How to fetch a word from a database using PHP? - php

Question

This is the PHP code for fetching a word that was entered by the user. May I ask on how to show the result of "jhon dewey" if the user entered the keyword "john" on my website as "jhon" only without the last name?
My database fields are:
first_name (contains the first name records)
last_name (contains the last name records)
This is my code. Please help me find the error. Can I change the %john% into a variable so that any word entered by the user can search to my database?
<?php
// Database connection
$sql = new mysqli("localhost", "root", "MYPassword", "test");
if (mysqli_connect_errno ()) {
die(mysqli_connect_error());
exit;
}
$query = "SELECT * FROM `usage` WHERE 'first_name' LIKE %john%";
$result = $sql->query($query);
if (!$result)
die($mysqli->error);
while ($row = $result->fetch_assoc()) {
$rows[] = $row;
}
$result->close();
$sql->close();
print("<pre>");
print_r($rows);
print("</pre>");
?>

Answer 1

Patterns are string literals so they must be wrapped with single quotes.
Second, column names are identifiers so they shouldn't be wrap with single quotes as single quotes are for string literals.
Snippet:
$query = "SELECT * FROM `usage` WHERE `first_name` LIKE '%john%'"
Since your column name is not a reserved keyword, backticks are optional on this case
$query = "SELECT * FROM `usage` WHERE first_name LIKE '%john%'"

Answer 2

Check this one
<?php
// Database connection
$sql = new mysqli("localhost", "root", "MYPassword", "test");
if (mysqli_connect_errno ()) {
die(mysqli_connect_error());
exit;
}
$query = "SELECT * FROM `usage` WHERE `first_name` LIKE '%john%'";
$result = $sql->query($query);
if (!$result)
die($mysqli->error);
$rows = array();
while ($row = $result->fetch_assoc()) {
$rows[] = $row;
}
$result->close();
$sql->close();
print("<pre>");
print_r($rows);
print("</pre>");
?>