does not work request

Question

What am I doing wrong in my query? thank you in advance for your help

new - not work

$query  = "SELECT * "; 
$query .= "FROM photographs "; 
$query .= "WHERE `caption` LIKE '%".$query."%' ";
    $query .= "OR `caption2` LIKE '%".$query."%' ";
//$query .= "WHERE visible = 1 ";
$query .= "ORDER BY $order_by LIMIT $start, $display ";     
$result = mysqli_query ($connection, $query);

old query - work

//$query = ("SELECT * FROM photographs WHERE (`caption` LIKE '%".$query."%') OR (`caption2` LIKE '%".$query."%')");
//$result = mysqli_query($connection, $query);

Maybe it wasn't such a good idea to reuse the variable. I also see that you don't use query parameters. I just hope you have sanitized your input. http://xkcd.com/327/ — Klas Lindbäck, Oct 21 '13 at 11:35
echo the $query value before passing it to mysqli_query function , and take the output and try it on any mysql client that you use such as phpMyAdmin ,and monitor the results — Leo Bali, Oct 21 '13 at 11:42

score 3 · Answer 1 · answered Oct 21 '13 at 11:32

3

You are overwriting the $query variable with parts of your query. :-)

answered Oct 21 '13 at 11:32

mkf

700
3
12

score 1 · Accepted Answer · answered Oct 21 '13 at 11:33

1

LIKE '%".$query."%' ";

should be replaced with

LIKE '%".$yourTerm."%' ";

where $yourTerm is what you are trying to search in your database

answered Oct 21 '13 at 11:33

tCode

444
3
8

does not work request

2 Answers2