You can do like this. Test data: create table tbl_cat (id int, cat_name varchar(32)); insert into tbl_cat(id, cat_name) values(1, “category1”), (2, “category2”); create table tbl_subcat (id int, cat_id int, subcat_name varchar(32)); insert into tbl_subcat(id, cat_id, subcat_name) values(1, 1, “subcat11”), (2, 1, “subcat12”), (3, 1, “subcat13”), (4, 2, “subcat21”); Then, PHP script can be written like … Read more
You can still use having, but test for each product — and for the absence of any other product: SELECT o.order_id FROM orders o GROUP BY o.order_id HAVING SUM( product_id = 222 ) > 0 AND SUM( product_id = 555 ) > 0 AND SUM( product_id NOT IN (222, 555) ) = 0 ; Note … Read more
This is how you have to add the data into a table. if (mysqli_num_rows($result) > 0) { echo “<table>”; echo “<tr> <th>First Name</th> <th>Last Name</th> <th>Appointment Time</th> </tr>”; while($row = mysqli_fetch_assoc($result)) { echo “<tr> <td>{$row[‘FirstName’]}</td> <td>{$row[‘LastName’]}</td> <td>{$row[‘AppTime’]}</td> </tr>”; } echo “</table>”; } else { echo “No results, please try again”; } If the issue is … Read more
The syntax for selecting values into variables in MySQL is select … into. For example you could write: SELECT POSITION, SecPosition FROM orderdetails WHERE OrderID = orderDetID INTO strPos, strPosOtherRes; The message “Not allowed to return a result set from a function” means that the select statements as they stand now would be returning a … Read more
after couple of hours i got my solution…thanks for your support..its works perfectly for me … SELECT * FROM table_name WHERE STR_TO_DATE(created_datetime, ‘%d/%m/%Y’) BETWEEN DATE_SUB(CURDATE(), INTERVAL 15 DAY) AND DATE_SUB(CURDATE(), INTERVAL 15 DAY) solved How did fetch records inserted on exactly on 15th day before from current date in mysql [closed]
If the empty fields have NULL value, you can use SELECT * FROM sometable ORDER BY ISNULL(price_1) + ISNULL(price_2) + ISNULL(price_3) DESC; But a more sensible solution would be: You have one table, which contains the products You have another table, which contains the product’s ID, the price and a value which indicates which website … Read more
Seems like a candidate for conditional aggregation: SELECT `Dia_Semana`, `Fetcha_Deposte` , IFNULL(ROUND(`Suma_Bondiolas s/hueso`, 2), 0) AS `Suma_Bondiolas s/hueso` , IFNULL(ROUND(`Suma_huesos_Bondiola`, 2), 0) AS `Suma_huesos_Bondiola` , `Suma_Bondiolas s/hueso` / `Suma_huesos_Bondiola` AS `Huesos Bondiola/Bondiolas sin huesos` FROM ( SELECT ELT(WEEKDAY(fecha) + 1, ‘Lunes’, ‘Martes’, ‘Miercoles’, ‘Jueves’, ‘Viernes’, ‘Sabado’, ‘Domingo’) AS `Día_Semana` , fecha AS `Fecha_Desposte` , SUM(CASE … Read more
yes, this is valid as google will tell you http://dev.mysql.com/doc/refman/5.0/en/select.html For your actual error, from the php docs: For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error. // Perform Query $result = mysql_query($query); // Check result // This shows the actual query sent to … Read more
There is one problem your not setting value and its empty. By which the values you select wont be added or updated to your database. The second is you have not given the name to your select dropdown by which even if you select the values it wont be posted to your action. <form action=”update_student.php” … Read more
$query = $this->db->query(“SELECT * FROM table-one JOIN table-two ON table-1-id = table-2-id”); return $query->result(); Note:- JOIN and ON are keywords to get data of both tables AND table-one and table-2 are your required tables AND table-one-id and table-two-id are the column names of both tables for the join solved How to Join Two Table In … Read more
You are doing nothing with $check, Make id like this $ids=””; if(!empty($_POST[‘id’])) { foreach($_POST[‘id’] as $id) { $ids[] = $id; } } $check = implode(“,”, $ids); 0 solved how to update multiple ids with different value
Write a Query in MySql server that prints a list of employee names who have been employed for less than 10 months having salary>2000 solved Write a Query in MySql server that prints a list of employee names who have been employed for less than 10 months having salary>2000
Without a better understanding of the rules and data this is the best I can come up with. It is based on your first array example – SELECT `r`.* FROM `rule_attribute_value` `rav` INNER JOIN `rule` `r` ON `rav`.`rule_id` = `r`.`rule_id` INNER JOIN `rule_attribute` `ra` ON `rav`.`attribute_id` = `ra`.`attribute_id` WHERE (`rav`.`store_id` = 0 AND `ra`.`attribute_code` = … Read more
store the date time format in YYYY-MM-DD H:I:S, so you can easily SELECT * FROM table_name WHERE start_datetime BETWEEN value1 AND value2 AND end_datetime BETWEEN value1 AND value2; 3 solved Between And Greater Than Date Should Be Implement In Same Query [closed]
Changed <input src=”https://stackoverflow.com/questions/49816990/tick.png” title=”Mark as read” type=”image” alt=”submit” name=”mark_read”> to <button type=”submit” name=”mark_read” style=”padding: 0; border: none;”><img title=”Mark as read” src=”https://stackoverflow.com/questions/49816990/tick.png” /></button> Input type failed to submit data because of the image type. works fine now. 1 solved PHP query not updating in db throught form