query result array in loop for another query - php

Question

I Queried Database Table 'users' for 'user_id'. and get an array of ids.
$sel = "SELECT user_id FROM users WHERE status='Approved'";
$result = #mysqli_query ($dbcon, $sel);
Then i inserted values into another table income for all those user ids.
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
{
$ins = "INSERT INTO income (user_id, income_amount) VALUES ('$row', '100')";
$giv = #mysqli_query ($dbcon, $ins);
}
Notice: Array to string conversion in E:\xampp\htdocs\project\t.php on line 109
Can anyone help me resolve this issue.

Answer 1

$sel = "SELECT user_id FROM users WHERE status='Approved'";
$result = #mysqli_query ($dbcon, $sel);
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
{
$ins = "INSERT INTO income (user_id, income_amount) VALUES ('" . $row['user_id'] . "', '100')";
$giv = #mysqli_query ($dbcon, $ins);
}

Answer 2

First , Check if $results is in array ..you can put some error handling checked is_array($result).
If it is fine then pass it to mysqli_fetch_array().
Do't add suppress # error ,while developing.

Answer 3

i would like to suggest you a single query for that so after that you need not to use while loop to insert your data in income table:
Just try it :
INSERT INTO income (user_id,income_amount) SELECT user_id,'100' AS income_amount FROM users WHERE status = 'Approved';
You can use it like that way :
$sel = "INSERT INTO income (user_id,income_amount) SELECT user_id,'100' AS income_amount FROM users WHERE status = 'Approved'";
$result = #mysqli_query ($dbcon, $sel);