Efficiency can mean many things. Maintainability and scalability are as important as pure performance when you design a solution. Method 2 depends on files on disk which means that in a production environment with multiple servers (which every prod env should have) you will have to keep the files in sync between all servers in … Read more
If I understand your question correctly you could try something like the following. Edit: taken into account you already have a connection in config.php In config.php make sure you have the following. <?php $con=mysqli_connect(“hostname”,”username”,”password”,”database”); if (mysqli_connect_errno()) { echo “Failed to connect to MySQL: ” . mysqli_connect_error(); } ?> And then on the page where you … Read more
Try file hash code. You can store the original file hash code in the DB with the data and compare it against the input image. If it’s the same image they should have the same hash value. 1 solved How to compare an input image from user with a stored image on the database in … Read more
This worked the magic for me. <?php $dbLink = mysqli_connect(‘localhost’, ‘usr’, ‘passd’); mysqli_select_db(‘edudata’, $dbLink); $mytableselect=$_POST[‘mytableselect’]; $sql = “SELECT * FROM $mytableselect”; $result = mysqli_query($sql) or die(mysqli_error()); // Print the column names as the headers of a table echo “<table><tr>”; for($i = 0; $i < mysqli_num_fields($result); $i++) { $field_info = mysqli_fetch_field($result, $i); echo “<th>{$field_info->name}</th>”; } // … Read more
Consider the following data set… DROP TABLE IF EXISTS results; CREATE TABLE results (id_competitor INT NOT NULL ,score INT NOT NULL ,id_route INT NOT NULL ,PRIMARY KEY(id_competitor,id_route) ); INSERT INTO results VALUES (1,100,2), (2,100,2), (3,60,2), (4 ,60,2), (1,70,1), (2,80,1), (3,70,1), (4,100,1); SELECT * FROM results; +—————+——-+———-+ | id_competitor | score | id_route | +—————+——-+———-+ | … Read more
Please, don’t mix mysqli and mysql as they are different modules. In your second code block you are using mysql_select_db and mysql_error, the first one requires mysql connection, not mysqli connection. 1 solved use mysqli_connect and mysql_select_db [duplicate]
You currently have a function that defaults to showing 15 rows sorted by added. You can change this one of two ways: Change the function itself. This will make the DEFAULT value be site_views: function list_videos($how = ‘site_views’, $limit=”15″) { // newest, top views, etc etc etc Now if you call list_videos(), you will get … Read more
select Status, COUNT(Status) from Count() — please be sure to use the right name of the table after From. group by Status order by COUNT(Status) desc 🙂 1 solved I am new to SQL, and trying to learn what I can through books and websites
Familiarize with the data that you have: The first key thing is to understand what data you have. Here in this case, you have four tables InsuranceCompanies Patient Doctors Visits Your goal: Find the list of all the patients who visited all orthopedists (specialty) associated with their Insurance Companies. Let’s take a step back and … Read more
You will need to log the change in another table. Like changes with the attributes: id, table, attribute, old_value, new_value, datetime. Then, whenever something changes you add e.g. id, TESTING, last_name, Doe, DoKnow, 2015-03-06 12:34:56 Then you can use this table to show the latest changes. But in the long run and for big databases … Read more
$query = “SELECT * FROM ‘users’”; <?php $servername = “localhost”; $username = “username”; $password = “password”; $dbname = “myDB”; // Create connection $conn = new mysqli($servername, $username, $password, $dbname); // Check connection if ($conn->connect_error) { die(“Connection failed: ” . $conn->connect_error); } $sql = “SELECT * FROM users”; $result = $conn->query($sql); if ($result->num_rows > 0) { … Read more
Try die() like following in your php file: $connection = mysqli_connect($host,$user,$password,$database); /* check connection */ if (mysqli_connect_errno()) { die(“Connect failed: “, mysqli_connect_error()); } $query=”insert into users(firstname, lastname)VALUES(‘”.$_REQUEST[‘nome’].”‘,'”.$_REQUEST[‘cognome’].”‘)”; if ($result = mysqli_query($connection ,$query)) { print(“Success!”); } 3 solved Connect to Database and Insert data – Php & MySQL –
You have to create your service (which you can see by clicking right start menu bar and task manager). To create your service you have to open cmd as an adminstrator. Then write cd where/your/ sql/bin/. After that, write mysqld -install – you should see service succesfully installed. Go to services from task manager, find … Read more
UPDATE table1 a CROSS JOIN lookup b SET a.ADRES = REPLACE(a.ADRES, b.`WRONG`, b.`RIGHT`), gender=”$sex”, dob = ‘$dob’, reg_date=”$reg_date” WHERE a.ADRES LIKE CONCAT(‘%’, b.`WRONG`, ‘%’) OR id = ‘$id’ the query is vulnerable with SQL Injection, please see the article below to lear how to protect from it, How can I prevent SQL injection in PHP? … Read more
open “config.inc.php” on your PMA folder. add this line to prompt the authentication before login $cfg[‘Servers’][$i][‘auth_type’] = ‘cookie’; /* $cfg[‘Servers’][$i][‘user’] = ‘root’; */ /* $cfg[‘Servers’][$i][‘password’] = ”; */ OR manually define your authentication by this line $cfg[‘Servers’][$i][‘auth_type’] = ‘config’; $cfg[‘Servers’][$i][‘user’] = ‘root’; $cfg[‘Servers’][$i][‘password’] = ‘yourpassword’; solved Access denied with localhost [closed]