zachatk1 Posted September 3, 2010 Share Posted September 3, 2010 Alright so I created a MySQL database that has 5 tables each named a brand of a dirt bike. In each table their are 2 fields, one INDEX_ID and MODELS. Under that for rows I have every model bike named. A quick question before I get onto what I want to do: Can data be added underneath the rows? For example, users will be able to submit information about each bike model. If each bike model is a row, can there be a category past that row or do I need to make each field a model name and just have a ton of fields and have the rows be the information users submit. To make it easier to understand I'll post the SQL code for the brand Honda: CREATE TABLE `Honda` ( `INDEX_ID` int(3) NOT NULL auto_increment, `MODELS` varchar(20) collate latin1_general_ci NOT NULL, PRIMARY KEY (`INDEX_ID`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 COLLATE=latin1_general_ci AUTO_INCREMENT=23 ; -- -- Dumping data for table `Honda` -- INSERT INTO `Honda` VALUES(1, 'CR85'); INSERT INTO `Honda` VALUES(2, 'CR125'); INSERT INTO `Honda` VALUES(3, 'CR250'); INSERT INTO `Honda` VALUES(4, 'CRF100'); INSERT INTO `Honda` VALUES(5, 'CRF150'); INSERT INTO `Honda` VALUES(6, 'CRF230'); INSERT INTO `Honda` VALUES(7, 'CRF250X'); INSERT INTO `Honda` VALUES(8, 'CRF250R'); INSERT INTO `Honda` VALUES(9, 'CRF450X'); INSERT INTO `Honda` VALUES(10, 'CRF450R'); INSERT INTO `Honda` VALUES(11, 'CRF50'); INSERT INTO `Honda` VALUES(12, 'CRF70'); INSERT INTO `Honda` VALUES(13, 'CRF80'); INSERT INTO `Honda` VALUES(14, 'XR650'); INSERT INTO `Honda` VALUES(15, 'CR500'); INSERT INTO `Honda` VALUES(16, 'XR100'); INSERT INTO `Honda` VALUES(17, 'XR200'); INSERT INTO `Honda` VALUES(18, 'XR250'); INSERT INTO `Honda` VALUES(19, 'XR400'); INSERT INTO `Honda` VALUES(20, 'XR50'); INSERT INTO `Honda` VALUES(21, 'XR70'); INSERT INTO `Honda` VALUES(22, 'XR80'); Anyway I still need to figure out how to have this under a form that a user can use to select the bike they want to submit information about. A code like this perhaps?: <? $connection = mysql_connect("localhost","user","pass"); $fields = mysql_list_fields("dbname", "table", $connection); $columns = mysql_num_fields($fields); echo "<form action=page_to_post_to.php method=POST><select name=Field>"; for ($i = 0; $i < $columns; $i++) { echo "<option value=$i>"; echo mysql_field_name($fields, $i); } echo "</select></form>"; ?> Thanks. Quote Link to comment Share on other sites More sharing options...
pengu Posted September 3, 2010 Share Posted September 3, 2010 Alright so I created a MySQL database that has 5 tables each named a brand of a dirt bike. In each table their are 2 fields, one INDEX_ID and MODELS. Under that for rows I have every model bike named. A quick question before I get onto what I want to do: Can data be added underneath the rows? For example, users will be able to submit information about each bike model. If each bike model is a row, can there be a category past that row or do I need to make each field a model name and just have a ton of fields and have the rows be the information users submit. To make it easier to understand I'll post the SQL code for the brand Honda: CREATE TABLE `Honda` ( `INDEX_ID` int(3) NOT NULL auto_increment, `MODELS` varchar(20) collate latin1_general_ci NOT NULL, PRIMARY KEY (`INDEX_ID`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 COLLATE=latin1_general_ci AUTO_INCREMENT=23 ; -- -- Dumping data for table `Honda` -- INSERT INTO `Honda` VALUES(1, 'CR85'); INSERT INTO `Honda` VALUES(2, 'CR125'); INSERT INTO `Honda` VALUES(3, 'CR250'); INSERT INTO `Honda` VALUES(4, 'CRF100'); INSERT INTO `Honda` VALUES(5, 'CRF150'); INSERT INTO `Honda` VALUES(6, 'CRF230'); INSERT INTO `Honda` VALUES(7, 'CRF250X'); INSERT INTO `Honda` VALUES(8, 'CRF250R'); INSERT INTO `Honda` VALUES(9, 'CRF450X'); INSERT INTO `Honda` VALUES(10, 'CRF450R'); INSERT INTO `Honda` VALUES(11, 'CRF50'); INSERT INTO `Honda` VALUES(12, 'CRF70'); INSERT INTO `Honda` VALUES(13, 'CRF80'); INSERT INTO `Honda` VALUES(14, 'XR650'); INSERT INTO `Honda` VALUES(15, 'CR500'); INSERT INTO `Honda` VALUES(16, 'XR100'); INSERT INTO `Honda` VALUES(17, 'XR200'); INSERT INTO `Honda` VALUES(18, 'XR250'); INSERT INTO `Honda` VALUES(19, 'XR400'); INSERT INTO `Honda` VALUES(20, 'XR50'); INSERT INTO `Honda` VALUES(21, 'XR70'); INSERT INTO `Honda` VALUES(22, 'XR80'); Anyway I still need to figure out how to have this under a form that a user can use to select the bike they want to submit information about. A code like this perhaps?: <? $connection = mysql_connect("localhost","user","pass"); $fields = mysql_list_fields("dbname", "table", $connection); $columns = mysql_num_fields($fields); echo "<form action=page_to_post_to.php method=POST><select name=Field>"; for ($i = 0; $i < $columns; $i++) { echo "<option value=$i>"; echo mysql_field_name($fields, $i); } echo "</select></form>"; ?> Thanks. <?PHP $con=mysql_connect(DB_HOSTNAME,DB_USERNAME,DB_PASSWORD); mysql_select_db(DB_DATABASE, $con); $sql = "SELECT field1 FROM table"; $result = mysql_query($sql,$con); ?> <form> <select> <?PHP while($row = mysql_fetch_assoc($result)) { echo "<option>" .$row['field1]. "</option>"; } ?> </select> </form> Something like this would work. Quote Link to comment Share on other sites More sharing options...
zachatk1 Posted September 3, 2010 Author Share Posted September 3, 2010 Ok it displays a dropdown box, but no text or anything is in it. Quote Link to comment Share on other sites More sharing options...
dgoosens Posted September 3, 2010 Share Posted September 3, 2010 make sure you replace the instances of "field1" with the column titles of your DB (INDEX_ID & MODEL). Also, you might want to pass the ID into the form echo "<option value=\"" . $row['INDEX_ID'] . "\">" . $row['MODEL'] . "</option>" . PHP_EOL; Quote Link to comment Share on other sites More sharing options...
pengu Posted September 5, 2010 Share Posted September 5, 2010 echo "<option>" .$row['field1]. "</option>"; Should be.. echo "<option>" .$row['field1']. "</option>"; Quote Link to comment Share on other sites More sharing options...
zachatk1 Posted September 11, 2010 Author Share Posted September 11, 2010 Still having issues. I viewed the page source and discovered something that wasn't displaying on the webpage. It says: PHP Error Message Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/a2986016/public_html/modform.php on line 22 What does it mean? Quote Link to comment Share on other sites More sharing options...
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