PHP website database search - mySQL query results not being displayed
Question
I am designing a website with a search bar to enter queries. However, whenever I enter a query, I get a blank page.
In index.php I have:
<?php
include_once('search.php');
?>
<head>
<title>Search</title>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<link rel="stylesheet" type="text/css" href="style.css"/>
</head>
<body>
<form method="POST" action="search.php">
<input type="text" name="q" placeholder="Enter query"/>
<input type="submit" name="search" value="Search" />
</form>
</body>
In search.php I have
<?php
include_once('db.php'); //Connect to database
if(isset($_POST['search'])){
$q = $_POST['q'];
$query = mysqli_query($conn, "SELECT * FROM 'words' WHERE 'englishWord' LIKE '%$q%'");
$count = mysqli_num_rows($query);
if($count == "0"){
$output = '<h2>No result found</h2>';
}else{
while($row = mysqli_fetch_array($query)){
$s = $row['yupikWord']; //column of results
$output .= '<h2>'.$s.'</h2><br>';
}
}
}
echo $output;
mysqli_close($conn);
?>
In db.php I have
<?php
$servername = "localhost";
$username = "qaneryararvik";
$password = "PASSWORD";
$database = "yupik";
// Create connection
$conn = new mysqli($servername, $username, $password, $database);
$dbcon = mysqli_select_db($conn,$database);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully<br>";
I am able to enter text into the search box and click the 'Search' button. When I click 'Search', all I get is a blank page that says 'Connected Successfully'. The query results do not show. What am I doing wrong?
2 answers
solution1
1 2018-11-05 06:02:49
Quotations are wrong. Should be ` instead of ' for tables and rows:
"SELECT * FROM `words` WHERE `englishWord` LIKE '%$q%'"
solution2
0 2018-11-05 06:24:07
In search.php should be below
include_once('db.php'); //Connect to database
$output = "";
if(isset($_POST['search'])){
$q = $_POST['q'];
$sql ="SELECT * FROM words WHERE englishWord LIKE '%$q%'";
if ($result=mysqli_query($conn,$sql))
{
while($row = mysqli_fetch_array($result)){
$s = $row['yupikWord']; //column of results
$output .= '<h2>'.$s.'</h2><br>';
}
mysqli_free_result($result);
}else{
$output = '<h2>No result found</h2>';
}
mysqli_close($conn);
}else{
$output = '<h2>No result found</h2>';
}
echo $output;
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