php, my sql and database help

ianhaney · April 15, 2012

Hi

I am in the middle of designing a website and am trying to get a webpage to connect and display results from the database to my webpage when a customer does a search but is not working

The coding is below

Any ideas how to get it to work, I have been trying for hours today

<?php

$db=mysql_connect("2up2downhomes.com.mysql", "userxxx", "passxxx");
mysql_select_db("2up2downhomes_c", $db);

$result = mysql_query("SELECT * FROM properties");   

while($row = mysql_fetch_assoc($result)){
   echo "ID: ".$row['id'].", Type of Property:".$row['typeProperty']
       .", bedrooms:".$row['bedrooms'].", garden:.$row['garden'].", description:.$row['description'].", price:.$row['price'].", location:.$row['location']."<br/>";
}

?>

<html>

<body>
You have searched for a <?php echo $_POST["propertytype"]; ?>
<br>
You have searched in the location of <?php echo $_POST["location"]; ?>
<br />
You have searched for <?php echo $_POST["bedrooms"]; ?> bedrooms
<br />
You have searched for <?php echo $_POST["bathrooms"]; ?> bathrooms
<br />
You have searched for £<?php echo $_POST["minprice"]; ?>
<br />
You have searched for £<?php echo $_POST["maxprice"]; ?>

</body>
</html>

Thank you in advance, appreciate it

Ian

EDIT (KP): Removed credz & please use code tags next time

ianhaney · April 15, 2012

Hi

Thank you so much for removing the data and I will do, I am so sorry, bit new to this forum

Muddy_Funster · April 16, 2012

1 - not sure what this has to do with MS SQL Server? :shrug:

2 - more information than "it's not working" is always appreciated

3 - the following three lines should all have an or die statement at the end to display any errors coming back from the database:

$db=mysql_connect("2up2downhomes.com.mysql", "userxxx", "passxxx") or die (mysql_error());
mysql_select_db("2up2downhomes_c", $db) or die (mysql_error());

$result = mysql_query("SELECT * FROM properties") or die (mysql_error());

ianhaney · April 16, 2012

Hi Muddy Funster

Thank you for the reply

I put in the coding you said and am still getting the following error

Server error

The website encountered an error while retrieving http://www.2up2downhomes.com/saleresults.php. It may be down for maintenance or configured incorrectly.

Here are some suggestions:

Reload this web page later.

HTTP Error 500 (Internal Server Error): An unexpected condition was encountered while the server was attempting to fulfil the request.

The code I am using is below

<?php
ini_set('display_errors', 1);
error_reporting(E_ALL);
//$connection =  mysql_connect($host, $user, $password) or  die ("Couldnt connect to server");

$db=mysql_connect("$host", "$user", "$password") or die (mysql_error());
mysql_select_db("$database", $db) or die (mysql_error());

//Program: getdata.php
//Desc: Gets data from a database using a function
?>
<html>
<head>
<title>Properties</title>
</head>
<body>
<?php
$typeProperty = gettypeProperty("House"); //call function
$f_price = number_format ($typeProperty['price',2);
echo "<p><b>{$typeProperty['typeProperty']}</b><br>\n;
Description: {$typeProperty['Description']}<br>\n;
Price: \${$typeProperty['price']}\n";
?>
</body>
</html>
<?php
function gettypeProperty($typeProperty)
{
$db = mysql_select_db("properties") or die ("Couldnt select database");
$result = mysql_query("SELECT * FROM properties") or die (mysql_error()); 
//$query = "SELECT * FROM properties WHERE typeProperty='typeProperty'";
$result = mysql_query($query) or die ("Couldnt execute query.");
return mysql_fetch_array($result,MYSQL_ASSOC);
}
?>

<html>
<body>
You have searched for a <?php echo $_POST["propertytype"]; ?>
<br>
You have searched in the location of <?php echo $_POST["location"]; ?>
<br />
You have searched for <?php echo $_POST["bedrooms"]; ?> bedrooms
<br />
You have searched for <?php echo $_POST["bathrooms"]; ?> bathrooms
<br />
You have searched for £<?php echo $_POST["minprice"]; ?>
<br />
You have searched for £<?php echo $_POST["maxprice"]; ?>
</body>
</html>

Muddy_Funster · April 16, 2012

do you have custom error messages set up?

ianhaney · April 16, 2012

Hi

I have this in my coding and have got the php error messages turned on in my control panel with my host but that's it

ini_set('display_errors', 1);

error_reporting(E_ALL);

PFMaBiSmAd · April 16, 2012

Your code is actually using mysql, so the MS SQL form section where you posted it is not where this belongs at all. In any case, your current problem is the php code, so moving thread to the php help section...

PFMaBiSmAd · April 16, 2012

You have a fatal parse (syntax) error in -

Parse error: syntax error, unexpected ',', expecting ']' in your_file.php on line 19

For fatal parse errors to be reported in your main file, the error_reporting/display_errors setting must be set before your script is requested (your code never runs, so putting the settings into your script won't ever turn the settings on.)

You should have the error_reporting/display_errors setting set in your master php.ini on your development system. You should also not be attempting to develop and debug php code on a live server. Set up a localhost development system, you will save a TON of time.

