Let me explain fast what i want to do! I want to show similar rows from my database by a PHP term. I have a table called “games” and a column called “title” that titles are looks like “Rockstar – GTA V”. So i want to remove all words after dash and use new string as keyword to search in
Tag: mysqli
Follow Unfollow system
I am working on this site that follow and unfollow users just like Twitter follow. The query below works then I notice it doesn’t show me users I don’t follow because the user has been followed by somebody or is following somebody Below is my table structure In the user_follow_list Ben is following betty and tom. Now john follows a
Why does PHP 7.4 with mysqli only run in root directory? [closed]
Closed. This question needs debugging details. It is not currently accepting answers. Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question. Closed last year. Improve this question Recently my hosting site (HostGator) upgraded my site’s PHP from 5.4 to 7.4. Pages
How can I add a parameter to a callback function like in JavaScript?
I wanted to make a small function that let’s me connect to the database, do something in a callback function and then automatically close the connection like so: Usage would be like this: Is it possible to achieve this somehow? Because my mysqli_conn doesn’t seem to work. How to do it? Answer Supposing the $c_… are globals / defined in
Do we need to test the return value of bind_param when building prepared statements in PHP? [closed]
Linked question is here. I have upgraded the code in the linked question to use a prepared statement. I now have: $stmt = $conn->prepare(“INSERT INTO `workbook-data` (`workbook-language`, `gui-…
populating my dropdown menu using procedural prepared statement
I would like to ask on how do I populate my dropdown/ via retrieving my data in MySQL database. I am using procedural prepared statement to add a security or avoid SQL injection. Problem:…
How do I convert this mysqli array statment into a mysqli prepared statement?
How do I rewrite this mysqli statement into a prepared statement? I tried converting the code provided below into the prepared statement. But, only the first row of values get inserted not the rest. I am trying to integrate dynamic text fields into my form. So, user can easily add or remove input fields. Original code: action.php my try: Answer
Show only user from the same company
I want ask how it’s possible when an user Type=Chef logs into the website, in the Dashboard it only shows the user from his company Foreign key (fk_FirmaID). I don’t how process it to only show a specific company. Here is my code that I have for the moment. When an user is logging in I’m saving his user details
Use more placeholders in WHERE statement after IN statement with arrays
I need more placeholders to be added to the following query. Is this possible? I have no idea where to begin and I can not find any options on the Internet. The query now: $in2 = str_repeat(‘?,’, …
Check availability of appointment slot
I am creating an area for a user where he can make an appointment. The available appointments are inside a table called: bigo_appuntamento This table has these 5 fields bigo_appto_id –> ID …