Displaying message when no results found in PHP MySQL search

Question

I have a PHP search script which queries a MySQL database. Currently, when no results are displayed the script shows and error. How can I make it display a message like \"No

Answer 1

<?php

mysql_connect("localhost","username","password");
mysql_select_db("database");

if(!empty($_GET['q'])){
    $query          =   mysql_real_escape_string(trim($_GET['q']));
    $searchSQL      =   "SELECT * FROM links WHERE `title` LIKE '%{$query}%'  LIMIT 8";
    $searchResult   =   mysql_query($searchSQL);

    // the query was run successfully 
    // and it returned at least a result
    if(is_resource($searchResult) && mysql_num_rows($result) > 0){
        while ($row=mysql_fetch_assoc($searchResult)){
            $results[]="<div class='webresult'><div class='title'><a href='{$row['url']}'>{$row['title']}</a></div><div class='desc'>{$row['description']}</div><div class='url'>{$row['url']}</div></div>";
        }

        echo implode($results);
    } else{
        echo 'No Results were found';
    }
}
?>

Answer 2

You could count the number of elements in the array, and either continue with your implode or display the message you mentioned.

<?php
     if(count($results) > 0){
         echo implode($results);
     }
     else {
         echo "No results were found.";
     }
?>

You also really should not be using the mysql_* functions. Use either the improved version (mysqli_*) or PDO.

Answer 3

if (mysql_num_rows($searchResult) == 0) {
   echo "some error message"; 
} else { 
  ... process data
}

Answer 4

You could also use the function mysql_num_rows, which will tell you the number of rows returned by your query.

$rows = mysql_num_rows($searchResult);
if($rows <= 0){
    /* Code if there are no rows */
}
else{
    /* At least one row has been found */
}

Answer 5

How about a simple counter in the while statement:

$i = 0;

while (condition) {
   $i++;
   do stuff;
}

if ($i ==0) {echo 'No results found';}

Answer 6

Create a MYSQL Connection and paste this code below

$sql="SELECT * FROM tablename WHERE columnname LIKE your variable or constant ";
$result=mysql_query($sql);

$count=mysql_num_rows($result);

if($count>=1){if result was found}

else {if result was not found}

?>