PHP database call problems

Question 1. in the process of creating a web page, you want to write the hyperlink in the Header of the website by calling the database, so that you can easily modify the later Management. The problem is: how to call the form content stored in the database to make it a hyperlink.
2. there are many images in the "partners" column of the website. these images are also hyperlink to other Internet and stored in the database. how can they be called.


Thank you for your great help. I don't know if 100 points will be missing ...... New guy, please help !!

Reply to discussion (solution)

Learn basics first
$ SQL = "select * from your database"
$ Result = mysql_query ($ SQL );
While ($ rs = mysql_fetch_array ($ result )){
Echo "". $ rs [title]. "";
}

You store the hyperlink address in the database and assemble the hyperlink yourself when displaying it.
Images are the same. You save the image address and hyperlink address to the database.

The display is assembled during output.

This is indeed a basic problem.
1. first, simple database operations should be used to add, delete, modify, and query
2. simple database design. for example, you have two functions. In fact, there are mainly two fields in the data group. one is the hyperlink link content and the other is the hyperlink link address!