MYSQL Results; no records found statement [duplicate] - php

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Display Data From MYSQL; SQL statement error
I have the code below displaying data from a MYSQL database (currently looking into sql injection issue) I need to insert an error message when no results are found...not sure where to position this! I have tried the code if( mysql_num_rows($result) == 0) {
echo "No row found!" but keep on gettin syntax errors, does anyone know the correct position in the code for this?
--
require 'defaults.php';
require 'database.php';
/* get properties from database */
$property = $_GET['bedrooms'] ;
$sleeps_min = $_GET['sleeps_min'] ;
$availability = $_GET['availability'] ;
$query = "SELECT * FROM `properties` WHERE bedrooms = '{$bedrooms}' AND sleeps_min = '{$sleeps_min}' AND availability = '{$availability}'";
$row=mysql_query($query);
$result = do_query("SELECT * FROM `properties` WHERE bedrooms = '{$bedrooms}' sleeps_min = '{$sleeps_min}' AND availability = '{$availability}'", $db_connection);
while ($row = mysql_fetch_assoc($result))
{
$r[] = $row;
}
?>

Answer 1

I have found few errors in your code that in line
$query = "SELECT * FROM `properties` WHERE bedrooms = '{$bedrooms}' AND sleeps_min = '{$sleeps_min}' AND availability = '{$availability}'";
$row=mysql_query($query);
You use bedrooms = '{$bedrooms}' but $bedrooms is not variable in whole cod it must be $preopery. I have made a few changes in your code given below please try it.
<?php
require 'defaults.php';
require 'database.php';
/* get properties from database */
/*if get $_GET['bedrooms'] value else ''*/
if (isset($_GET['bedrooms'])) {
$property = $_GET['bedrooms'];
} else {
$property = '';
}
/*if get $_GET['sleeps_min'] value else ''*/
if (isset($_GET['sleeps_min'])) {
$sleeps_min = $_GET['sleeps_min'];
} else {
$sleeps_min = '';
}
/*if get $_GET['availability'] value else ''*/
if (isset($_GET['availability'])) {
$availability = $_GET['availability'];
} else {
$availability = '';
}
$query = "SELECT * FROM `properties` WHERE bedrooms = '" . $property . "' AND sleeps_min = '" . $sleeps_min . "' AND availability = '" . $availability . "'";
$result = mysql_query($query) or die(mysql_error());
if ($result) {
while ($row = mysql_fetch_assoc($result)) {
$r[] = $row;
}
}
?>

Answer 2

Do var_dump($GET_) to debug whether you are getting valid strings. If any of these are blank, the query will try to match blank values instead of NULL. You should prevent this by doing:
if(!$_GET['bedrooms'] || $_GET['bedrooms'] == ''){
$property = 'NULL';
}//repeat for all three
$query = "SELECT * FROM `properties` WHERE 'bedrooms' = '$bedrooms' AND 'sleeps_min' = '$sleeps_min' AND 'availability' = '$availability'";
Instead of:
while ($row = mysql_fetch_assoc($result)) {
$r[] = $row;
}
You can simply do:
$r = mysql_fetch_array($query);
But enclose that in a conditional to see if your query found anything:
if(mysql_affected_rows() > 0){
//your code here will execute when there is at least one result
$r = mysql_fetch_array($query);
}
else{//There was either nothing or an error
if(mysql_affected_rows() == 0){
//There were 0 results
}
if(mysql_affected_rows() == -1) {
//This executes when there is an error
print mysql_error(); //not recommended except to debug
}
}