How to create a search using PHP, mysqli and a html form

Question

I want to create a form which allows the user to type in a search and have it pick up the right values from a database and display them, for some reason I can't get my query to work it just displays "could not search"

Here is my php code

 <?php

  include "connect.php";

  $output = '';

  if(isset($_POST['search'])) {
    $search = $_POST['search'];
    $search = preg_replace("#[^0-9a-z]i#","", $search);

    $query = mysqli_query("SELECT * FROM house WHERE town LIKE '%$search%'") or die ("Could not search");
    $count = mysqli_num_rows($query);
    
    if($count == 0){
      $output = "There was no search results!";

    }else{

      while ($row = mysqli_fetch_array($query)) {

        $town = $row ['town'];
        $street = $row ['street'];
        $bedrooms = $row ['bedrooms'];
        $bathroom = $row ['bathrooms'];

        $output .='<div> '.$town.''.$street.''.$bedrooms.''.$bathrooms.'</div>';

      }

    }
  }

  ?>

Here is my form

 <form action ="home.php" method = "post">
  
          <input name="search" type="text" size="30" placeholder="Belfast"/>

          <input type="submit" value="Search"/>

          </form> 

          <?php print ("$output");?>

echo your query SELECT * FROM house WHERE town LIKE '%$search%' and try the query from phpmyadmin or mysql... — Nishant Solanki
– Nishant Solanki, Commented Mar 30, 2015 at 13:24
This might help. askseeker.com/code-snippets/search-form-php-using-mysqli — smzapp
– smzapp, Commented Feb 13, 2018 at 8:34

Funk Forty Niner · Accepted Answer · 2015-03-30 13:40:07Z

2

You're not connecting to your DB in your query:

$query = mysqli_query("SELECT
                      ^ missing connection variable

there is no connection variable (unknown what you are using to connect with)

$query = mysqli_query($connection, "SELECT ...
                      ^^^^^^^^^^^^

From the manual http://php.net/manual/en/mysqli.query.php

Object oriented style mixed mysqli::query ( string $query [, int $resultmode = MYSQLI_STORE_RESULT ] )

Procedural style mixed mysqli_query ( mysqli $link , string $query [, int $resultmode = MYSQLI_STORE_RESULT ] )

figuring you are using mysqli_ to connect with. If you're using mysql_ or PDO to connect with, that won't work. Those different MySQL APIs do not intermix with each other.

Plus, instead of or die ("Could not search") do or die(mysqli_error($connection)) to catch any errors, if any.

Add error reporting to the top of your file(s) which will help find errors.

<?php 
error_reporting(E_ALL);
ini_set('display_errors', 1);

// rest of your code

Sidenote: Error reporting should only be done in staging, and never production.

Example mysqli connection:

$connection = mysqli_connect("myhost","myuser","mypassw","mybd") 
                or die("Error " . mysqli_error($connection));

For more information, visit:

http://php.net/manual/en/function.mysqli-connect.php

edited Mar 30, 2015 at 13:40

answered Mar 30, 2015 at 13:19

Funk Forty Niner

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6 Comments

Keep Coding Over a year ago

Can you please explain it for my knowledge? Error reporting should only be done in staging, and never production.

Funk Forty Niner Over a year ago

@Testing Should there be any errors that will display system paths, db tables/column names the person does not want to show, connection credentials, those should not be displayed to the user.

Rebekah Over a year ago

I have a connection to the database in a connect file which I already have included

Funk Forty Niner Over a year ago

@Rebekah You still need to pass the connection variable to your query. I don't know which variable you're using to connect with, which is why I used $connection as an example. You will have to replace it with the one you're using.

Funk Forty Niner Over a year ago

Just change your query to $query = mysqli_query($con, "SELECT * FROM house WHERE town LIKE '%$search%'") or die ("Could not search"); - Here I used $con, seeing that's what you're using as a variable in your other question. You need to pass the connection to your query, just reload my answer also. @Rebekah

|

Bista · Accepted Answer · 2016-09-14 15:12:52Z

0

$query = mysqli_query("SELECT * FROM house WHERE town LIKE '%$search%'") or die ("Could not search");

$result = mysqli_query($connection,$query);

$count = mysqli_num_rows($result);

The $connection is the variable declared in your connect.php to connect to your database .

You should have out the $result before the $count.

edited Sep 14, 2016 at 15:12

Bista

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answered Sep 14, 2016 at 14:49

reyven

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Comments

jalazbe · Accepted Answer · 2020-01-13 08:42:17Z

-1

Try to remove the space between if($count==0) it will start working!!

if($count==0){
      $output = "There was no search results!";}

edited Jan 13, 2020 at 8:42

jalazbe

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answered Jan 13, 2020 at 7:50

butter bee

1

1 Comment

Paflow Over a year ago

Despite the fact that this might be solved or not of relevance after four years spaces in this context have to effect.

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