MySQL 8
Create a PHP program that will read in each table of the database you created in Assignment 8 and display the values into a good looking table.
When I run it, I'm having this error " Parse error: syntax error, unexpected '$dbName' (T_VARIABLE) in C:\xampp\htdocs\assignment9\includes\dbh.inc.php on line 6'
Here is my index.php
<?php
include_once 'includes/dbh.inc.php';
?>
<!DOCTYPE html>
<html>
<head>
<title></title>
</head>
<body>
<?php
$sql = "SELECT * FROM comic_book.books;";
$result = mysqli_query($conn, $sql);
if ($resultCheck > 0) {
while ($row =
mysqli_fetch_assoc($result)) {
echo
$row['book_name'] . "<br>";
}
}
?>
</body>
</html>
Here is my dbh.inc.php
<?php
$dbServername = "localhost";
$dbUsername = "root";
$dbPassword = "Password"
$dbName = "comic_book";
$conn = mysqli_connect($dbServername, $dbUsername, $dbPassword, $dbName);
Dear Student ,
As per the requirement submitted above , kindly find the below solution.
Error Description :
Below is the corrected files.
dbh.inc.php :
<?php
$dbServername = "localhost";
$dbUsername = "root";
$dbPassword = "Password";
$dbName = "comic_book";
$conn = mysqli_connect($dbServername, $dbUsername, $dbPassword, $dbName);
?>
*********************
index.php :
<?php
include_once 'dbh.inc.php';
?>
<!DOCTYPE html>
<html>
<head>
<title></title>
</head>
<body>
<?php
$sql = "SELECT * FROM comic_book.books;";
$result = mysqli_query($conn, $sql);
if ($result > 0) {
while ($row = mysqli_fetch_assoc($result)) {
echo $row['book_name'] . "<br>";
}
}
?>
</body>
</html>
NOTE : PLEASE FEEL FREE TO PROVIDE FEEDBACK ABOUT THE SOLUTION.
MySQL 8 Create a PHP program that will read in each table of the database you...
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