This happens when your result is not a result (but a “false” instead). You should change this line to this: because the ” can interprete $variables while ‘ cannot. Works fine with integers (numbers), for strings you need to put the $variable in single quotes, like If you want / have to work with single … Read more
I’m getting the above error when running the below code to display bookings made from a database. I only started to receive the error when i started hosting it live. It works fine on my personal computer, the databse connection works fine also.
You are mixing mysqli and mysql function. If your are using mysql function then instead mysqli_real_escape_string($your_variable); use If your using mysqli_* function then you have to include your connection to database into mysqli_real_escape function : Note : Use mysqli_* function since mysql has been deprecated. For information please read mysqli_*
I think, there is something wrong with PHP configration.First, debug your database connection using this script at the end of ./config/database.php : Then see what the problem is.
Have you checked the return value of mysqli_query()? It returns FALSE if an error occurred. In that case mysqli_error() gives you more information about the error.
That’s because there was an error in your query. MySQli->query() will return false on error. Change it to something like:: That should throw an exception if there’s an error…
The mysqli_query() method returns an object resource to your $result variable, not a string. You need to loop it up and then access the records. You just can’t directly use it as your $result variable.
i just update my server. it showing an error today Deprecated: mysql_query(): The mysql extension is deprecated and will be removed in the >future: use mysqli or PDO instead in C:\wamp\www\work\db\dbfields – Copy.php on line 33 my dbfields – Copy.php page is mysql_query(“insert into user(name,address) values(‘$name’,’$address’)”); i create 2 columns (name&address), need to insert the … Read more
I made a PHP page that is supposed to select two names from a database and displays them. It just says: Warning: mysqli_query() expects at least 2 parameters, 1 given in /home/tdoylex1/public_html/dorkhub/index.php on line 4 Warning: mysqli_query() expects at least 2 parameters, 1 given in /home/tdoylex1/public_html/dorkhub/index.php on line 8 My code is:
yo need create the user “pma” in mysql or change this lines(user and password for mysql): Linux: /etc/phpmyadmin/config.inc.php