Database test content statistics userName=" fight"a total of several items.
:
:
data shows SQL COUNT Grammar:
SQL COUNT (DISTINCT column_name) Syntax
COUNT (DISTINCT column_name) function returns the number of different values for the specified column:
SELECT COUNT (DISTINCT column_name) FROM table_name
Why can"t I display the data I need? I want to get the statistical data back to the front end for judgment. Which experienced friend can advise me as a beginner in sql and PHP
$sql = "select count (*) as userNum from ajax_test1 where userName='$uName'";
$result = mysqli_query ($con, $sql);
$res = mysqli_fetch_array ($result);
/ / print_r ($res);
echo $res ['userNum'];
there is a problem with your sql splicing. UserName='$uName' is not parsed
single quotation marks do not resolve variables
now there are two tables, as shown in the figure the question is: how do I write a sql statement to query the data that contains a field (user_id) in Table 2 (m_top_user_list)? Thank you ~ ...
it was fine before rebooting, but after rebooting, it will never be able to start again root@iZuf6acp86oa3jizjzwaqdZ:~-sharp service mysql start Job for mysql.service failed because the control process exited with error code. See "systemctl status...