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$_POST Help


Attila

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Hello I am trying to have someone enter somthing into a text block and submit it and have the results found in the database be shown.  I am not sure what I am doing wrong here.  You can see my site at: http://thaczero.com/search.php

 

My coding is like this:

<?
  	$top_form = "<form method=\"post\" action=\"$_SERVER[php_SELF]\">
     <table border=\"2\" cellpadding=\"2\" cellspacing=\"2\" bordercolor=\"#333333\" bgcolor=\"#CCCCCC\">
        <tr align=\"center\" valign=\"middle\">
          <td colspan=\"1\">";
$bottom_form = "</td>
        </tr>
	<tr>
          <td>Enter your search word, name, or phrase</td>
          <td><input type=\"text\" value=\"$_POST[search]\"></td>
        </tr>
        <tr align=\"center\" valign=\"middle\">
          <td colspan=\"2\">
	  <input type=\"hidden\" name=\"op\" value=\"ds\">
	  <input type=\"submit\" name=\"Submit\" value=\"Search\"></td>
        </tr>
      </table>
  </form>";
  
if ($_POST[op] != "ds") 
{ 
	echo "In the if";
	echo "$top_form"."$bottom_form";
} 
else
	{
		echo "in the else";
		echo "$_POST[search]";
		$item = $_POST[search];
		echo "$item";
		$sql = 'SELECT * FROM `ddoitems` WHERE `ItemDescription` LIKE "$item"';
		database_connect();
		$result = mysql_query($sql) or die(mysql_error());
		echo "$top_form"."$bottom_form";
   		while($row = mysql_fetch_array($result)) {
			echo '<tr>';
        		echo '<td><div align="center" class="style1">'.$row['ItemName'].'</div></td>';
			echo '<td><div align="center" class="style1">'.$row['ItemDescription'].'</div></td>';
			echo '<td><div align="center" class="style1">'.$row['Type'].'</div></td>';
			echo '<td><div align="center" class="style1">'.$row['QuestName'].'</div></td>';
			echo '<td><div align="center" class="style1">'.$row['lvl'].'</div></td>';
			echo '<td><div align="center" class="style1">'.$row['Notes'].'</div></td>';
        		echo '</tr>';

		}
	}

?>


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Try putting your $item variable inside %'s like so:

 

<?php
  $sql = 'SELECT * FROM `ddoitems` WHERE `ItemDescription` LIKE "%$item%"';
?>

 

 

Also, This isn't necessary but I recommend putting quotes around your $_POST, $_GET, etc. variables.

 

Example

<?php
  $item = $_POST['search'];
  instead of
  $item = $_POST[search];
?>

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I made some changes to your code

@Goldeneye it is necessary imagine a constant named search having a value foobar, would take a long time to debug

 

<?

  $top_form = "<form method=\"post\" action=\"{$_SERVER['PHP_SELF']}\">

    <table border=\"2\" cellpadding=\"2\" cellspacing=\"2\" bordercolor=\"#333333\" bgcolor=\"#CCCCCC\">

       <tr align=\"center\" valign=\"middle\">

         <td colspan=\"1\">";

$bottom_form = "</td>

       </tr>

<tr>

         <td>Enter your search word, name, or phrase</td>

         <td><input type=\"text\" value=\"{$_POST['search']}\"></td>

       </tr>

       <tr align=\"center\" valign=\"middle\">

         <td colspan=\"2\">

 <input type=\"hidden\" name=\"op\" value=\"ds\">

 <input type=\"submit\" name=\"Submit\" value=\"Search\"></td>

       </tr>

     </table>

 </form>";

 

if ($_POST['op'] != "ds")

{

echo "In the if";

echo "$top_form"."$bottom_form";

}

else

{

echo "in the else";

echo "$_POST['search']";

$item = $_POST['search'];

echo "$item";

$sql = 'SELECT * FROM `ddoitems` WHERE `ItemDescription` LIKE "%$item%"';

database_connect();

$result = mysql_query($sql/*, $dbconnection check php.net for the right syntax */) or die(mysql_error());

echo "$top_form"."$bottom_form";

  while($row = mysql_fetch_array($result)) {

echo '<tr>';

        echo '<td><div align="center" class="style1">'.$row['ItemName'].'</div></td>';

echo '<td><div align="center" class="style1">'.$row['ItemDescription'].'</div></td>';

echo '<td><div align="center" class="style1">'.$row['Type'].'</div></td>';

echo '<td><div align="center" class="style1">'.$row['QuestName'].'</div></td>';

echo '<td><div align="center" class="style1">'.$row['lvl'].'</div></td>';

echo '<td><div align="center" class="style1">'.$row['Notes'].'</div></td>';

        echo '</tr>';

 

}

}

 

?>

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That helped a lot but it does not apear to be getting what I type into the text block?  So what is wrong with that portion of the code.  Do I need the page to reload to post that into the $_POST['search'];

 

Thanks,

 

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