[SOLVED] Warning...

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teckyan888
Forum Commoner
Posts: 40
Joined: Tue May 11, 2004 10:46 am

[SOLVED] Warning...

Post 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]
User avatar
harsha
Forum Contributor
Posts: 103
Joined: Thu Jul 11, 2002 1:35 am
Location: Bengaluru (Bangalore) > Karnataka > India

Post 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
teckyan888
Forum Commoner
Posts: 40
Joined: Tue May 11, 2004 10:46 am

thanks

Post by teckyan888 »

thanks a lot...i know the solution...thanks... :D
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