mysql_fetch_array() error on submit form to self - php

Question

I have a page called index.html which takes in a few variable from the user, when the user submits them, it goes to the next page (result.php) On results.php the variables from the form on the index.html are posted. The results.php connects to database runs a query etc. All of this works fine. the results.php looks something like this:
<?php
$h = "localhost";
$u = "root";
$p = "******";
$d = "********";
//connectipn to mysql
$conn = mysql_connect( $h, $u, $p );
//connection to database
mysql_select_db($d);
$code = $_POST['code'];
$usage = $_POST['usage'];
$days = $_POST['days'];
$value = $_POST['value];
$sql_runners_up = "SELECT code
from some_table
where number = $value;"
$plans['p_h10'] = array();
$rs2 = mysql_query( $sql_runners_up, $conn );
while ( $row = mysql_fetch_array( $rs2 ))
{
$plans['p_h10'][] = $row["p_h10"];
}
?>
The results.php also has a form that the user can update some fields that are used in the query. the form that I currently have looks like this...
<form action="results.php" name="filter" method="post">
usage:<input name="" type="search" value="" /><br/>
days:<input name="" type="search" value="" /><br/>
<input name="" type="checkbox" value="" />value<br />
<input type="submit" value="submit" name="submit" />
</form>
Even with a blank as soon as the user submits the form I get an error (i have tried filling out the form to use real variable but still get the same error)
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL
result resource
the error refers to the line around the while statement. my guess is that some variable/s are not being passed after the form is submitted or the entire results.php form is not being submitted / processed by the server. But to be honest I am really not sure. Any help will be greatly appreciated.

Answer 1

That's because mysql_query returned false.
Also, in your code, you've got a syntax error
$value = $_POST['value'];
Also, your code is vulnerable. Don't trust user's input!
so, correct code would be
$value = intval($_POST['value']);

Answer 2

Try:
$value = (int) $value;
$sql_runners_up = "SELECT code from some_table where number = $value";
$plans['p_h10'] = array();
$rs2 = mysql_query( $sql_runners_up);

Answer 3

try with this:
$sql_runners_up = "SELECT * from some_table where number = '$value';"
and also check that the $value has containing the value..
EDIT: in the above code i use * which select all the fields, cause that is see in your code you did:
$sql_runners_up = "SELECT code from some_table where number = $value;"
and in fetch query you are calling to field name p_h10:
while ( $row = mysql_fetch_array( $rs2 ))
{
// didn't selected $row["p_h10"] in your query
$plans['p_h10'][] = $row["p_h10"];
}

Answer 4

First of all, this row is invalid:
$value = $_POST['value];
It should be $_POST['value'];
But, this error occurs usually when there is an error in your query, or the connection is bad. Check your mysql_connect() & mysql_select_db() settings, it is probably bad.

Answer 5

looks like you did typos...
<?php
$h = "localhost";
$u = "root";
$p = "******";
$d = "********";
//connectipn to mysql
$conn = mysql_connect( $h, $u, $p );
//connection to database
mysql_select_db($d);
$code = $_POST['code'];
$usage = $_POST['usage'];
$days = $_POST['days'];
$value = $_POST['value']; // here
$sql_runners_up = "SELECT code
from some_table
where number = $value"; // and here
$plans['p_h10'] = array();
$rs2 = mysql_query( $sql_runners_up, $conn );
while ( $row = mysql_fetch_array( $rs2 ))
{
$plans['p_h10'][] = $row["p_h10"];
} ?>
also your form is not POSTing parameters... because name attributes are not set
<form action="results.php" name="filter" method="post">
usage:<input name="usage" type="search" value="" /><br/>
days:<input name="days" type="search" value="" /><br/>
<input name="value" type="checkbox" value="" />value<br />
<input type="submit" value="submit" name="submit" />
</form>