I have about 5 million records in a database table (MySQL 5.6), and I wanted to get the last 2 dates, grouped by ID. Looking around the web, I found examples that allowed me to cobble together the following… Running this code from MySQL Workbench, returned 5,700 rows, which is what I expected. I then tried to call this SQL
Tag: php
SQL Like Operator not working as expected in PHP
I am trying to fetch all the data from the database with a certain content tags value. If I run the code below, it only returns the row with id 1, even though the word tag is part of the tags value in both rows. I also added %or% operator but it is fetching the exact row. When I search
update sql table from two tables Using a single query
i want to update a table column by taking i/p from two tables in a single query. This is what i have tried. But this is not updating the table. Where Table-all_stores is: And table-i_v is: Answer if I understand correctly , you want to do this :
Data from 2 tables is not displaying on the same page
I’m a newbie in PHP and mySQL, I’m currently working on a profile page where I will display the logged in user’s personal information from one table called users and also display the tours that they will book in the future from my website from this table: booking I’m fetching the data from users table with prepared statement which will
How to add escape $ in Zend framework query
I need to add escape ‘$’ in zend framework. expected query: SELECT * FROM V_HOME_SEARCH sv WHERE sv.NAME LIKE ‘%gar$_%’ ESCAPE ‘$’ I tried this: public function selectData($str){ $select = $this->…
Laravel/MariaDB: errno 150 “Foreign key constraint is incorrectly formed”
I’m on Laravel 5.4, PHP 5.6, Ubuntu 18.04, MariaDB 10.4.8. When I run php artisan migrate, I get: I’m trying to use https://github.com/klisl/laravel-comments. Before trying to perform a migration with this package I had created DB at phpMyAdmin, had configured .env by adding DB name and stuff, had successfully run php artisan migrate, php artisan make:auth and php artisan make:controller
How to replace Over clause in sql version 4.9.5
I am trying to run a query to find the count and percentage of artisan based on gender. The query is running absolutely fine on the SQL server 8. But not on my live server which is 4.9.5. The below is my query. Answer In any database, you should be able to use:
PHP select query print out numerous items from database [closed]
My Table So I want it to print off like Coffee Name: ColumbianPrice: 7.99 Total Sold: 3 Total earnings: 23.9 And then to repeat for each type of coffee Is there any web pages that allow me to do this …
Laravel (8.x) Is there a better Eloquent query for this many-to-many filtering problem?
What I have movies many-to-many stars movies many-to-many tags studio one-to-many movies I pass studio name, star name and/or tag name as optional query strings What I need Select all movies with tag and star and studio if all 3 are passed as params Select all movies with tag and star if those 2 are passed as params Select all
Javascript won’t work properly with my form
So I am currently doing a project for our class, in which I have to display something that I previously stored in a database on a separate page. I chose to display every dataset, as it is the easiest to do. With PHP I implemented a foreach loop which is supposed to show each dataset with a button “EDIT” underneath