Form submit POST & php on one file problem.

[Dave Siegel]

For some reason the submit POST and php on one form doesn't work for me.

Here's my form code (it's dutch)

[html] <form method="POST">
<select name="locatie" size="1">
<option value="">Kies de locaties</option>
<option>Parijs</option>
<option>Lyon</option>
<option>Marseile</option>
<option>Lile </option>

</select>
<br>
<br>
<select name="prijs" size="1">
<option value="">Kies de prijzen</option>
<option value="1">0-100.000 </option>
<option value="2">100.00-200.00 </option>
<option value="3">200.000-300.000 </option>
<option value="4">300.000-400.000 </option>
<option value="5">400.000-500.000 </option>
<option value="6">500.000-600.000 </option>
<option value="7">600.000 en meer </option>

</select>
<br>
<br>
<input type="submit" name="button" id="button" value="Zoeken" />
</form>[/html]


And here's my php code :

[PHP]<?
ERROR_REPORTING(E_ALL);
include("config.php");
if(isset($_POST['button']))
{
$location = $_POST['locatie'];
$price = $_POST['prijs'];

//check what price it is
if($locatie == ''){
echo"Foute locatie gekozen, Kies aub een goede locatie.";
}

if($price == ''){
echo"Foute prijs gekozen. Kies aub een goede prijs.";
}

elseif($price == '1'){


$query = "SELECT * FROM houses WHERE location = '" . $location . "' AND price >= 0 AND price <= 100000";


}

elseif($price == '2'){
$query = "SELECT * FROM houses WHERE location = '" . $location . "' AND price >= 100000 AND price <= 200000";
}

elseif($price == '3'){
$query = "SELECT * FROM houses WHERE location = '" . $location . "' AND price >= 200000 AND price <= 300000";
}

elseif($price == '4'){
$query = "SELECT * FROM houses WHERE location = '" . $location . "' AND price >= 300000 AND price <= 400000";
}

elseif($price == '5'){
$query = "SELECT * FROM houses WHERE location = '" . $location . "' AND price >= 400000 AND price <= 500000";
}

elseif($price == '6'){
$query = "SELECT * FROM houses WHERE location = '" . $location . "' AND price >= 500000 AND price <= 600000";
}

elseif($price == '7'){
$query = "SELECT * FROM houses WHERE location = '" . $location . "' AND price <= 600000";
}

while($results = mysql_fetch_array($query))
{

echo "<table width='434' height='72' border='0'>";

echo "<tr><td width='133' height= '130'>";

echo "<img width='132' height='129' src='" . $results['Foto1'] . "'></img>";

echo "</td><td class='style5'>";

echo "Type: " . $results['type'] . "<br>Descriptie:" . $results['description'];

echo "</td><td class='style5'>";

echo "prijs: " . $results['price'] . ".-";

echo "</td></tr>";

echo "</table>";

}
}
?>[/PHP]


Any idea why it doesn't work?

Help is much appreciated.

[Atli]

Hi Dave.

First of all, I'd recommend changing the <? tag in your PHP code into <?php. The short-tags version you are currently using is disabled by default in PHP and can cause problems on other servers.

As for your problem. The <form> tag is missing the 'action' property. Without that the form won't know where to send the data. So even if the data is meant to be posted to the current page, set the 'action' property.

For example:

<form action="thisPage.php" method="POST">
 

[Dave Siegel]

Hi Dave.

First of all, I'd recommend changing the <? tag in your PHP code into <?php. The short-tags version you are currently using is disabled by default in PHP and can cause problems on other servers.

As for your problem. The <form> tag is missing the 'action' property. Without that the form won't know where to send the data. So even if the data is meant to be posted to the current page, set the 'action' property.

For example:

<form action="thisPage.php" method="POST">
 

Thanks, That worked.

