[Resolved] searching database

[sallystudent]

Hi, I am trying to create a search form but when i try it out the following error message appears:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:\program files\easyphp1-7\www\find_customer_results.php on line 41

Is not NULL and the database connection works

any ideas???😕

[mrhappiness]

change

$some_var = mysql_query('SELECT whatever');

to

$some_var = mysql_query('SELECT whatever') or die(mysql_error());

and provide us with some code please...

[sallystudent]

sorry - here u go:

<?php
include_once ("../sallysurfconnection/header.html");
$page_title = 'find_customer_results.php';

?>
<!--Space layout image positioning-->
<td colspan="8" valign="top">
<img src="C:\Program Files\EasyPHP1-7\www\Images\1pix.jpg" width="561" height="9">

	<h2>Find Customer search results</h2>

<?php
//create short variable names

$searchtype=$POST["searchtype"];
$searchterm=$POST["searchterm"];
$searchterm= trim($searchterm);

if (!$searchtype || !$searchterm)
{
echo '<p><font color="red">You have not entered search details. Please go back and try again.</font></p>';
exit;
}

$searchtype = addslashes($searchtype);
$searchterm = addslashes($searchterm);

@ $db = mysql_pconnect("localhost","Admin","admin");

if (!$db)
{
echo '<p><font color="red">Error: Could not connect to database. Please try again later.</font></p>';
exit;
}

$query = "SELECT * FROM customer_table WHERE ".$searchtype." like '%".$searchterm."%'";
$result = mysql_query($query);

$num_results = mysql_num_rows($result);

echo '<P> Number of items found: '.$num_results.'</p>';

for ($i=0; $i <$num_results; $i++)
{
$row = mysql_fetch_array($result);
echo '<p><strong>'.($i+1).'.Surname: ';
echo htmlspecialchars(stripslashes($row['Surname']));
echo '</strong><br />First_Name: ';
echo stripslashes($row['First_Name']);
echo '<p><strong><br />Cust_Id: ';
echo stripslashes($row['Cust_Id']);
echo '</p>';
}

include_once ("../sallysurfconnection/footer.html");

?>

heres the search Form:

<?php
include_once ("../sallysurfconnection/header.html");
$page_title = 'Find_customer.php';

?>
<!--Space layout image positioning-->
<td colspan="8" valign="top">
<img src="C:\Program Files\EasyPHP1-7\www\Images\1pix.jpg" width="561" height="9">

	<h2>Customer search</h2>

	<form action="find_customer_results.php" method="post">
	Enter Search Type:<br />
	<select name="searchtype">
	<option value="Surname">Surname</option>
	<option value="First_Name">First_Name</option>
    <option value="Cust_Id">Cust_Id</option>

    </select>
    <br />
    Enter search term:<br />
    <input name="searchterm" type="text">
    <br />
    <input type="submit" value="Search">
    </form>

<?php

include_once ("../sallysurfconnection/footer.html");

?>

thanks

[mrhappiness]

Originally posted by mrhappiness
change

$some_var = mysql_query('SELECT whatever');

to

$some_var = mysql_query('SELECT whatever') or die(mysql_error());

[/B]

where's the output this gives?

[sallystudent]

Hi,

I have made the changes u suggested but it still displays the same error message:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:\program files\easyphp1-7\www\find_customer_results.php on line 41

Number of items found:

[mrhappiness]

you have

$query = "SELECT * FROM customer_table WHERE ".$searchtype." like '%".$searchterm."%'";
$result = mysql_query($query) or die(mysql_error());

$num_results = mysql_num_rows($result); 

?

[sallystudent]

I had put it in wrong place!🙁 , i Have now put in correct place - but get error 'no database selected'

Where should I state which database to use??

[mrhappiness]

below mysql_pconnect

http://www.php.net/mysql-select-db is the function you have to use

[sallystudent]

Cheers, Thanx very much - it all seems to work now😃