Do it like this example: try to run this set @fulfillStatus = 1; select CASE when @fulfillStatus = 0 then ‘open’ when @fulfillStatus = 1 then ‘processing’ when @fulfillStatus = 2 then ‘complete’ when @fulfillStatus = 5 then ‘Shipped’ when @fulfillStatus = 9 then ‘void’ else “None” END as “Order Status” 3 solved MySql case … Read more
You almost for sure will need a sub query to achieve this one way or another. select movie.title, actor.name from movie inner join casting on movie.id = casting.movieid and casting.ord = 1 –Assuming this is the cast order flag inner join actor on casting.actorid = actor.id where movie.id in ( select movie.id from movie inner … Read more
SELECT users.ID, ownership_company_managers.*, company_user.*, phonebook_list.*, ownership_phonebook.* FROM users LEFT JOIN phonebook_list ON users.ID = ownership_company_managers.USER_ID LEFT JOIN ownership_phonebook ON … , LEFT JOIN … LEFT JOIN … WHERE users.ID=’4′ solved MySQL Query LEFT JOIN 5 tables
Hi you can try like this ALTER TABLE table_name ADD email varchar(100) not null you can change column name ALTER TABLE tableName RENAME COLUMN “oldcolname” TO “newcolname” datatype(length); you can modify also ALTER TABLE table_name MODIFY COLUMN column_name datatype; solved Why this sql statement doesn’t working —-“ALTER TABLE ALTER COLUMN email varchar(100) not null;”—-?
throw new Exception(“id not supplied”); This line throws the exception that causes the fatal error you’re seeing. It’s run under this condition: if(!isset($id)){ So obviously, the condition matches, which means that the $id variable is not set. Also, extract($_REQUEST) is extremely bad practice. Simple scope example: function foo($a) { $a = 5; echo $a; //5 … Read more
As suggested, prepared statements are the best way to achieve good protection from SQL injection. Shortened Example You will need to add entries to fill in all columns you wish to insert. $email = $_POST[‘e-mail’]; $fn = $_POST[‘firstname’]; $ln = $_POST[‘lastname’]; if ($stmt = $mysqli->prepare(“INSERT INTO `newcartdb`.`orders`(Email,Firstname,Lastname) values(?,?,?)”) { $stmt->bind_param(“sss”, $email, $fn, $ln); “sss” – … Read more
I think this should work. I used :start and :end as placeholders for your input values. The query selects all employees with their working dates when the working date is at least partially part of your input range. Visualization (I: input range, y: selected, n: not selected): ——–IIIIIIIIIIIIIIIIIIII———- –nnn-yyyy—yyyy———yyyyy—-nn- Query: SELECT * FROM workdates WHERE … Read more
It’s called Polymorphic relationship (association). And you can’t have a column (added_by in your case) that references two parent tables simultaneously. But what you can do to be able to use foreign key constraints is to have two nullable columns added_by_super_admin and added_by_admin only one of which will hold the value per record. CREATE TABLE … Read more
2014-02-18 05:25:38 is a very very specific time to query. You’re only using one of your variables. You probably mean something like this: $from=$_POST[“datefrom”]; $to=$_POST[“dateto”]; SELECT * FROM ss_orders where order_time >= ‘”.$from.”‘ AND order_time <= ‘”.$to.”‘ limit 60 Also, switch to mysqli or PDO and sanitize those inputs. If this is a public form … Read more
You query is returning false on the following line $dn = mysql_query(‘select user_name, user_email from users where user_id=”‘.$id.'”‘); You will need to find out why, you should not attempt anything untill it returns true: $sql = sprintf(“SELECT user_name, user_email FROM users WHERE user_id = %s”, mysql_real_escape_string($id)); $dn = mysql_query($sql); if(!$dn){ echo “mysql_query() failed”; exit(); } … Read more
I found concatenating must be done in where condition select table_date from table_name where concat(DATE_FORMAT(now(),’%Y-%m’),’-‘,table_date) < curdate() solved How can I use MySQL alias in Where condition
Use this CSS on your display element: .lineBreaks { white-space: pre-line; } For example in your page: <div style=”white-space:pre-line”> <?php echo $mysqlData ?> </div> 2 solved Php mysql tabs don’t output [closed]
Use PHP function utf8_encode Try: <input type=”text” id=”de” value=”<?php echo utf8_encode($row->de); ?>” class=”input” size=”50″/> Make sure that the IDE uses is configured as the default UTF-8. This is spot-checking, ie the entire return must place this function. To correct definitive in check as below: In every PHP output header, specify UTF-8 as the encoding: header(‘Content-Type: … Read more
What ever you have written is a JOIN clause that is used to combine rows from two or more tables, based on a related column between them. Including outer join or left join depends on the table structure and the requirement to pull data from those tables. IF you are looking for what is left … Read more
If your records are ordered by ID, you could use a query like this: (SELECT * FROM yourtable WHERE ID<5 AND catID=(SELECT catID FROM yourtable WHERE ID=5) ORDER BY ID DESC LIMIT 2) UNION ALL (SELECT * FROM yourtable WHERE ID>5 AND catID=(SELECT catID FROM yourtable WHERE ID=5) ORDER BY ID LIMIT 2) Please see … Read more