illuz1on Posted March 27, 2007 Share Posted March 27, 2007 hey whats wrong here? getting this error: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/capetown/public_html/feature-edit.php on line 134 Cant figure it out! code of edit-feature.php:: <? //connect to mysql require_once("db.php"); //If cmd has not been initialized if(!isset($cmd)) { //display all the news $result = mysql_query("select * from featured"); //run the while loop that grabs all the news scripts while($r=mysql_fetch_array($result)) { //grab the title and the ID of the news $ft=$r["ft"];//take out the title $id=$r["id"];//take out the id //make the title a link echo "<a href='feature-edit.php?cmd=edit&id=$id'>$ft - Edit</a>"; echo "<br>"; } } ?> <? if($_GET["cmd"]=="edit" || $_POST["cmd"]=="edit") { if (!isset($_POST["submit"])) { $id = $_GET["id"]; $sql = "SELECT * FROM featured WHERE id=$id"; $result = mysql_query($sql); $myrow = mysql_fetch_array($result); ?> <form action="feature-edit.php" method="post"> <input type="hidden" name="id" value="<?php echo $myrow["id"] ?>"> Name:<INPUT TYPE="TEXT" NAME="ft" VALUE="<?php echo $myrow["ft"] ?>" SIZE=30><br> Picture:<INPUT TYPE="TEXT" NAME="fpic" VALUE="<?php echo $myrow["fpic"] ?>" SIZE=30><br> Text:<INPUT TYPE="TEXT" NAME="fw" VALUE="<?php echo $myrow["fw"] ?>" SIZE=30><br> More Link:<INPUT TYPE="TEXT" NAME="flink" VALUE="<?php echo $myrow["flink"] ?>" SIZE=30> <br> <input type="submit" name="cmd" value="edit"> </form> <? } ?> <? if ($_POST["$submit"]) { $ft1 = $_POST["ft"]; $fpic1 = $_POST["fpic"]; $fw1 = $_POST["fw"]; $flink1 = $_POST["flink"]; mysql_query("UPDATE featured SET fpic='$fpic1', ft='$ft1', fw='$fw1', flink='$flink1' WHERE id='$id'") or die (mysql_error()); echo "Thank you! Information on SPECIAL FEATURES updated."; } } ?> Quote Link to comment Share on other sites More sharing options...
MadTechie Posted March 27, 2007 Share Posted March 27, 2007 Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/capetown/public_html/feature-edit.php on line 134 well the code is only 61 lines so error on 134 is weird however this may reveal the problem <?php $id = $_GET["id"]; $sql = "SELECT * FROM featured WHERE id=$id"; echo $sql; $result = mysql_query($sql); $myrow = mysql_fetch_array($result); ?> Quote Link to comment Share on other sites More sharing options...
hitman6003 Posted March 27, 2007 Share Posted March 27, 2007 Change: $result = mysql_query("select * from featured"); to $result = mysql_query("select * from featured") or die(mysql_error()); Quote Link to comment Share on other sites More sharing options...
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