You are talking about two different concepts here: Data input: Relationships is one of the methods used to ensure data integrity. It restricts what data can be inserted to the database. Data retrieval: The LEFT JOIN that you mentioned is on of the methods used to retrieve the data from the database. It, kind of, … Read more
This looks like a terrible database design, so I would highly recommend changing it! Having duplicate IDs hurts me deeply. But, here’s a MySQL query that could get the result you are after: SELECT `ID`, `Date`, CONCAT(MIN(`Time`), ‘ ‘, MAX(`Time`)) AS `Time` FROM `test` GROUP BY `ID`, `Date`; 7 solved How to compare the current … Read more
Do it on command line: mysql -u USER -p DBNAME < SQLFILE.SQL Enter password when prompted and there you go. solved How long should a 75mb sql file take to import in PHPMyAdmin?
$query = “SELECT id FROM words where word =’abc'”; $result2 = $conn1->query($query); //fetch the data from the database while ($row = $result2->fetch_assoc() ) { echo “id is:”.$row[‘id’]; } where $conn1 is the connection variable $conn1 = new mysqli($servername, $username, $password, $dbname); solved How to get value of cell knowing neighboring cell from same row in … Read more
/* * SQL CREATE TABLE `NewTable` ( `id` int NOT NULL AUTO_INCREMENT , `col1` varchar(255) NOT NULL , `col2` varchar(255) NOT NULL , `col3` varchar(255) NOT NULL , PRIMARY KEY (`id`) ) ; * * /SQL */ $mysqli = new mysqli(“localhost”, “DB_USERNAME”, “DB_PASSWORD”, “DB_NAME”); if($mysqli->connect_error) { die(‘Connect Error’ . $mysqli->connect_error); } $mysqli->query(“SET NAMES ‘utf8′”); $mysqli->query(“SET … Read more
Try this: mysql_query(” INSERT INTO m_select (id, id_m, hello, bye) SELECT ‘$U’ AS id, ‘$$t_id’ AS id_m, hello, bye FROM test1 WHERE id=’$test_id’ “); I’m not sure whether the double-dollar sign on $$t_id (rather than $t_id) is intentional or not, but I thought I’d at least make you aware of it. 4 solved MySQL INSERT … Read more
Error parsing dataorg.json.JSONException: Value <br of type java.lang.String cannot be converted to JSONObject solved Error parsing dataorg.json.JSONException: Value
There’s a lot going wrong here, but I don’t mean to blame you specifically. Many of the things that make PHP popular, like it’s long history of support on shared hosting providers and how it’s easy to pick up and get going, are also a huge liability: There’s a lot of dangerously old advice out … Read more
Here is one way to solve the problem. It doesn’t hardcode any of the child node names (TI, MO, AU, etc.) so you could potentially use it on similar documents. I’ve put comments in the code so please read them and ask if you don’t understand what the code is doing. $txt=”your XML string goes … Read more
If you have a DATE MySQL object, you could use the DAYNAME function, and search by that: SELECT * FROM mytable WHERE DAYNAME(date) = ‘Monday’ In PHP/PDO-land, that would be: foreach($db->query(“SELECT * FROM mytable WHERE DAYNAME(date) = ‘Monday'”) as $row) { // do stuff with $row } 0 solved How to grab bunch of dates … Read more
Your update query need to change like this, hardwareID='”.$id.”‘ instead of hardwareID='”.$id[‘id’].”‘ 1 solved Warning: Illegal string offset ‘id’ in
you can try sth like this SELECT customers FROM orders WHERE price > (select AVG(price) from orders) ; solved I want to select customers whose price is higher than the average price
Without your actual database schema/more information it is difficult to give you the best advice possible. Add a field to the table radio so that it looks something like this: RADIO (*user*, *questionId*, Option1, Option2, Option3) (due to stack overflow formatting stars represent the primary key(s)) where questionID is a foreign key that references QUESTION(id). … Read more
In SQL-query replace all entries of ‘$out[]’ by {$out[]} And try to use IDE: NetBeans or PhpStorm. Also, don’t forget to sanitize your data in SQL, consider to use PDO and don’t use closing ?> tag. Your fixed code: <?php $link = mysql_connect(‘localhost’, ‘name’, ‘password’); if (!$link) { die(‘Could not connect: ‘.mysql_error()); } echo ‘Connected … Read more
$query = “SELECT a.*,b.mobilenumber,b.firstname,b.lastname FROM user_address a left join user b on a.user_id=b.user_id WHERE a.user_id IN (SELECT id FROM user WHERE email=””.$email.””)”; 1 solved Joining 2 fields into another table