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Displaying data from MySQL DB problems


MartinaL

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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 - Heading
Content

I 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]
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https://forums.phpfreaks.com/topic/4399-displaying-data-from-mysql-db-problems/
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[!--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.
[!--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?

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