PHP : Search with drop down list

Question

i'm creating a search function with a drop down list where the user can search based on the chosen drop down list and display it on the same page. When I press the submit button, it didn't display any result.

Sorry if this has been asked before, but I suspect there is something that I did wrong at the PHP code but I could not figure out why. Can someone please point it out for me?

This is my search form:

<h2>Search</h2>
<form name="search" method="POST" action="<?php echo $_SERVER['PHP_SELF']; ?>">
  Keyword &nbsp;
  <input type="text" name="searchq" id="searchq" size="30" placeholder="Enter keyword..">&nbsp; By &nbsp;
  <select name="searchopt" id="searchopt">
    <option value=""></option>
    <option value="Members">Members</option>
    <option value="Journal">Journal</option>
    <option value="Conference">Conference</option>
    <option value="Awards">Awards</option>
    <option value="Grants">Grants</option>
    <option value="Patents">Patents</option>
    <option value="Research Grants">Research Grants</option>
    <option value="Book Chapter">Book Chapter</option>
    <option value="Book Publications">Book Publications</option>
    <option value="Other Publications">Other Publications</option>
  </select>
  &nbsp;
  <input type="submit" id="submit" name="submit" value="Search">
</form>
<br>
<p>
  <h4>Results</h4>
</p>

And this is my PHP code:

include ("includes/dbcon.php");

if(isset($_POST['searchq']) && $_POST['searchq'] != "")
{
    if(isset($_POST['searchopt']) && $_POST['searchopt'] != "")
    {
        $escsearch = mysqli_real_escape_string($conn, $_POST['searchq']);
        $searchval = preg_replace('#[^a-z 0-9]#i', '', $escsearch);
        $opt = $_POST['searchopt'];

        switch($opt)
        {
            case "Members":
            $query = "SELECT * FROM members where memberName LIKE '%$searchval%'";
            $res = mysqli_query($conn, $query) or die(mysqli_error());
            $count = mysqli_num_rows($res);
            if($count > 1)
            {
                $output .= "$count results for <strong>$escsearch</strong>.";

                while($row = mysqli_fetch_array($res))
                {
                    $name = $row['memberName'];
                    $pos = $row['position'];
                    $fac = $row['faculty'];
                    $email = $row['email'];
                    $int = $row['research_interests'];

                    echo "<table>
                            <tr>
                                <td>Name</td>
                                <td>Position</td>
                                <td>Faculty</td>
                                <td>E-mail</td>
                                <td>Research Interests</td>
                            </tr>
                            <tr>
                                <td>$name</td>
                                <td>$pos</td>
                                <td>$fac</td>
                                <td>$email</td>
                                <td>$int</td>
                            </tr>
                         </table>";
                }
            }else{
                $output = "<p>No records found.</p>"; 
            }
            break;
        }
    }
}

$query = "SELECT * FROM members where memberName= '$searchval'" Why using LIKE operator.. Try this query — Nana Partykar, Sep 03 '15 at 12:38
var_dump($_POST['searchq']); to check the value send to server — Vishu238, Sep 03 '15 at 12:40
@Vishu238 i put the var_dump code after the first if statement, and when i tried to search for member's name (lau) , it display this: string 'lau' (length=3) — Azie, Sep 03 '15 at 12:47
@Vishu238 I got this when I selected 'Members' - string 'Members' (length=7) — Azie, Sep 03 '15 at 12:52
Sorry for so many questions.. I gues you should use var_dump($query), It will print your query that you can run in phpmyadmin — Vishu238, Sep 03 '15 at 12:54
@Vishu238 oh it's okay. btw i got this result - string 'SELECT * FROM members where memberName = 'lau'' (length=46) — Azie, Sep 03 '15 at 13:03
It seems like your code is OK. is there a row in your table in which membareName = lau? — Vishu238, Sep 03 '15 at 13:05
If there is only one record with membername like lau then you will get nothing output to the page. This line says there has to be more than one result for the script to continue... `if($count > 1)` You might want it to be `if($count >= 1)` — Kuya, Sep 03 '15 at 13:07
Why do you have this: `$output .=` ? at that instance `$output` is being defined, it should be `$output =` — CodeGodie, Sep 03 '15 at 13:09
@Kuya ah no, there are also other records than this. and i tried change it to if($count >= 1), it display the data but i got an error 'Undefined variable: output on line 27' — Azie, Sep 03 '15 at 13:13
ah thank you @CodeGodie i finally able to get rid of the error. :) — Azie, Sep 03 '15 at 13:15

score 1 · Accepted Answer · answered Sep 03 '15 at 13:18

Numerous problems here:

Your submit button should not have a name attribute since you are not doing anything with it.

Change this:

<input type="submit" id="submit" name="submit" value="Search">

To this:

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

your if conditional, checking for results will return false if one result is found.

Change this:
```
if($count > 1)
```
To this:
```
if($count > 0)
```
Lastly, your $output variable is defined the wrong way. Using .= assumes that $output has been previously defined.

Change this:
```
$output .= "$count results for <strong>$escsearch</strong>.";
```
To this:
```
$output = "$count results for <strong>$escsearch</strong>.";
```

score 0 · Answer 2 · answered Sep 03 '15 at 13:09

0

Correct you condition

if($count > 1) => if($count >= 1)

It seems like you have only one record in database

Also define $output variable

answered Sep 03 '15 at 13:09

Vishu238

673
4
17

Kuya · Answer 3 · 2015-09-03T13:16:20.563

0

If there is only one record with membername LIKE lau then you will get nothing output to the page.

This line says there has to be more than one result for the script to continue... if($count > 1) You might want it to be if($count >= 1)

First you have to define $output as an empty string... $output = '';

Hope this helps

edited Sep 03 '15 at 13:16

answered Sep 03 '15 at 13:10

Kuya

7,280
4
19
31

PHP : Search with drop down list

3 Answers3