mysql-error-1064

I have a table which has 2 columns which I copied from two different tables.What I want to do now is give a foreign key constraint on both the column names email and id shown below. ALTER TABLE users_role_map ADD CONSTRAINT FK_users_role_map FOREIGN KEY (email) REFERENCES usert(email), FOREIGN KEY (id) REFERENCES rolet(id) ON UPDATE CASCADE ON DELETE CASCADE; I get the following error: ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'FOREI GN KEY (id) REFERENCES rolet(id) ON UPDATE CASCADE

i got some problem during open my old website. My dataTable show: DataTables warning: JSON data from server could not be parsed. This is caused by a JSON formatting error. After that, I tried to debug my script and found error in mysql: Error occuered during query execution: (<small>SELECT SQL_CALC_FOUND_ROWS ID,name,remark,avrusepmonth , CONCAT('<input type=''checkbox''id=''cb' , ID ,''' name=''check[]'' value=''',ID,''' >','<label class=''lbcb'' for=''cb', ID,'''><=update=</label>') as checkb ,monthavrage(ID,12) as latestavr , moq, leadtime FROM test_media WHERE nowuse=1 and monthavrage(ID

I have a Table and Rows in table like below CREATE TABLE Areas(AreaName VARCHAR(255), PinCode VARCHAR(255)) INSERT INTO Areas(AreaName, PinCode) VALUES('Teynampet', '6000018'), ('Ramapuram', '6000089'), ('TNagar', '6000017'), ('Mylapore', '6000014'), ('Gopalapuram', '6000087') I Wrote a SQL Procedure as Below DROP PROCEDURE IF EXISTS mp_test; CREATE PROCEDURE mp_test(IN pArea VARCHAR(255)) BEGIN SET @Query = 'SELECT PinCode FROM Areas'; IF pArea != '' THEN SET @City = CONCAT(' WHERE AreaName = ', pArea); END IF; SET @Query = CONCAT(@Query, @City); PREPARE stmt FROM @Query; EXECUTE stmt; END

I would like to check if the website can connect to mySQL. If not, I would like to display an error saying that the user should try to access the page again in a few minutes... I really do not know how to do this ;) Any help would be greatly appreciated! string mysql_error ([ resource $link_identifier ] ) But how do I use this? This just gives me the error, but I want the message to display with any error. Thanks Try this: <?php $servername = "localhost"; $database = "database"; $username = "user"; $password = "password"; // Create connection $conn = new mysqli($servername, $username,

I have a program with a MySQL insert query: $sql = "INSERT INTO people (person_id, name, username, password, email, salt) VALUES ($person_id, $name, $username, $password, $email, $salt)"; if ($con->query($sql) === TRUE) { successful(); } else { unsuccessful("Error: " . $sql . "<br>" . $con->error); } When I run the code I get this error: Error: INSERT INTO people (person_id, name, username, password, email, salt) VALUES (2, Tom Jack, tjack95, 8a01fc598a676a249c5844bd3baf4f4d83cecdf2, tom@tommy.com, eb694539989fda2e17f79e70fb306f8911c3dee5) You have an error in your SQL syntax; check the manual