I have the following code:
$surname=$_POST['surname'];
$sql2="SELECT * FROM andriana WHERE surname LIKE '$surname%'";
if (!mysql_query($sql2,$con)){
die('Error: ' . mysql_error());
}
$result2 = mysql_query($sql2);
echo "<table>";
while ($data = mysql_fetch_array($result2)) {
echo "<tr>";
echo "<td style='width:100px;height:40px'>".$data['name']."</td>";
echo "<td style='width:100px;height:40px'>".$data['surname']."</td>";
echo "<td style='width:100px;height:40px'>".$data['checkIN']."</td>";
echo "</tr>";
}
echo "</table><br><br>";
and let's say the following records in my table:
- Surname -
Greyjoy
Lannister
Stark
What happens is that if I won't type the full surname, it throws error that that surname doesn't exist. As a result the LIKE "%" is not working.
I have tried LIKE '".$surname."$' or LIKE '{$surname}%', but nothing happens too.
I searched here in Stack a lot, and it seems that the above tryouts should be working.
What am I missing?
- post-comments-editing -
To be more understood, I am sure that the variable contains the actual surname as a string, because if I type the whole surname, my application works normally. However, if I type the first 3 letters (or 4...) the application returns my homemade message that the surname typed is wrong.
Also, to go over the problem with case sensitive, my testing is done with a surname which has only small characters.
Thank you all for your effort, still havinf the issue!