[SOLVED] Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result

[H]

I have a problem, below is my code and I get this error

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

 

<p><font color="#999999"><span style="font-size: 15pt">Question Asked<br>
</span></font><font size="2">
<?php
$question = $_POST['question'];
echo $question;
?>
</font></p>
<p><font color="#999999"><span style="font-size: 15pt">Answer<br>
</span></font><font size="2">
<?php
$host = 'localhost';
$user = '?';
$pass = '?';
$db = '?';
$connect = mysql_pconnect($host, $user, $pass);
$selectdb = mysql_select_db($db);

mysql_connect($host, $user, $pass);

$sql = mysql_query("SELECT * FROM search WHERE question LIKE $question"); 
$fetch = mysql_fetch_array($sql);
$answer = $fetch['answer'];
echo $answer;
?> 
</font></p>

Please Please Help

[H]

I have taken out the usernames and passwords from the code because there is no problem there otherwise i will be getting unable to connect error

[MadTechie]

try

 

<p><font color="#999999"><span style="font-size: 15pt">Question Asked<br>
</span></font><font size="2">
<?php
$question = addslashes($_POST['question']); //<-Changed
echo $question;
?>
</font></p>
<p><font color="#999999"><span style="font-size: 15pt">Answer<br>
</span></font><font size="2">
<?php
$host = 'localhost';
$user = '?';
$pass = '?';
$db = '?';
$connect = mysql_pconnect($host, $user, $pass);
$selectdb = mysql_select_db($db);

mysql_connect($host, $user, $pass);
$SQLq = "SELECT * FROM `search` WHERE `question` LIKE '%$question%'";
echo $SQLq;
$sql = mysql_query($SQLq)or die(mysql_error()); 
$fetch = mysql_fetch_array($sql);
$answer = $fetch['answer'];
echo $answer;
?> 
</font></p>

 

any errors ?

 

 

EDIT: updated the SQLq

 

[H]

Question Asked

what is vb

 

Answer

SELECT * FROM search WHERE question LIKE what is vbYou have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'vb' at line 1

[MadTechie]

try with the changes (my first post) i updated it

 

try the WHOLE thing their a few changes

[H]

Same

 

Question Asked

what is vb

 

Answer

SELECT * FROM search WHERE question LIKE what is vbYou have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'vb' at line 1

 

 

[H]

Oh sorry, its coming up with

 

SELECT * FROM `search` WHERE `question` LIKE '%what is vb%'No database selected

[MadTechie]

check the database name is being selected!

 

change

$db = '?';
$connect = mysql_pconnect($host, $user, $pass);
$selectdb = mysql_select_db($db);

to

$db = '?';
$connect = mysql_pconnect($host, $user, $pass) or die(mysql_error());
$selectdb = mysql_select_db($db) die(mysql_error());

[H]

I made a few changes and finally fixed it, i couldent have done it without you, heres the new code

 

<p><font color="#999999"><span style="font-size: 15pt">Question Asked<br>
</span></font><font size="2">
<?php
$question = addslashes($_POST['question']); //<-Changed
echo $question;
?>
</font></p>
<p><font color="#999999"><span style="font-size: 15pt">Answer<br>
</span></font><font size="2">
<?php
$host = 'localhost';
$user = '?';
$pass = '?';
$db = '?';
$connect = mysql_pconnect($host, $user, $pass);
$selectdb = mysql_select_db($db);

mysql_connect($host, $user, $pass);
mysql_select_db($db);
$SQLq = "SELECT * FROM `search` WHERE `question` LIKE '$question'";
$sql = mysql_query($SQLq)or die(mysql_error()); 
$fetch = mysql_fetch_array($sql);
$answer = $fetch['answer'];
echo $answer;
?> 
</font></p>

[MadTechie]

Cool, well done

 

as a note

$question = addslashes($_POST['question']);

cleans up the user input

 

can you click solved (bottom left)