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

edmaunders

😕

Hey,
I having a real big problem with the following error

Warning: mysql_fetch_array (): supplied argument is not a valid MySQL result resource in /home/a8567431/public_html/viewpassanger.php on line 23

I have a page called show.php which passes two variables called flightnum and flightdate. The second page called viewinfo.php uses the $_GET function to store flightnum in var $a and flight date into $b.

For example the show.php will pass 1895 as flightnum and 05/04/09 as flightdate. The viewinfo.php then store these value two $a and $b I know this Is working because when I just echo out $a and $b I get the correct results.

Also when I query the database using the following code

$sql = 'SELECT * FROM passanger WHERE Flightnum = 1895 and date = "05/04/09"';

it bring up the correct results. I only get the error when I replace the value with the variable $a and $b

$q = 'SELECT * FROM passanger WHERE Flightnum = $a and date = $b';

here is the code for the viewinfo.php page

<?php
$a = $GET["flightnum"];
$b = $GET["flightdate"];

$conn = mysql_connect("mysql7.000webhost.com","a8567431_aviance","xxxx");
mysql_select_db("a8567431_aviance");

$q = 'SELECT * FROM passanger WHERE Flightnum = $a and date = $b';

$result = mysql_query($q);

<tr><th width = "20%">passanger name</th><</tr>

<?php

while($row = mysql_fetch_array($result))
{

echo "<tr><td>";
echo $row[Pname];
echo "</td><td>";
echo $b;
echo "</td></tr>";
}

Please can some one help
Ed

djjjozsi

if you see this message then your mysql query is wrong.

$sql = 'SELECT * FROM `passanger` WHERE `Flightnum` = 1895 and `date` = "05/04/09"';

change to:

$sql = 'SELECT * FROM `passanger` WHERE `Flightnum` = 1895 and `date` = \'05/04/09\'';

if you add a string in mysql you should use an apostrofe not a double-qoute.

How you should use the variables if you use a single quote:

$q = 'SELECT * FROM `passanger` WHERE `Flightnum` = \''.$a.'\' and `date` = \''.$b.'\'';

there is a better way to build your query:

$q = sprintf("SELECT * FROM `passanger` WHERE `Flightnum` = '%s' and `date` = '%s'" , $a, $b);

If you use apostrofe be careful if you add a variable:

$a="value";
echo 'This is $a';  //output is:    This is $a

echo 'This is '.$a.' ';  //output is:  This is value 

echo "This is $a";    //output is:  This is value

edmaunders

Thank you so much that worked grate, You have helped me out a grate deal

ed 🙂

djjjozsi

Your're welcome.