Solved

do I need to use an array?

Posted on 2009-07-10
17
196 Views
Last Modified: 2012-05-07
Lets see if I can explain this correctly...

I am building an administration area.  I am trying to display ALL items in a table.  Table 1 is called service_locations.  Table 2 is called doctors. Table 3 is called locations.  I have setup an INNER JOIN on all three tables.

I have a <?php do loop to display all data in service_locations such as

id          city          doctor
1          5               12
2          3                9
3          19              3

The values here are exactly as they are in the service_locations table, however, what I actually want displayed on the screen is the city Name and doctor Name instead of their id's.  My INNER JOIN works fine, meaning I can display the city and name but the problem is that it ONLY displays the last record.  So if the id values equal the following:

city table
5 = Georgetown
3 = Young
19 = Freshman

doctor table
12 = Hughes
9 = Smith
3 = Thompson

my display shows

id          city                doctor
1           Freshman     Thompson
2           Freshman     Thompson

What I need to do, is possibly a for each loop in addition to my do while loop?  I'm just a bit lost here on how to display the city name and doctor name for each result.
0
Comment
Question by:rbudj
  • 8
  • 5
  • 4
17 Comments
 
LVL 6

Expert Comment

by:ahmad2121
Comment Utility
you may have an error in your inner join. can you please post it?
0
 
LVL 16

Author Comment

by:rbudj
Comment Utility
SELECT locations.city, physicians.PhysicianFirstName, physicians.PhysicianLastName
FROM service_locations INNER JOIN locations ON locations.id = Service_LocationsServiceID INNER JOIN physicians ON PhysicianID = Service_LocationsDocID

0
 
LVL 6

Expert Comment

by:ahmad2121
Comment Utility
do you really need an inner join for that?
SELECT 

  locations.city, physicians.PhysicianFirstName, physicians.PhysicianLastName

FROM 

  service_locations,locations,physicians

WHERE

  Service_LocationsDocID = PhysicianID 

 AND 

  Service_LocationsServiceID = locations.id

Open in new window

0
 
LVL 6

Expert Comment

by:ahmad2121
Comment Utility
i think there is a mistake in the above query.. hold on.
0
 
LVL 108

Expert Comment

by:Ray Paseur
Comment Utility
Is this by any chance a homework assignment?  We are prohibited from providing answers for homework by the TOS of EE, but we can provide guidance to learning resources.  Just checking, thanks, ~Ray
0
 
LVL 16

Author Comment

by:rbudj
Comment Utility
no this is not a homework assignment :/
0
 
LVL 108

Expert Comment

by:Ray Paseur
Comment Utility
Great.  Please post the CREATE TABLE statements so we can see what the field names are.  Thanks.
0
 
LVL 16

Author Comment

by:rbudj
Comment Utility
here is the part that matters.  Again, I want to display the city and doctor names instead of the id's
<?php do { ?>

            <tr class="<?php echo $WARRT_AltClass1->getClass(true); ?>">

              <td class="WADAResultsTableCell"><a href="service_locations_Detail.php?Service_LocationsID=<?php echo(rawurlencode($row_WADAservice_locations['Service_LocationsID'])); ?>" ><?php echo($row_WADAservice_locations['Service_LocationsID']); ?></a></td>

              <td class="WADAResultsTableCell"><a href="service_locations_Detail.php?Service_LocationsID=<?php echo(rawurlencode($row_WADAservice_locations['Service_LocationsID'])); ?>" ><?php echo $row_WADAservice_locations['Service_LocationsServiceID']; ?> - <?php echo $row_rsJoin['city']; ?></a></td>

              <td class="WADAResultsTableCell"><a href="service_locations_Detail.php?Service_LocationsID=<?php echo(rawurlencode($row_WADAservice_locations['Service_LocationsID'])); ?>" ><?php echo($row_WADAservice_locations['Service_LocationsDocID']); ?></a></td>

Open in new window

0
What Security Threats Are You Missing?

Enhance your security with threat intelligence from the web. Get trending threat insights on hackers, exploits, and suspicious IP addresses delivered to your inbox with our free Cyber Daily.

 
LVL 108

Expert Comment

by:Ray Paseur
Comment Utility
Please post the CREATE TABLE statements, thanks.
0
 
LVL 16

Author Comment

by:rbudj
Comment Utility
i dont know what you mean by create table statements.
0
 
LVL 108

Expert Comment

by:Ray Paseur
Comment Utility
When these data base tables were created, you (or someone) used CREATE TABLE and in that, you laid out the names of the fields, their characteristics, whether they were indexed or NULL, etc.  That is what I am looking for.  It will help us show you the correct queries and the best way to iterate over the query results set.
0
 
LVL 16

Author Comment

by:rbudj
Comment Utility
i used phpmyadmin
-- phpMyAdmin SQL Dump

-- version 2.11.9.5

-- http://www.phpmyadmin.net

--

-- Host: localhost

-- Generation Time: Jul 10, 2009 at 02:04 PM

-- Server version: 4.1.22

-- PHP Version: 5.2.6
 

SET SQL_MODE="NO_AUTO_VALUE_ON_ZERO";
 
 

/*!40101 SET @OLD_CHARACTER_SET_CLIENT=@@CHARACTER_SET_CLIENT */;

