PHP PDO Display message if no results are found in search

Question

I am trying to set an 'No Results Found' message when no search results are found after executing a MySQL 'LIKE' Query

I have the following code below:
I have an if statement just to test to see if an error message will work but I seem to get the output of the else statement 'found'

    <table class="center">          <!-- Creating a table with the class of 'center' -->
<!-- SEARCH FORM -->    
<?php 

    $KEYWORD = $_POST['keyword'];
    $stmt = $conn->prepare("SELECT DISTINCT dog_park_name FROM dog_park.items WHERE dog_park_name LIKE '%{$KEYWORD}%'");
    $stmt->execute();
    for($i=0; $row = $stmt->fetch(); ){ 
        $_SESSION["KEYWORD".$i] = $row[0];  


if(empty($stmt))
{
    echo 'Nothing found';
}
else
{
echo 'found';
}



?>
<!-- DISPLAY RESULTS -->
    <tr>            <!-- Adding the first table row -->
        <th>Dog Park</th>   <!-- Adding the second table header -->
    </tr>
    <tr>            <!-- Adding the second table row -->
        <td><a href="individual_item_page.php?keyword='<?php echo $row[$i] ?>' " ><?php echo $row[$i] ?></a></td>        <!-- Add the second cell on the second row -->
    </tr>
        <?php } ?>

</table>

Example:
If a user searches a keyword, and no results are found from that keyword, I am trying to get a message saying 'No Results found'

$stmt->rowcount() is your friend. php.net/manual/en/pdostatement.rowcount.php — P. Gearman
– P. Gearman, Commented May 18, 2016 at 14:49
You really need to look at how prepared statements are suppose to be used. php.net/manual/en/pdo.prepared-statements.php — chris85
– chris85, Commented May 18, 2016 at 14:52
If you are going through the effort of using prepared statements, you should not be concatenating input variables into the query string, ESPECIALLY user-provided variables like from $_POST. — Mike Brant
– Mike Brant, Commented May 18, 2016 at 14:54

vp_arth · Accepted Answer · 2016-05-18 15:54:07Z

1

To make a Mike Brant's answer a proper one:

$stmt = $conn->prepare("SELECT DISTINCT dog_park_name FROM items WHERE dog_park_name LIKE ?");
$stmt->execute(array("%".$_POST['keyword']."%"));
$_SESSION['KEYWORD'] = $stmt->fetchAll(PDO::FETCH_COLUMN);

if($_SESSION['KEYWORD']) {
?>
<table>
  <tr><th>Dog Park</th></tr>
 <?php foreach($_SESSION['KEYWORD'] as $word):?>
    <tr><td><?php echo $word?></td></tr>
 <?php endforeach?>
</table><?php
} else {
    echo 'Nothing found';
}

So in other words, you always have the query results to tell whether you have any results.

edited May 18, 2016 at 15:54

vp_arth

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answered May 18, 2016 at 15:08

Your Common Sense

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16 Comments

vp_arth Over a year ago

it will be bit better if you replace % in $_POST['keyword'] by [%]

Your Common Sense Over a year ago

I don't really get it. You want to escape %?

vp_arth Over a year ago

Sure, if user send 50% width to your script it match 50 shades of width text too ;)

user6324650 Over a year ago

Got follwoing error message PHP Parse error: syntax error, unexpected '[', expecting ')' in D:\users\n9692703\KEYWORD_SEARCH_RESULTS.php at code $stmt->execute(["%".$_POST['keyword']."%"]);

vp_arth Over a year ago

Oh, no) just str_replace('%', '[%]', $_POST['keyword'])

|

Majid Abbasi · Accepted Answer · 2016-05-18 14:51:04Z

1

Your if structure should show 'found' because your query executed successfully, in this cases you can count rows for decide about this issue:

if($stmt->rowCount() == 0)
{
     echo 'Nothing found';
}
else
{
     echo 'found';
}

answered May 18, 2016 at 14:51

Majid Abbasi

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Comments

Mike Brant · Accepted Answer · 2016-05-18 15:02:24Z

I am not going to get into the SQL injection problem you have with your current code. You need to fix it, likely using parameters with your prepared statement, but that is another topic.

I would also say your KEYWORD.$i approach is an antipattern. Why not just have numerically-indexed array under $_SESSION['KEYWORD']?

$_SESSION['KEYWORD'] = array();
$stmt = $conn->prepare("SELECT DISTINCT dog_park_name FROM dog_park.items WHERE dog_park_name LIKE '%{$KEYWORD}%'");
if($stmt) {
    $result = $stmt->execute();
    if($result) {
        while($row= $stmy->fetch() {
            $_SESSION['KEYWORD'][] = $row[0];
        }
    }
}  
if(count($_SESSION['KEYWORD']) === 0) {
    echo 'Nothing found';
} else {
    echo 'found';
}

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