phpretard

I get the same error without the single quotes and using: $result = mysql_query("SELECT * FROM URLS WHERE id<='".$GetId."' "); if (!result){die(mysql_error());} while($row = mysql_fetch_array($result)) { mysql_query("DELETE FROM URLS WHERE id<='".$delete."' "); }

I am trying to delete about 10K rows. The script works without the form. With the from I get the error: Warning: mysql_query() [function.mysql-query]: Unable to save result set in ... on line 20 Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in ... on line 24 if (isset($_POST['deleterows'])){ include ("../connect.php"); $GetId=$_POST['rowsNum']; $delete=$_POST['rowsNum']; echo $GetId; $result = mysql_query("SELECT * FROM URLS WHERE id<='$GetId'"); //<<<< LINE 20 if (!result){die(mysql_error());} while($row = mysql_fetch_array($result)) //<<<< LINE 24 { mysql_query("DELETE FROM URLS WHERE id<='$delete' "); } mysql_close($con); } // END IF else{ echo" <form action='' method='post'> <font face=arial>Delete All Rows And Lower Starting With:</font><br> <input type=text name='rowsNum' size='40' autocomplete=off> <input type=submit name='deleterows' value='Delete'> </form> "; } // END ELSE I can't seem to find the problem. Any help today?

It still shows all even with PaidID='0'

I can't figure out why this query still shows results where PaidID='1'. The url calling this would be: ...?page=listing/results&State=FL OR $SEARCH="FL"; << Session if (!ereg('[^0-9]', $SEARCH)){$searchResult = mysql_query("SELECT * FROM members WHERE ZipCodes LIKE '%$SEARCH%' AND PaidID!='1' ");} elseif (!ereg('[^A-Z]', $SEARCH) && (strlen($SEARCH) == 2)){ $searchResult = mysql_query("SELECT * FROM members WHERE PaidID!='1' AND LicState LIKE '%$SEARCH%' OR LicState2 LIKE '%$SEARCH%' OR LicState3 LIKE '%$SEARCH%' OR LicState4 LIKE '%$SEARCH%' "); } elseif (isset($_GET['State'])){ $searchResult = mysql_query("SELECT * FROM members WHERE PaidID!='1' AND (LicState='$State' AND Counties LIKE '%$SEARCH%' ) OR (LicState2='$State' AND Counties2 LIKE '%$SEARCH%') OR (LicState3='$State' AND Counties3 LIKE '%$SEARCH%') OR (LicState4='$State' AND Counties4 LIKE '%$SEARCH%') "); } elseif ($SEARCH=="EXAMPLE"){ $searchResult = mysql_query("SELECT * FROM members WHERE PaidID!='1' ORDER BY RAND() LIMIT 5"); } else{ $searchResult = mysql_query("SELECT * FROM members WHERE PaidID!='1' AND Counties LIKE '%$SEARCH%' OR Counties2 LIKE '%$SEARCH%' OR Counties3 LIKE '%$SEARCH%' OR Counties4 LIKE '%$SEARCH%' "); } Could anyone tell me? Thanks.

OH Well...

I have come up with this... onKeyDown=\"if(event.which==13||event.keyCode==13){alert('Please Click The Search Button')}\" I was hope for a real solution.

I have been searching the internet for 2 hours and nothing refers to an image as the submit obj. So should I $rephrase the question? if ($rephrase=='Yes'){$question="How can I force the focus?";} else {echo"beat it with this question!";}

Andy-H: I don't get it? Yes I am using <input... There was a lot of useless info to post (my bad). I should mention it works great when the image is "clicked". It does work in firefox but...I wish everyone used firefox. Any work around for IE?

Thanks for the quick response. // FORM PAGE <form action='SUBMITPAGE' method=post> <input type='text' name='PrimSearch' /> <type='image' src='/images/search.gif' name='MainSearch' /></td> </form> // SUBMIT PAGE if (isset($_POST['MainSearch_y'])){ echo "Enter Button Won't Work"; }

I have form using an image as a submit button. It won't submit when I press enter...only if I click the button. This may not be a PHP help but I figured I throw it out here and see if anyone could point me in the right direction. I've been all over Google and the only reference I could find was standard type='submit'. ...and yes if (isset(I_Know_y)){ blah

I took your advise on both post (THANK YOU!) I now get: mysql_num_rows(): supplied argument is not a valid MySQL result resource...on line 17 if (!ereg('[^A-Z]', $SEARCH) && (strlen($SEARCH) == 2)){ $searchResult = mysql_query(" SELECT * FROM members WHERE LicState LIKE '%$SEARCH%' OR LicState2 '%$SEARCH%' OR LicState3 LIKE '%$SEARCH%' OR LicState4 LIKE '%$SEARCH%' ");} if (!searchResult){die(mysql_error());} $num_rows= mysql_num_rows($searchResult); << -------- LINE 17 echo $num_rows." FOUND"; } Any ideas on this?