/*!40101 SET @OLD_CHARACTER_SET_RESULTS=@@CHARACTER_SET_RESULTS */;

/*!40101 SET @OLD_COLLATION_CONNECTION=@@COLLATION_CONNECTION */;

/*!40101 SET NAMES utf8 */;
 

--

-- Database: `mydatabase`

--
 

-- --------------------------------------------------------
 

--

-- Table structure for table `locations`

--
 

CREATE TABLE IF NOT EXISTS `locations` (

  `id` int(11) NOT NULL auto_increment,

  `cat_city_id` varchar(255) NOT NULL default '',

  `name_1` varchar(255) NOT NULL default '',

  `name_2` varchar(255) NOT NULL default '',

  `address` varchar(255) NOT NULL default '',

  `city` varchar(255) NOT NULL default '',

  `state` varchar(15) NOT NULL default '',

  `zip` varchar(10) NOT NULL default '',

  `phone_appt` varchar(15) NOT NULL default '',

  `phone` varchar(15) NOT NULL default '',

  `fax` varchar(15) NOT NULL default '',

  `hours` varchar(255) NOT NULL default '',

  `more_info` varchar(255) NOT NULL default '',

  PRIMARY KEY  (`id`)

) ENGINE=MyISAM  DEFAULT CHARSET=utf8 AUTO_INCREMENT=14 ;
 

-- --------------------------------------------------------
 

--

-- Table structure for table `physicians`

--
 

CREATE TABLE IF NOT EXISTS `physicians` (

  `PhysicianID` int(11) NOT NULL auto_increment,

  `PhysicianFirstName` varchar(255) NOT NULL default '',

  `PhysicianLastName` varchar(255) NOT NULL default '',

  `PhysicianUndergraduate` longtext NOT NULL,

  `PhysicianMedical_School` longtext NOT NULL,

  `PhysicianInternship` longtext NOT NULL,

  `PhysicianFellowship` longtext NOT NULL,

  `PhysicianBoard` longtext NOT NULL,

  `PhysicianHonors` longtext NOT NULL,

  `PhysicianResearch` longtext NOT NULL,

  `PhysicianMembership` longtext NOT NULL,

  `PhysicianPersonal` longtext NOT NULL,

  `PhysicianPhoto` varchar(255) NOT NULL default '',

  PRIMARY KEY  (`PhysicianID`)

) ENGINE=MyISAM  DEFAULT CHARSET=utf8 AUTO_INCREMENT=3 ;
 

-- --------------------------------------------------------
 

--

-- Table structure for table `service_locations`

--
 

CREATE TABLE IF NOT EXISTS `service_locations` (

  `Service_LocationsID` int(11) NOT NULL auto_increment,

  `Service_LocationsServiceID` varchar(255) NOT NULL default '',

  `Service_LocationsDocID` varchar(255) NOT NULL default '',

  PRIMARY KEY  (`Service_LocationsID`)

) ENGINE=MyISAM  DEFAULT CHARSET=utf8 AUTO_INCREMENT=3 ;

Open in new window

0
 
LVL 6

Expert Comment

by:ahmad2121
Comment Utility
try this one please:
SELECT locations.city, physicians.PhysicianFirstName, physicians.PhysicianLastName

FROM physicians

INNER JOIN service_locations ON Service_LocationsDocID = PhysicianID

INNER JOIN locations ON Service_LocationsID = locations.id

Open in new window

0
 
LVL 16

Author Comment

by:rbudj
Comment Utility
ahmad, that does not work.  My original statement pulls all the correct data, I just cannot get it to display properly.
0
 
LVL 6

Accepted Solution

by:
ahmad2121 earned 500 total points
Comment Utility
ah ok, that wasn't clear from your first post.

below is a sample display of the fields you want. that should work.
 $SQL = "YOUR SQL STATEMENT";

$result = mysql_query($SQL);
 

while ($db_field = mysql_fetch_assoc($result)) {

print "<td>" . $db_field['city'] . "</td>";

print "<td>" . $db_field['PhysicianFirstName'] . "</td>";

print "<td>" . $db_field['PhysicianLastName'] . "</td>";

}

Open in new window

0
 
LVL 16

Author Comment

by:rbudj
Comment Utility
im still working on this... i will report back
0
 
LVL 16

Author Comment

by:rbudj
Comment Utility
still working on this, i will report back.
0

Featured Post

How your wiki can always stay up-to-date

Quip doubles as a “living” wiki and a project management tool that evolves with your organization. As you finish projects in Quip, the work remains, easily accessible to all team members, new and old.
- Increase transparency
- Onboard new hires faster
- Access from mobile/offline

Join & Write a Comment

Suggested Solutions

Things That Drive Us Nuts Have you noticed the use of the reCaptcha feature at EE and other web sites?  It wants you to read and retype something that looks like this.Insanity!  It's not EE's fault - that's just the way reCaptcha works.  But it is …
Since pre-biblical times, humans have sought ways to keep secrets, and share the secrets selectively.  This article explores the ways PHP can be used to hide and encrypt information.
This video teaches users how to migrate an existing Wordpress website to a new domain.
The viewer will learn how to look for a specific file type in a local or remote server directory using PHP.

772 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

10 Experts available now in Live!

Get 1:1 Help Now