Muddy_Funster · April 16, 2012

this looks to be the offending line from PFM's info:

$f_price = number_format ($typeProperty['price',2);

ianhaney · April 16, 2012

I took that line out Muddy but am getting this now

php_network_getaddresses: getaddrinfo failed: Name or service not known

This is doing my head in now

I have included my coding in below

<?php

//$connection =  mysql_connect($host, $user, $password) or  die ("Couldnt connect to server");

$db=mysql_connect("$host", "$user", "$password"); mysql_select_db("$database", $db);

$db=mysql_connect("$host", "$user", "$password"); mysql_select_db("$database", $db) or die (mysql_error());
mysql_select_db("$database", $db) or die (mysql_error()) or die (mysql_error());

//Program: getdata.php
//Desc: Gets data from a database using a function
?>
<html>
<head>
<title>Properties</title>
</head>
<body>
<?php
$typeProperty = gettypeProperty("House"); //call function
echo "<p><b>{$typeProperty['typeProperty']}</b><br>\n;
Description: {$typeProperty['Description']}<br>\n;
Price: \${$typeProperty['price']}\n";
?>
</body>
</html>
<?php
function gettypeProperty($typeProperty)
{
$db = mysql_select_db("properties") or die ("Couldnt select database");
$query = "SELECT * FROM properties WHERE typeProperty='typeProperty'";
$result = mysql_query($query) or die ("Couldnt execute query.");
return mysql_fetch_array($result,MYSQL_ASSOC);
}
?>

<html>
<body>
You have searched for a <?php echo $_POST["propertytype"]; ?>
<br>
You have searched in the location of <?php echo $_POST["location"]; ?>
<br />
You have searched for <?php echo $_POST["bedrooms"]; ?> bedrooms
<br />
You have searched for <?php echo $_POST["bathrooms"]; ?> bathrooms
<br />
You have searched for £<?php echo $_POST["minprice"]; ?>
<br />
You have searched for £<?php echo $_POST["maxprice"]; ?>
</body>
</html>

Muddy_Funster · April 16, 2012

I dont see any code therre that would bear any relevance to that error message

ianhaney · April 16, 2012

I am thinking of starting from scratch as is going wrong somewhere

ianhaney · April 16, 2012

Ok I am winning sort of, I have got connection the mysql database, how do I get the data from the database to display on the webpage

ianhaney · April 16, 2012

Hi

I have managed to connect to mysql database and managed to display results but now how I want exactly

I want the results to be displayed by what the customer searches as its a property website

so for example, in my search form I have the following

Property Type

Location

Number of bedrooms

Number of Bathrooms

Min Price

Max Price

for example, I put in the following

House

Benfleet

4

2

£100,000

£300,000

but when I click on search, it returns all the results that are in mysql database and not just what I am looking for

Please help

I have included the coding below

<html>
<body>
You have searched for a <?php echo $_POST["propertytype"]; ?>
<br>
You have searched in the location of <?php echo $_POST["location"]; ?>
<br />
You have searched for <?php echo $_POST["bedrooms"]; ?> bedrooms
<br />
You have searched for <?php echo $_POST["bathrooms"]; ?> bathrooms
<br />
You have searched for £<?php echo $_POST["minprice"]; ?>
<br />
You have searched for £<?php echo $_POST["maxprice"]; ?>
<br><br>
Please find your results below
</body>
</html>
<br><br>
<?php

$db=mysql_connect("host", "user", "password");
mysql_select_db("database", $db);

?>

<?php

$db = mysql_select_db("database") or die ("Couldnt select database");
?>

<?php

// Query database
$result = mysql_query("SELECT * FROM properties");

if (!$result)
{
    echo "Error running query:<br>";
    trigger_error(mysql_error());
}
elseif(!mysql_num_rows($result))
{
    // no records found by query.
    echo "No records found";
}
else
{
    $i = 0;
    echo '<div class="container" style="float:left;">';
  while($row = mysql_fetch_array($result)) {     // Loop through results
        $i++;
echo '</div>';		
echo '<div class="textholder" style="font-family:helvetica; font-size:13px; float:left; padding-top:10px;">';
        echo "<span style=\"color:green;\"><b>Displaying record $i<br>\n</b><br></span>";
        echo "<b>" . $row['id'] . "</b><br>\n";      // Where 'id' is the column/field title in the database
	echo "Property Type: ". $row['typeProperty'] . "<br>\n";       // as above
	echo "Bedrooms: ". $row['bedrooms'] . "<br>\n";  // ..
	echo "Bathrooms: ". $row['bathrooms'] . "<br>\n";  // ..
	echo "Garden: ". $row['garden'] . "<br>\n";  // ..
	echo "Description: ". $row['description'] . "<br>\n";  // ..
	echo "Price: ". $row['price'] . "<br>\n";         // ..
        echo "Location: ". $row['location'] . "<br>\n";            // Where 'location' is the column/field title in the database
echo '</div>';
echo '<div style="clear:both"></div>';				
    }
echo '</div>';	}

?>

Kind regards

Ian

ianhaney · April 17, 2012

Any ideas any one please

I am well and truly stuck

Kind regards

Ian

Muddy_Funster · April 17, 2012

you need to have a WHERE condition in your query :

$result = mysql_query("SELECT * FROM properties WHERE field1='$type' AND field2='$area' AND......");

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