Does anyone know why I would get this? Fatal error: Can't use function return value in write context $SEARCH="AB"; if (!ereg('[^A-Z]', $SEARCH) && (strlen($SEARCH) = 2))){ $searchResult = mysql_query(" SELECT * FROM members WHERE LicState LIKE '%$SEARCH%' OR LicState2 '%$SEARCH%' OR LicState3 LIKE '%$SEARCH%' OR LicState4 LIKE '%$SEARCH%' ");} if (!searchResult){die(mysql_error());} $num_rows= mysql_num_rows($searchResult); echo $num_rows." FOUND"; I am looking for: "1 FOUND" ...and I get: Fatal error: Can't use function return value in write context Thank again!

Can anyone see why this would return error: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource $searchResult = mysql_query("SELECT * FROM members WHERE LicState LIKE '%$SEARCH%' OR LicState2 '%$SEARCH%' OR LicState3 LIKE '%$SEARCH%' OR LicState4 LIKE '%$SEARCH%' "); if (!sql){die(mysql_error());} $num_rows = mysql_num_rows($searchResult); echo $num_rows." FOUND"; Thanks!

So if my conection was var $con then it would be: global $con ?

Thank you...I have much to learn. PS. What is global $db;

How can I convert the code below in a funtion. I then need to call it out and see the results. echo"<select name='SiteArea'>"; $resultshape = mysql_query("SELECT * FROM shape"); while($row = mysql_fetch_array($resultshape)) { $id=$row['id']; $name=$row['name']; echo "<option value='$name'>$name</option>"; } echo "<select>"; Any help today? -Anthony

Not a good question?

I know how to cnovert files to PDF. Does anyone know how to convert files to .VSS Any tutorials, pointers, etc... anything would be helpful. It's a Visual SafeSource file. http://msdn.microsoft.com/en-us/library/aa302175.aspx#vssmap_topic2 Thanks! -Anthony

What kind of files?

Put this before session start and you're good to go! header('P3P: CP="CAO PSA OUR"');

I am losing my session $vars after redidecting within a frameset? Any suggestions. This code works great without frames involved so I was hoping there was a trick to a frame. $result = mysql_query("SELECT * FROM table WHERE this='$that"); while($row = mysql_fetch_array($result)) { $MemberID=$row['MemberID']; $Type=$row['Type']; $UserID=$row['UserID']; $CLFirstName=$row['CLFirstName']; $CLLastName=$row['CLLastName']; } $_SESSION['MemberID']=$MemberID; $_SESSION['Type']=$Type; $_SESSION['UserID']=$UserID; $_SESSION['LastLog']="$DATE_TIME_STAMP"; echo "<pre>". print_r($_SESSION,true) ."</pre>"; <<< I KNOW THEY GET SET HERE header("Location: ?page=orderform"); <<< THEY"RE GONE WHEN YOU GET TO THIS PAGE } This is all done inside of a frame. Does anyone know what's wrong or the "trick" to passing session within frames? PS. Yes I hate frames...it's my only option.

I see what you mean... <?php session_start(); $_SESSION['PoolID']="Stuff"; if (!empty($_SESSION['PoolID'])) { setcookie('session', 'active', time()+20); } else { header ('Location: http://www.google.com'); exit; // exit the current script } if (!isset($_COOKIE['session'])) { session_destroy(); // if not, destroy session header ('Location: http://www.google.com'); exit; } echo "<pre>". print_r($_SESSION,true) ."</pre>"; ?> So how do check for the cookie on the same page?

I just do not want people sitting Idle on my site (logged in)

// THERE SHOULD BE SOME CONDITION HERE <<<< IT MISSING SURLEY........... //LIKE YOU BEEN TOLD YOUR GOTO GOOGLE REGARDLESS, UNLESS YOU ADD ANOTHER CONDITION This is the part I need help with.

I am... any suggestion?