Sorting Results Problem

[russot00]

I have 3 drop down menus that are used in a search to locate restaurants in a db. All of the drop down menus function, a search can be submitted with any combination of drop downs and the results are sorted by restaurant name. The problem occurs when you do not select anything from a drop down ("All" is submitted for all drop downs) and submit the search, the following error occurs: "Can't connect because You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'BY `restaurant Name` ASC' at line 1." The error shows up after I add in "$restaurant_re sults = conn($find_rest aurants."ORDER BY `restaurant Name` ASC"); " line 93. Without this line, the reults are returned by index number and will not be sorted by restaurant name. I need to have all the results sorted by restaurant name. I have included the code below. Any help would be greatly appreciated.

<?php 
/*------------------------------------------------------------------------ 
            control codes 
------------------------------------------------------------------------*/ 
 
if (isset($_POST['submit'])) 
{ 
 
  search();       //call the search function 
 
}else{ 
 
  show_form();   //call the show form function 
 
}//end if 
 
/*------------------------------------------------------------------------ 
            show the search form 
------------------------------------------------------------------------*/ 
 
function show_form() 
{ 
  //call the dropdown function which creates an html string to build a select box for each element 
  $neighborhood_dropdown   = dropdown('neighborhood','restaurant'); 
  $cuisine_dropdown  = dropdown('cuisine1','restaurant'); 
  $type_dropdown     = dropdown('type','restaurant');
 
 
 
 
 
 
  echo "<form name='search' action='".$_SERVER['PHP_SELF']."' method='post'> 
        <table width='50%' align='center' valign='center'> 
        <tr> 
          <td colspan='2' align='center'>Search Form</td> 
        </tr> 
        <tr> 
          <td align='right'>Neighborhood:</td><td>$neighborhood_dropdown</td> 
        </tr> 
        <tr> 
          <td align='right'>Cuisine:</td><td>$cuisine_dropdown</td> 
        </tr> 
        <tr> 
          <td align='right'>Type:</td><td>$type_dropdown</td> 
        </tr> 
        <tr> 
          <td colspan='2' align='center'> </td> 
        </tr> 
        <tr> 
          <td colspan='2' align='center'><input type='submit' name='submit' value='Go!'></td> 
        </tr> 
        </table> 
        </form>"; 
 
}//end function 
 
 
/*------------------------------------------------------------------------ 
            run the search and show the results 
------------------------------------------------------------------------*/ 
 
function search() 
{ 
 
//base sql 
$find_restaurants = "select * FROM `restaurant` WHERE 1";
 
//get the values from the form 
//NOTE: You should do way more valdation on the values before you attempt to process anything 
 
 
  if ((!empty($_POST['neighborhood']))&&($_POST['neighborhood'] != 'all')) 
  { 
$find_restaurants .= " and neighborhood like '". addslashes($_POST['neighborhood']). "%'"; 
  }
 
  if ((!empty($_POST['cuisine1']))&&($_POST['cuisine1'] != 'all')) 
  { 
$find_restaurants .= " and cuisine1 like '". addslashes($_POST['cuisine1'])."%'"; 
  } 
 
  if ((!empty($_POST['type']))&&($_POST['type'] != 'all')) 
  { 
$find_restaurants .= " and type like '". addslashes($_POST['type'])."%' "; 
  }
 
  //add more elements (or take away) as you desire...follow the same code structure as above 
 
 
  //run query
 
    $restaurant_results = conn($find_restaurants."ORDER BY `restaurant Name` ASC"); 
 
  if (!$restaurant_results){ die("No results due to database error.<br>".mysql_error());  } 
 
  if (mysql_num_rows($restaurant_results)==0) 
  { 
    echo "No Results found!"; 
  }else{ 
 
    echo "<table border='1'><th>Restaurant Name</th><th>neighborhood</th><th>cuisine</th><th>type</th></tr>"; 
 
    while ($rows= mysql_fetch_array($restaurant_results)) 
    { 
      echo "<tr>"; 
      echo "<td>". $rows['restaurant Name'] ."</td>";
      echo "<td>". $rows['neighborhood']  ."</td>"; 
      echo "<td>". $rows['cuisine1'] ."</td>"; 
      echo "<td>". $rows['type']    ."</td>";
      echo "</tr>"; 
 
    }  
    echo "</table>"; 
  }//end if 
 
?>
<p align="center"><a href="test3.php">refine search</a></p>
<?
 
}//end function 
 
/*------------------------------------------------------------------------ 
            create the drop downs 
------------------------------------------------------------------------*/ 
 
function dropdown($field, $table) 
{  
  //initialize variables 
  $oHTML  = ''; 
  $get_dropdown_values = ''; 
 
  //check to see if the field is passed correctly 
  if (($field == "")||($table == "")) 
  { 
    die("No column or table specified to create drop down from!"); 
  } 
 
  $get_dropdown_values = "select distinct($field) from $table ORDER BY $field"; 
 
  //call the db function and run the query 
  $result_drop_down = conn($get_dropdown_values); 
 
  //if no results are found to create a drop down return a textbox 
  if ((!$result_drop_down) ||(mysql_num_rows($result_drop_down)==0)) 
  { 
    $oHTML .= "<input type='text' name='$field' value='' size='15'>"; 
  }elseif (($result_drop_down)&&(mysql_num_rows($result_drop_down)>0)){ 
 
    //build the select box out of the results 
    $oHTML .= "<select name='$field'>\n<option value='all'>All</option>\n"; 
    while ($rows = mysql_fetch_array($result_drop_down)) 
    { 
      $oHTML .= "<option value='".$rows[$field]."'>".$rows[$field]."</option>\n"; 
    } 
    $oHTML .= "</select>\n"; 
  } 
 
  //send the value back to the calling code 
  return $oHTML; 
}//end function 
 
/*------------------------------------------------------------------------ 
            database connection function 
------------------------------------------------------------------------*/ 
 
function conn($database) 
{    
include 'config.php';
include 'opendb.php';
 
    $result = mysql_query($database) or die ("Can't connect because ". mysql_error()); 
 
    return $result; 
 
}//end function      
 
?>
 

[helimeef]

I have 3 drop down menus that are used in a search to locate restaurants in a db. All of the drop down menus function, a search can be submitted with any combination of drop downs and the results are sorted by restaurant name. The problem occurs when you do not select anything from a drop down ("All" is submitted for all drop downs) and subm...

Change line 93 to

$restaurant_results = die($find_restaurants."ORDER BY `restaurant Name` ASC");
 

And tell me what it says on the page. FYI die() is a PHP function that ends the script and displays a message, so this will let me see what is getting queried.
I'm almost 100% sure that it not valid mysql to have a 2 word name for a column (or field as some peeps say), and maybe the random "1" placed at the end of the query on line 67.