MartinaL Posted March 8, 2006 Share Posted March 8, 2006 I want to display some rows from a MySQL database in a php page in the following format;Each row should be displayed one after the other like this. column 1 is called Date, 2 is called Heading and 3 is called Content-----Date - HeadingContentI have tried setting this up with the following piece of code but nothing is being displayed on the screen at all;[code]<?php $dbhost = 'localhost'; $dbuser = 'dhopec'; $dbpass = '<<MY PASSWORD>>'; $dbname = 'dhopec_com_-_content'; $conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql'); mysql_select_db($dbname); $query_string = "select Date, Heading, Content from news"; $result = mysql_query($query); $num=mysql_numrows($result); mysql_close(); echo "<table><tr><td colspan="2"><b>Database Output</b></td></tr>"; $i=0; while ($i < $num) { $Date=mysql_result($result,$i,"Date"); $Heading=mysql_result($result,$i,"Heading"); $Content=mysql_result($result,$i,"Content"); echo "<tr><td><b>$Date</td><td>$Heading</td></tr><tr><td colspan="2">$Content</td></tr></table>"; $i++; } ?>[/code] Link to comment https://forums.phpfreaks.com/topic/4399-displaying-data-from-mysql-db-problems/ Share on other sites More sharing options...
greycap Posted March 8, 2006 Share Posted March 8, 2006 [!--quoteo(post=352717:date=Mar 7 2006, 09:02 PM:name=MartinaL)--][div class=\'quotetop\']QUOTE(MartinaL @ Mar 7 2006, 09:02 PM) [snapback]352717[/snapback][/div][div class=\'quotemain\'][!--quotec--][/quote]$result = mysql_query($query);Shouldn't that be mysql_query($query_string) in your example?Also, any one of those mysql functions can fail. You should be checking for that and handling it every time. Link to comment https://forums.phpfreaks.com/topic/4399-displaying-data-from-mysql-db-problems/#findComment-15266 Share on other sites More sharing options...
MartinaL Posted March 8, 2006 Author Share Posted March 8, 2006 I updated but still no luck, nothing at all is displayed on the page.I'm fairly new to PHP so to check each line to see if it fails how do I do this? Link to comment https://forums.phpfreaks.com/topic/4399-displaying-data-from-mysql-db-problems/#findComment-15299 Share on other sites More sharing options...
greycap Posted March 8, 2006 Share Posted March 8, 2006 [!--quoteo(post=352757:date=Mar 7 2006, 10:51 PM:name=MartinaL)--][div class=\'quotetop\']QUOTE(MartinaL @ Mar 7 2006, 10:51 PM) [snapback]352757[/snapback][/div][div class=\'quotemain\'][!--quotec--]I updated but still no luck, nothing at all is displayed on the page.I'm fairly new to PHP so to check each line to see if it fails how do I do this?[/quote]The other problem is this:$num=mysql_numrows($result);Theres no such thing as mysql_numrows() its mysql_num_rows(). This will make $num = null which is why that loop is never entered.Dont you have error messages turned on? Link to comment https://forums.phpfreaks.com/topic/4399-displaying-data-from-mysql-db-problems/#findComment-15302 Share on other sites More sharing options...
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