getting the following error in mysql when i run the query

Question

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mysql_fetch_array() expects parameter 1 to be resource, boolean given in select

$username="root";
$password="webmaster";
$database="demo1";
mysql_connect("localhost","root","webmaster");
@mysql_select_db($database) or die( "Unable to select database");
/*query the database*/
$query = "select field1_name, field2_name, field3_name, field4_name, 
                 field5_name, field6_name, field7_name, field8_name,
                 field9_name, field10_name, field11_name, field12_name,
                 field13_name, image, fieldl1_name, fieldl2_name,
                 fieldl3_name, fieldl4_name, fieldl5_name, fieldl6_name,
                 fieldl7_name
           from tablenamevehicle 
           full join members on members.member_id=tablenamevehicle.field13_name 
           where field4_name like '%".$searchterm4."%' 
             and field5_name like '%".$searchterm5."%'
             and field1_name like '%".$searchterm1."%'
             and field2_name like '%".$searchterm2."%'";

$result = mysql_query($query);
/*number of rows found*/
$num = mysql_numrows($result);

mysql_close();

the error which i ma getting is:

Warning: mysql_numrows() expects parameter 1 to be resource, boolean given in 
F:\xampp\htdocs\xampp\Login1\vehicle\search.php on line 171

is there anything wrong in the query???

I'm getting the feeling this code was copied from somewhere else...could you provide some details on the database table structure? — Chris Kempen, Jan 25 '12 at 17:49
$result = mysql_query($query) or die(mysql_error()); to check error from your query — Zul, Jan 25 '12 at 17:49
sir i have not copied the code.first table name is "tablenamevehicle" and it has 15 fields. Another table name is "members" . members table has a primary key as member_id and the same field is present as field15_name which acts as a foreign key to connect both tables.i have to print contents of all from "tablenamevehicle" and other details corresponding to the table members — user1150362, Jan 25 '12 at 17:55

score 3 · Accepted Answer · answered Jan 25 '12 at 17:57

3

FULL JOIN isn't something you can do in MySQL (see 12.2.8.2. JOIN Syntax). You can confirm this by inspecting mysql_error()

You probably want INNER JOIN or CROSS JOIN

answered Jan 25 '12 at 17:57

Conrad Frix

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yes only inner join works. thanks – user1150362 Mar 07 '12 at 11:57

score 0 · Answer 2 · answered Jan 25 '12 at 17:50

0

Assuming your data type's for the two joining columns are the same which the naming convention from this perspective suggests otherwise the SQL looks good.

members.member_id=tablenamevehicle.field13_name

I also don't see where you are defining your search terms...

answered Jan 25 '12 at 17:50

Mathew A.

1,421
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sir i have defined search term as $searchterm1 = $_POST['searchterm1']; trim ($searchterm1); /*check if search term was entered*/ if (!$searchterm1){ echo 'Please enter a search term.'; } /*add slashes to search term*/ if (!get_magic_quotes_gpc()) { $searchterm1 = addslashes($searchterm1); } $searchterm2 = $_POST['searchterm2']; trim ($searchterm2); /*check if search term was entered*/ if (!$searchterm2){ echo 'Please enter a search term.'; } /*add slashes to search term*/ if (!get_magic_quotes_gpc()) { $searchterm2 = addslashes($searchterm2); } – user1150362 Jan 25 '12 at 18:02
As others have stated I think the error is with the FULL JOIN. – Mathew A. Jan 25 '12 at 18:10

getting the following error in mysql when i run the query

2 Answers2