Really need some help with this one Guyz??

[Modernvox]

Hi Guyz,

 

I am nearing the end of my project. A lot of guyz on here have helped me throughout the long process and I am very grateful-believe me!

 

Everything is working as expected and the final part is displaying values from database.

 

The layout i chose is that similar to craigslist, where the user can click on the ads title and it opens in it's own page.

 

What i need:

1. The' title' and 'actual_location' are displayed in the results page as a link

2.User clicks link and another page opens with the remaining info which is

$title

$location

$actual_location

$details

$email //I would like to make emails anonymous?

$flag// This is just a link to flag the listing

 

Here's what I have

/////////////////Test category selected/////////////////
require_once(db_cred_all.inc);
$conn = mysql_connect($dbhost,$dbuser, $password);
if (!$conn)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db($dbname, $conn);


$category= isset($_POST['category']);
if ($category== "venues")
$result = mysql_query(SELECT * FRom venue
WHERE state = '$st'");
While($venue = mysql_fetch_array($result))
{
echo $title['title'] . " " . $actual_location['actual_location'];
echo "<br />";  //////How to only display Title and actual_location initially as a link and than the rest on next page?
}

[Modernvox]

the lather

[wildteen88]

File path should be enclosed in quotes. This is wrong

require_once(db_cred_all.inc);

it should be

require_once('db_cred_all.inc');

 

This portion of code is incorrect

$category= isset($_POST['category']);
if ($category== "venues")
$result = mysql_query(SELECT * FRom venue
WHERE state = '$st'");
While($venue = mysql_fetch_array($result))
{
echo $title['title'] . " " . $actual_location['actual_location'];
echo "<br />";  //////How to only display Title and actual_location initially as a link and than the rest on next page?
}

 

It should be

if(isset($_POST['category']))
{
    $category = $_POST['category'];
    if ($category == "venues")
    {
        $result = mysql_query("SELECT * FROM venue WHERE state = '$st'");
        While($venue = mysql_fetch_array($result))
        {
            echo $venue['title'] . " " . $venue['actual_location'];
            echo "<br />";
        }
    }
}

 

 

[Modernvox]

File path should be enclosed in quotes. This is wrong

require_once(db_cred_all.inc);

it should be

require_once('db_cred_all.inc');

 

This portion of code is incorrect

$category= isset($_POST['category']);
if ($category== "venues")
$result = mysql_query(SELECT * FRom venue
WHERE state = '$st'");
While($venue = mysql_fetch_array($result))
{
echo $title['title'] . " " . $actual_location['actual_location'];
echo "<br />";  //////How to only display Title and actual_location initially as a link and than the rest on next page?
}

 

It should be

if(isset($_POST['category']))
{
    $category = $_POST['category'];
    if ($category == "venues")
    {
        $result = mysql_query("SELECT * FROM venue WHERE state = '$st'");
        While($venue = mysql_fetch_array($result))
        {
            echo $venue['title'] . " " . $venue['actual_location'];
            echo "<br />";
        }
    }
}

 

Cool thanks. But how about the link part?

 

Do i just use  echo "<a href=\"{$venue[title]}\" target=\"_blank\">{$venue[actual_location]}<font size=\"3\"></font></a><br />";

  print "<BR />\n";

 

How do i print the rest of the fields in the same statement?

[wildteen88]

Do i just use  echo "<a href=\"{$venue[title]}\" target=\"_blank\">{$venue[actual_location]}<font size=\"3\"></font></a><br />";

  print "<BR />\n";

Yes that would be the correct code. However not sure what '<font size=\"3\"></font>' is for. You should use CSS for formatting your text.

 

 

How do i print the rest of the fields in the same statement?

Just echo out the necessary variable, eg  $venue['field_name_here']

[Modernvox]

Do i just use  echo "<a href=\"{$venue[title]}\" target=\"_blank\">{$venue[actual_location]}<font size=\"3\"></font></a><br />";

  print "<BR />\n";

Yes that would be the correct code. However not sure what '<font size=\"3\"></font>' is for. You should use CSS for formatting your text.

 

 

How do i print the rest of the fields in the same statement?

Just echo out the necessary variable, eg  $venue['field_name_here']

 

Awesome, thanks wildteen. I don't know CSS yet. I have been using a WYSIWYG builder. I mean , I understand how css works , but it seems like a pain in the ass to me. Why bother?

[wildteen88]

I understand how css works , but it seems like a pain in the ass to me. Why bother?

Its probably a pain the ass to you because you don't understanding it. To me CSS is actually quite simple.

 

I have been using a WYSIWYG builder.

I wont say any more.

[Modernvox]

I understand how css works , but it seems like a pain in the ass to me. Why bother?

Its probably a pain the ass to you because you don't understanding it. To me CSS is actually quite simple.

 

I have been using a WYSIWYG builder.

I wont say any more.

LOL. Thanks dude!