Getting error in my search (php)

Question

I'm making a search field to search id, firstname, lastname and email from the table registration but getting error:

Notice: Undefined variable: valuetosearch in D:\Xampp\htdocs\Registration_system\dashboard.php on line 13

Notice: Undefined variable: query in D:\Xampp\htdocs\Registration_system\dashboard.php on line 14

Warning: mysqli_query(): Empty query in D:\Xampp\htdocs\Registration_system\dashboard.php on line 27

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in D:\Xampp\htdocs\Registration_system\dashboard.php on line 61

<style type="text/css">
    
    table, tr, th, td
    {

      border: 1px solid black;
    }
</style>
<?php

if (isset($_POST['valuetosearch'])) {
    $valuetosearch =$_POST['valuetosearch'];
    $valuetosearch = "SELECT * FROM `registration` WHERE CONCAT(`firstname`, `lastname`, `email`, `phonenumber`) LIKE '%".$valuetosearch."'"; 
    $search_result= filtertable($query);
}

else
{
$query= "SELECT * FROM `registration`";
$search_result= filtertable($query);

}

function filtertable($query){

require_once'config.php';
$filter_result= mysqli_query($CONN, $query);
return $filter_result;

}
?>


<h1>Welcome on dashboard </h1>
<ul>
<li><a href="dashboard.php">Home</a></li>
<li><a href="#">About</a></li>
<li><a href="dashboard.php">Profile</a></li>
<li><a href="login.php">Logout</a></li>
</ul>

<form action="dashboard.php" method="POST">

<input type="text" name="valuetosearch" placeholder="Search here..."> 
<input type="submit" name="search" value="submit"> <br><br>

<table>
    
<tr>
    
<th>id</th>
<th>First name</th>
<th>Last name</th>
<th>email</th>
<th>Phone number</th>

</tr>
<?php while($row= mysqli_fetch_array($search_result)): ?>
<tr>
    <td><?php echo $row['id']; ?></td>
    <td><?php echo $row['firstname']; ?></td>
    <td><?php echo $row['lastname']; ?></td>
    <td><?php echo $row['email']; ?></td>

</tr>

<?php endwhile; ?>
</table>

</form>

can u explain it further? – Zarah Hafeez Aug 15 '17 at 08:24 — Zarah Hafeez, Aug 15 '17 at 08:24

sndsabin · Answer 1 · 2017-08-15T09:00:51.400

2

Replace your if (isset($_POST['search'])) { ... } With

if (isset($_POST['search'])) {
$query = "SELECT * FROM `registration` WHERE CONCAT(`firstname`, `lastname`, `email`, `phonenumber`) LIKE '%".$_POST['valuetosearch']."'"; 
$search_result = filtertable($query);}

edited Aug 15 '17 at 09:00

answered Aug 15 '17 at 08:31

sndsabin

164
1
6

– Malek Zarkouna Aug 15 '17 at 08:38
@MélikZarkouna Sorry I didn't get you. Could you please clarify ? – sndsabin Aug 15 '17 at 08:55
the input of the value he wants to search it not 'search' look my answer you will get it – Malek Zarkouna Aug 15 '17 at 08:57
1

@MélikZarkouna Check Query bro. I am taking the value of $_POST['valuetosearch'] . if (isset($_POST['search']) just to check if the form has been posted or not since he has place submit type input – sndsabin Aug 15 '17 at 09:00
i thought he want to test if the input is empty or not .Thank you for the clarification – Malek Zarkouna Aug 15 '17 at 09:14
thankyou, errors not coming, but nothing appearing on search. – Zarah Hafeez Aug 15 '17 at 09:18
echo the $query and try executing it in mysql and see if it returns any result for your query. – sndsabin Aug 15 '17 at 09:22

score 1 · Answer 2 · answered Aug 15 '17 at 08:24

1

$valuetosearch is used two time in the same line where you define it ( to store your string and as a variable inside you sql)

$query wasn't defined

answered Aug 15 '17 at 08:24

sheplu

2,937
3
24
21

score 1 · Answer 3 · answered Aug 15 '17 at 08:26

There are many mistakes in your code First Mistake is in your code

if (isset($_POST['search'])) {
    $valuetosearch = "SELECT * FROM `registration` WHERE CONCAT(`firstname`, `lastname`, `email`, `phonenumber`) LIKE '%".$valuetosearch."'"; 
    $search_result= filtertable($query);
}

It Should be

if (isset($_POST['search'])) {
    $query = "SELECT * FROM `registration` WHERE CONCAT(`firstname`, `lastname`, `email`, `phonenumber`) LIKE '%".$valuetosearch."'"; 
    $search_result= filtertable($query);
}

Getting error in my search (php)

3 Answers3