Hi I have been struggling for nearly two months, I have tried many different statergies, but I just can’t seem to figure out what I am doing wrong. Could someone please look at my code and let me know how I can display images on my web page
-- Table structure for table `project_details`
--
CREATE TABLE `project_details` (
`project_id` int(10) unsigned NOT NULL auto_increment,
`project_name` varchar(50) NOT NULL,
`street` varchar(60) NOT NULL,
`CityOrTown` varchar(50) NOT NULL,
`county` varchar(30) default NULL,
`country` varchar(50) NOT NULL default '',
`pCode` varchar(30) NOT NULL,
`description` text NOT NULL,
`value` varchar(6) default NULL,
`client` varchar(90) default NULL,
`structEng` varchar(50) NOT NULL,
`mContractor` varchar(50) NOT NULL,
`projectType` varchar(25) default NULL,
`projectType2` varchar(60) default NULL,
`keywords` varchar(255) default NULL,
`projectDate` datetime default NULL,
`pImage_name` varchar(60) default NULL,
`image_name` varchar(60) default NULL,
PRIMARY KEY (`project_id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=30 ;
-- Table structure for table `project_image`
--
CREATE TABLE `project_image` (
`pImage_id` int(10) unsigned NOT NULL auto_increment,
`project_id` int(10) unsigned NOT NULL default '0',
`pImage_name` varchar(60) NOT NULL default '',
`iDescription` varchar(255) default NULL,
`image_name` varchar(30) default NULL,
`iKeywords` varchar(255) default NULL,
PRIMARY KEY (`pImage_id`),
KEY `project_id` (`project_id`),
KEY `pImage_name` (`pImage_name`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=0 ;
[php]
<?php require_once ('mysql_connect.php'); $data = mysql_query("SELECT * FROM project_details, project_image WHERE project_details.project_id = project_image.project_id LIMIT $from, $max_results") or die ('Cannot select database'); Print "
Project Name: ".$info['project_name'] . ", ";
print " ".$info['street'] .", ";
print " ".$info['CityOrTown'] .", ";
print "".$info['country'] .". "; print " Project Type: ".$info['projectType'] ." "; print " Client: ".$info['client'] ." "; print " Main Contractor: ".$info['mContractor'] ." "; print " Structural Engineer: ".$info['structEng'] ." "; print " Value: £".$info['value'] ."M "; print " Project Description: ".$info['description'] ." "; print " |
";
//print .info['pImage'] .
/* this is where I am stuck! I want to display an image which is stored on the project_image table where the project_details.project_id = project_image.project_id. Does anyone know how this can be done?
*/
" |
[/php]
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