Web Development and Design Foundations with HTML5 (9th Edition) (What's New in Computer Science)
Web Development and Design Foundations with HTML5 (9th Edition) (What's New in Computer Science)
9th Edition
ISBN: 9780134801148
Author: Terry Felke-Morris
Publisher: PEARSON
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Chapter 14, Problem 13FIB
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The jQuery method named “attr()” is to allow the user to configure the value of HTML attributes.

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If I have a table In database which called occupations  columns are: workid, workname, salary and sector  workid is a primary key and increase  I insert to the table  1, fireman , 500, 2 2, officer, 800, 1 the code in html / php  Will show a drop menu select the workname from the table in the database so the menu will show the fireman and the officer As soon the user choose one of them(onchange) will be a textbox display just the salary  how to write the code in php thanks in advance
If I have a table In database which called occupations  columns are: workid, workname, salary and sector  workid is a primary key and increase  I insert to the table  1, fireman , 500, 2 2, officer, 800, 1 the code in html / php  Will show a drop menu select the workname from the table in the database so the menu will show the fireman and the officer As soon the user choose one of them(onchange) will be a textbox display just the salary  do I need a JavaScript to create function to put for onchange ? how to write the code in php thanks in ad
If I have a table In database which called occupations  columns are: workid, workname, salary and sector  workid is a primary key and increase  I insert to the table  1, fireman , 500, 2 2, officer, 800, 1 the code in html / php  Will show a drop menu select the workname from the table in the database so the menu will show the fireman and the officer As soon the user choose one of them(onchange) will be a textbox display just the salary  The code will create a dropdown menu of all the occupations in the table in the database. When the user chooses one of the occupations, the corresponding salary will be displayed under the dropdown menu. I did the dropdown and it's connected to the database but when I click / chooce one of it they don't show the salary. would you complete the missing part in my code    <?php$servername = "localhost";$username = "root";$password = "";$dbname="occupation"; // Create connection$conn = new mysqli($servername, $username, $password, $dbname); // Check…
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