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[SOLVED] Warning...

Posted: Mon Nov 01, 2004 2:16 pm
by teckyan888

Code: Select all

<?php
  require("globals.php");
  require("common.php");
  
  
  // create short variable names
  $searchtype=$HTTP_POST_VARS['searchtype'];
  $searchtype1=$HTTP_POST_VARS['searchtype1'];
  $searchtype2=$HTTP_POST_VARS['searchtype2'];

  if (!$searchtype || !$searchtype1 || !$searchtype2)
  {
     echo 'You have not entered search details.  Please go back and try again.';
     ReturnToSearch();
     exit;
  }
  
  $searchtype = addslashes($searchtype);
  $searchtype1 = addslashes($searchtype1);
  $searchtype2 = addslashes($searchtype2);


  @ $db = mysql_pconnect($hostName, $userName, $password);

  if (!$db)
  {
     echo 'Error: Could not connect to database.  Please try again later.';
     exit;
  }

  if(isset($_POST["search"]))
  {
  mysql_select_db($databaseName);
  $query = "select * from Info where DepartureDate LIKE '%".$searchtype."%'-'%".$searchtype1."%'-'%".$searchtype2."%' ";
  $result = mysql_query($query);

  $num_results = mysql_num_rows($result);
  
	if(!$num_results)
	{
	echo 'No user can be found...Please search again!';
	ReturnToSearch();
	exit();
	}

	printf("<TABLE BORDER WIDTH="100%%">\n");
	printf("<TR>
         <TD><B>TripID</B></TD>
         <TD><B>DepartureDate</B></TD>
         <TD><B>Price</B></TD>
         <TD><B>Season</B></TD>
         <TD><B>DepartureTime</B></TD>
         <TD><B>FlightTime</B></TD>
         <TD><B>NoOfPassenger</B></TD>
         <TD><B>TypeOfPassenger</B></TD>
	   <TD><B>CompanyName</B></TD>
	   <TD><B>NoOfPassengerTurnUpDepartureTime</B></TD>
         </TR>\n");

	while(($row = mysql_fetch_object($result))){
	printf("<TR>
         <TD>%s</TD>
         <TD>%s</TD>
         <TD>%s</TD>
         <TD>%s</TD>
         <TD>%s</TD>
         <TD>%s</TD>
         <TD>%s</TD>
         <TD>%s</TD> 
	   <TD>%s</TD>
         <TD>%s</TD>
        </TR>\n",
	$row->TripID,$row->DepartureDate, $row->Price, $row->Season, $row->DepartureTime, $row->FlightTime,$row->NoOfPassenger,$row->TypeOfPassenger,$row->CompanyName, $row->NoOfPassengerTurnUpDepartureTime);}
	printf("</TABLE>\n");
	mysql_free_result($result);
 	}//end if statement

ReturnToSearch();

?>
It display the warning like that:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\Program Files\Apache Group\Apache2\htdocs\cb359\search.php on line 43
No user can be found...Please search again!


Anyone who know what the problem of this???Is it my sql statement got problem???

Below is my html page...

Code: Select all

<?php
<html>
<head>
<title>User's Detail Search</title>
<head><body>
<h1>User's Detail Search</h1>
<form action="search.php" method="post">
<table>

<tr>
Choose Search Type:
</tr>
<tr>

<tr><td>Year</td><td>Month</td><td>Day</td></tr>
<td>
<select name="searchtype">
<option value="1998">1998</option><option value="1999">1999</option>
<option value="2000">2000</option><option value="2001">2001</option>
<option value="2002">2002</option><option value="2003">2003</option>
</select>
</td>

<td>
<select name="searchtype1">
<option value="01">January</option><option value="02">February</option><option value="03">March</option>
<option value="04">April</option><option value="05">May</option><option value="06">Jun</option>
<option value="07">July</option><option value="08">Ogos</option><option value="09">September</option>
<option value="10">October</option><option value="11">Novermber</option><option value="12">December</option>
</select>
</td>

<td>
<select name="searchtype2">
<option value="01">1</option><option value="02">2</option><option value="03">3</option>
<option value="04">4</option><option value="05">5</option><option value="06">6</option>
<option value="07">7</option><option value="08">8</option><option value="09">9</option>
<option value="10">10</option><option value="11">11</option><option value="12">12</option>
<option value="13">13</option><option value="14">14</option><option value="15">15</option>
<option value="16">16</option><option value="17">17</option><option value="18">18</option>
<option value="19">19</option><option value="20">20</option><option value="21">21</option>
<option value="22">22</option><option value="23">23</option><option value="24">24</option>
<option value="25">25</option><option value="26">26</option><option value="27">27</option>
<option value="28">28</option><option value="29">29</option><option value="30">30</option>
<option value="31">31</option>
</select>
</td>

</tr>

</table>
</br>

<input type="submit" value="Search" name="search">
</form>
<form action="loginadmin.php" method=post>
<input type="submit" value="Return To Administrator Page" name="main">
</form>
</body>
</html>


?>
[/php_man]

Posted: Tue Nov 02, 2004 1:03 am
by harsha
select * from Info where DepartureDate LIKE '%".$searchtype."%' - '%".$searchtype1."%'-'%".$searchtype2."%' ";

echo this statement and check in PHPMyAdmin

YOu have to give correct spaces where ever necessary in the sql statement.

something like

SELECT * FROMTABLE NAME is WRONG
SELECT * FROM TABLE NAME is correct

thanks

Posted: Tue Nov 02, 2004 7:24 am
by teckyan888
thanks a lot...i know the solution...thanks... :D