Loosing it

[Slawomir Piwowarczyk]

two words from the form below and one simple query after, doesnt work, what
to do to make it work
<FORM ACTION=search.php METHOD=POST>
<INPUT TYPE=TEXT NAME=search size=20>
<INPUT TYPE=TEXT NAME=search1 size=20>
<INPUT TYPE="submit" VALUE="Proceed query">
</FORM>
<?
//this one works fine: "select * from table_1 where dane like '%$search%'"
//but i wanna the one below to work
$query = "select * from table_1 where dane (like '%$search%') or dane (like
'%search1%')";
//^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^
^^^^^^^^^^^^^^^^
//DOESNT WORK, WHY?

$result = mysql_query($query);
while($row = mysql_fetch_row($result)){
print_r($row[0]);
echo '<br>';
....

[George]

Hi,

$query = "select * from table_1 where dane (like '%$search%') or dane
(like
'%search1%')";

well first of all you dont have $ in front of the search1 variable.
also, why put the brackets there? this should be:

$query = "select * from table_1 where dane like '%$search%' or dane
like
'%search1%'";

not only that, but this query should be further improved to:

$query = "select * from table_1 where dane like '%".$search."%' or dane
like
'%".$search1."%'";

and you should always make sure that the variables you pass into the
query are escaped (the default setting of magic quotes varies from host
to host).

George

[Pedro Graca]

Slawomir Piwowarczyk wrote:

<?
//this one works fine: "select * from table_1 where dane like '%$search%'"
//but i wanna the one below to work
$query = "select * from table_1 where dane (like '%$search%') or dane (like
'%search1%')";
//^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^
^^^^^^^^^^^^^^^^
//DOESNT WORK, WHY?

$result = mysql_query($query);

Because you don't check the return from mysql_query(). Try

$result = mysql_query($query) or die('Bad query: ' . mysql_error());

You could also try to make $query have a valid syntax.
Hint: the parenthesis are badly positioned.

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