How I See my Result in Select from Where MYSQL using PHP?

Question

I have this code, i try to call my data from table database mysql , but didn't see any result. always go to else , not go to the process. what would i do?

<?php
require('connectDB.php');
                	$nama = $_GET['nama'];
                	echo $nama;
                    $query = "SELECT * FROM pesan
                    WHERE nama = '%" . mysqli_real_escape_string($connection, $nama) . "%'
                    
                    ";
            		$results = mysqli_query($connection, $query);
            		$baris = mysqli_num_rows($results);
            		

            		if (!$results) {
		             die('Invalid query: ' . mysql_error());
		            }

            		if ( $baris > 0) {
            			while($row = mysqli_fetch_assoc($results)) {
            			?>
            				<h3>Nama Mobil : <?php echo $row['mobil'] ?></h3>
            				<h3>ID Pembelian : <?php echo $row['id']; ?></h3>
            				<h3>Nama anda : <?php echo $row['nama']; ?></h3>
            				<h3>Alamat : <?php echo $row['alamat']; ?></h3>	
	            			<h3>Tanggal Masuk : <?php echo $row['tgl_masuk']; ?></h3>
	            	<?php
	            		}
	            	}else{
	            		echo "error";
	            	}
	            	?>

What wrong with my code? Thanks!

im sorry , this is my ConnectDB.php , i include in my html.

<?php
$connection = mysqli_connect('localhost', 'root', '', 'dealermobil');
if (!$connection){
    die("Database Connection Failed" . mysqli_error());
}

// $db = new PDO ('mysql:host=localhost;dbname=db_login;charset=utf8mb4','root','');
?>

Where is $connection defined? Try printing '$connection' variable. — user3025122
– user3025122, Commented Jul 9, 2017 at 9:44
... WHERE name = '%" ...? It should be ... WHERE name LIKE '%" ... — Rajdeep Paul
– Rajdeep Paul, Commented Jul 9, 2017 at 9:46
mysql_error() ? You're mixing mysql and mysqli APIs. RTM, http://php.net/manual/en/mysqli.error.php. Also, check how many rows are returned from the SELECT query, do var_dump($baris);. — Rajdeep Paul
– Rajdeep Paul, Commented Jul 9, 2017 at 9:51

Amir · Accepted Answer · 2017-07-09 11:37:33Z

1

Your connection check and result check is incorrect.

$connection = mysqli_connect('localhost', 'root', '', 'dealermobil');
if (!$connection){
    die("Database Connection Failed" . mysqli_connect_error());
}

Also

$results = mysqli_query($connection, $query);
$baris = mysqli_num_rows($results);


if (!$results) {
   die('Invalid query: ' . mysqli_error($connection));
 }

edited Jul 9, 2017 at 11:37

answered Jul 9, 2017 at 11:27

Amir

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2 Comments

Diky Ridhlo Over a year ago

thanks for the answer Mr. Amir , but i got answer by myself. can i ask how to connect a table database to b table database (ex mobil table to pesan table)

Amir Over a year ago

You should use join command SELECT ... FROM table1 JOIN table2 on ... WHERE ... please refer to documents

Kunal Awasthi · Accepted Answer · 2017-07-09 10:46:48Z

0

use this you are using = it should be LIKE when you are trying to search a field in database.

 $query = "SELECT * FROM `pesan` WHERE `nama` LIKE '%". mysqli_real_escape_string($connection, $nama) ."%'";

edited Jul 9, 2017 at 10:46

answered Jul 9, 2017 at 10:28

Kunal Awasthi

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1 Comment

Diky Ridhlo Over a year ago

i try it but no result mr. Kunal

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