rscott7706 Posted June 19, 2008 Share Posted June 19, 2008 Hello all, I have a problem with a table snippet. I have a page that works fine using the following code: <?php $db = mysql_connect("masterli.ipowermysql.com", "", ""); mysql_select_db("masterli_rates"); $query = "SELECT * FROM costs"; $result = mysql_query($query) or die(mysql_error()); ?> <?php echo "<font face=\"times new roman\">", "<size=\"1\">"; echo "<table>"; $query = mysql_query("SELECT Item, Price, Postage, Notes FROM costs"); while(list($Item, $Price, $Postage, $Notes)=mysql_fetch_row($query)){ echo " <tr> <td width=500, align: top, cell spacing: 3px>$Item</td> <td width=20, vertical-align: top, cell spacing: 3px align=right>$$Price</td> <td width=100, vertical-align: top, cell spacing: 3px>$Postage</td> <td width=400, vertical-align: top, cell spacing: 3px>$Notes</td> </tr>"; } echo "</table>"; ?> Here is the link: http://www.masterliens.com/prices-alone.php So, I copy this exact code to another site and re-address the necessary fields, now it won't work. And I get the following message: Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in /home/content/s/d/r/sdrental/html/automotive-page.php on line 92 Here is the code from the second site: <?php echo "<font face=\"verdana\">"; echo "<br />"; $db = mysql_connect("p50mysql99.secureserver.net", "", ""); mysql_select_db("tools_equip"); $query = "SELECT * FROM inventory"; $result = mysql_query($query) or die(mysql_error()); ?> <?php echo "<font face=\"arial\">", "<size=\"1\">"; echo "<table>"; $query = mysql_query("SELECT Desc, Model, Pic FROM inventory"); while(list($Desc, $Model, $Pic)=mysql_fetch_row($query)){ echo " <tr> <td width=50, align: top, cell spacing: 3px>$Desc</td></tr> <td width=200, vertical-align: top, cell spacing: 3px align=left>$Model</td> <td width=50, vertical-align: top, cell spacing: 3px>$Pic</td></tr> </tr>"; } echo "</table>"; ?> The first site is hosted by IpowerWeb, and is using: PHP Version 4.4.7 The second site is hosted by GoDaddy and is using: PHP Version 4.3.11 Any ideas? ??? Quote Link to comment Share on other sites More sharing options...
DarkWater Posted June 19, 2008 Share Posted June 19, 2008 $query = mysql_query("SELECT Desc, Model, Pic FROM inventory"); Change to: $query = mysql_query("SELECT Desc, Model, Pic FROM inventory") OR die(mysql_error()); Then post your error here. =/ Quote Link to comment Share on other sites More sharing options...
menriquez Posted June 19, 2008 Share Posted June 19, 2008 it looks like you don't have a userid and password for your mysql_connect function call... - mark Quote Link to comment Share on other sites More sharing options...
abdfahim Posted June 19, 2008 Share Posted June 19, 2008 you cannot write $query = mysql_query("SELECT Desc, Model, Pic FROM inventory"); because Desc is a mySQL command and so the above query will definitely gives an error. Either you should change the column name or re-write the query as follows $query = mysql_query("SELECT `Desc`, `Model`, `Pic` FROM inventory"); Quote Link to comment Share on other sites More sharing options...
DarkWater Posted June 19, 2008 Share Posted June 19, 2008 it looks like you don't have a userid and password for your mysql_connect function call... - mark He probably deleted them for this post. =) His error is caused by an invalid query. Quote Link to comment Share on other sites More sharing options...
rscott7706 Posted June 19, 2008 Author Share Posted June 19, 2008 DarkWater, menriquez and abdbuet, Thanks for your quick responses. menriquez - I left out log on info intentionally. abdbuet - your solution worked fine, thank you all so much!!! Ron Scott Quote Link to comment Share on other sites More sharing options...
DarkWater Posted June 19, 2008 Share Posted June 19, 2008 @rscott7706: I'd change the column name instead of relying on backtick (`) "escaping" (getting around a reserved word), because I think that only MySQL supports ` `. So change the column name. D: Quote Link to comment Share on other sites More sharing options...
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