But now its nagging with this error for some reason:

Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in /home/content/C/o/l/Collussus/html/school/ict/search.php on line 213

this is my php(i left the tags on <? ?> because i know my server allows it. :

[PHP]<?
ERROR_REPORTING(E_ALL);
include("config.php");
if(isset($_POST['button']))
{
$location = $_POST['locatie'];
$price = $_POST['prijs'];

//check what price it is
if($locatie == ''){
echo"Foute locatie gekozen, Kies aub een goede locatie.<br>";
}

if($price == ''){
echo"Foute prijs gekozen. Kies aub een goede prijs.";
}

elseif($price == '1'){


$query = "SELECT * FROM houses WHERE location = '" . $location . "' AND price >= 0 AND price <= 100000";


}

elseif($price == '2'){
$query = "SELECT * FROM houses WHERE location = '" . $location . "' AND price >= 100000 AND price <= 200000";
}

elseif($price == '3'){
$query = "SELECT * FROM houses WHERE location = '" . $location . "' AND price >= 200000 AND price <= 300000";
}

elseif($price == '4'){
$query = "SELECT * FROM houses WHERE location = '" . $location . "' AND price >= 300000 AND price <= 400000";
}

elseif($price == '5'){
$query = "SELECT * FROM houses WHERE location = '" . $location . "' AND price >= 400000 AND price <= 500000";
}

elseif($price == '6'){
$query = "SELECT * FROM houses WHERE location = '" . $location . "' AND price >= 500000 AND price <= 600000";
}

elseif($price == '7'){
$query = "SELECT * FROM houses WHERE location = '" . $location . "' AND price <= 600000";
}

$results = mysql_query($query);
while($results = mysql_fetch_object($results))
{

echo "<table width='434' height='72' border='0'>";

echo "<tr><td width='133' height= '130'>";

echo "<img width='132' height='129' src='" . $results['Foto1'] . "'></img>";

echo "</td><td class='style5'>";

echo "Type: " . $results['type'] . "<br>Descriptie:" . $results['description'];

echo "</td><td class='style5'>";

echo "prijs: " . $results['price'] . ".-";

echo "</td></tr>";

echo "</table>";

}
}
?>[/PHP]

[Atli]

These two lines are the source of your problem:

$results = mysql_query($query);
    while($results = mysql_fetch_object($results))
 

You are overwriting the database result resource with the return value of the mysql_fetch_object() function. So on the next loop, the $results is not a valid MySQL resource, causing the function to fail and a warning to be printed.

You need to change the name of the variable you use to store the current row of data.

[Dave Siegel]

These two lines are the source of your problem:

$results = mysql_query($query);
    while($results = mysql_fetch_object($results))
 

You are overwriting the database result resource with the return value of the mysql_fetch_object() function. So on the next loop, the $results is not a valid MySQL resource, causing the function to fail and a warning to be printed.

You need to change the name of the variable you use to store the current row of data.

Fixed that, still same problem.

[Atli]

Have you check out if there was an error in your SQL?
Your code does not validate that the query executed successfully.

Try adding something like this before you try to use the result from the database:

if(!$result) {
  echo "<pre><b>MySQL Query failed!</b>\n
           <b>Error:</b> ". mysql_error() ."\n
           <b>Query:</b> $query
        </pre>";
}
 

[Dave Siegel]

Have you check out if there was an error in your SQL?
Your code does not validate that the query executed successfully.

Try adding something like this before you try to use the result from the database:

if(!$result) {
  echo "<pre><b>MySQL Query failed!</b>\n
           <b>Error:</b> ". mysql_error() ."\n
           <b>Query:</b> $query
        </pre>";
}
 

Fixed the first error.

Now it just gives me this error:

MySQL Query failed!

Error:

Query: SELECT * FROM houses WHERE region = 'Parijs' AND price >= 0 AND price <= 100000

[Atli]

Note that the variable I used in my example was called $result, while the MySQL result in your code is called $results. (Force of habit for me. I always call mine $result :P)

[Dave Siegel]

That fixed it. Thanks.