Nothing obvious in the code, although you still need to parameterize the second query in the same way as the first, and you're still passing the password in the query-string to the registered.php page.
You could probably remove the else, since you've got an exit; in the if block.
Are you sure it's this page that's redirecting the user to the login page? Try tracing the network requests in your browser's developer tools, to see if either registered.php or dotnetba.php is doing the redirection instead.
"These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer
I don't know if php has something different but the easy way to do this in any web language is to use the querystring. index.php?var1=1&var2=2. Then on the next page you can read then from the querystring.
There are only 10 types of people in the world, those who understand binary and those who don't.
Hello there. I am trying to get data from 3 different tables based on simple join. One of the tables can have multiple values against one primary key. Here are the table designs
Table 1 - EmployeeDetails
EmployeeId INT, FirstName VARCHAR, SurName VARCHAR, SexId INT
Table 2 - EmployeeSex
SexId INT, Sex VARCHAR-- Values inserted-- 1, Male-- 2, Female
Table 3 - EmployeeContacts
EmployeeId INT, Contact VARCHAR-- may or may not be null AND can contain multiple values
I am using following query
SELECT ED.EmployeeId, ED.FirstName, ED.SurName, GROUP_CONCAT(EC.Contact)
FROM EmployeeDetails ED, EmployeeSex ES, EmployeeContacts EC
WHERE ED.SexId = ES.SexId AND ED.EmployeeId = EC.EmployeeId AND ED.EmployeeId = 'emp_password';
Now this query works fine if we have at least one contact number. But if there are not contacts, then this results in empty set. What is wrong with this query ? How can I improve so that it works in all scenarios (regardless of number of contacts in EmployeeContacts table). Thanks for any input.
Can anyone help me with the following two PHP scripts on the machine running Centos 7. Scripts written on a command line editor
1) a PHP script that will connect to the database and create the tables and populate default data, this script will be executed from command line using PHP -f createdatabase.php
2) the databasename will be the same as your username
3) a PHP script that will load from a browser and use HTML and CSS to display the populated information, the script will be loaded as http://localhost/myname/showme.php, you will get extra points for having fewer lines code, less SQL queries, and for commenting and indenting your code nicely.
The database contains the following tables with attributes:
Hai frnds, One of my frnd is searching for the vaccancy of fresher linux admin for last 1 & half year. till now he didnt get any job as trainee. He completed the b.tech in 2014. and he done the whole courses about that. kindly pls help me... any one know about that vaccancy..... pls tell me.. we are in banglore...
With Cloud Hosting, you are your own boss with your own mind and can go with anyone like Amazon, DigitalOcean, Google Compute etc. So any Cloud hosting company would be suitable for PHP 7. In my opinion, I'm personally using PHP 7 on Cloudways, they have 100% uptime and also have 24/7 awesome Customer Support.
If you need PHP7 in the shared hosting, then I would suggest you WebHostingHub, InMotion Hosting, and HostPapa. Usually, shared hosting providers are going a step behind the current software versions, but those companies must have the most recent software on the shared image sooner than everyone else.
I wanna display pictures from the database but the first picture of the album dont appear
Allô the pictures of the album are displayed only the first one
Dont understand please somebody help me!?
By the way i juste started coding