){     
$firstname = $_POST['firstname'][$i];
$lastname = $_POST['lastname'][$i];
mysqli_stmt_bind_param($schoolinfo,'ss', $firstname, $lastname)
mysqli_stmt_execute($schoolinfo);
}

I think this is due to the series of foreach within the foreach($id). I would write :

foreach($_POST['id'] as $index=>$id) {

to keep track of the index of your record, and then do not do any other foreach but rather :

$query = "UPDATE suppliers SET name = '".$_POST['n'][$index]."' where id ='".$id."'";

so you keep the link between the $id value and the other variables.

You need to use prepared statements in order to avoid errors and vulnerabilities of all sorts and also to get some minor performance gain

$stmt = $db->prepare("UPDATE site_email_templates SET Content=? WHERE ID = ?");
$stmt->bind_param("ss", $content, $id);
foreach ($postdata as $id => $content)
{
    $stmt->execute();
}

Reference: How can I prevent SQL injection in PHP?

Use this query:

UPDATE <TABLENAME> SET <COLUMNNAME>=REPLACE(<COLUMNNAME>, '.Z', 'OZ') WHERE <COLUMNNAME> LIKE '%.Z'

• is the name of the table you want to update (remove the <> chars)
• is the name of the column in the table you want to update (remove <>)

Replace is a MySQL function to replace characters in a string (https://www.w3resource.com/mysql/string-functions/mysql-replace-function.php)

LIKE is the search operator. You are only looking for values ending with .Z. The % is a wildcard to have anything at the beginning.