Switch statement seems to work ...somewhat :S

[cyandi_man]

Hi fellas...i took some advice in using a switch statement to accomplish what i wanted.

I basically just want a specific list to be displayed depending on the office type and location chosen from a dropdown list.  Only now the listing that is displayed does not seem to co-inside with the

selected choices from the drop-down.  ex. if choosing a chiropractor from toronto, i get an optometrist showing up in the list!  I checked the switch code and it seems to be correct!? help please!

 

 

thanks for the help!

only now it seems that there is something wrong with my switch statement!

--database statements--
  $db = mysql_select_db($database,$connection)
       or die ("Couldn't select database");

if ($city=="-Any City-" && $industry=="-Any Discipline-") {
    $i==0;
}
elseif ($city!="-Any City-" && $industry!="-Any Discipline-") {
    $i==1;
} 
elseif ($city=="-Any City-" && $industry!="-Any Discipline-") {
    $i==2;
} 
elseif($industry=="-Any Discipline-" && $city !="-Any City-") {
   $i==3;
}

switch ($i) {
    case 0:
$query = "SELECT * FROM Proffesional_list";
    break;
    case 1:
$query = "SELECT * FROM Proffesional_list WHERE industry='$industry' AND city='$city'"; 
    break;
    case 2:
     $query = "SELECT * FROM Proffesional_list WHERE industry='$industry'";
     break;
case 3:
    $query = "SELECT * FROM Proffesional_list WHERE city='$city'";
    break;
}


$result= mysql_query($query)
or DIE("unable to retrieve database info");
--echo statements pas here--

 

I do get listings that show up - but not the proper selections!

[RussellReal]

  $i==3;

 

you're using the comparison operator, not to set it

 

$i = 3;

 

is correct

[cyandi_man]

  $i==3;

 

you're using the comparison operator, not to set it

 

$i = 3;

 

is correct

 

Thanks - i corrected that .  however it seems that i am not getting the values for industry and city from my form page when i choose "-any city-" and/or "-any discipline-"

 

Im new to coding...so - im not sure if i coded my drop down menu correctly.

i coded it as follows.

 

<?php 
-database connection code-


  $connection = mysql_connect($host,$user,$password)
       or die ("couldn't connect to server");
  $db = mysql_select_db($database,$connection)
       or die ("Couldn't select database");


  $query = "SELECT DISTINCT industry FROM Proffesional_list ORDER BY industry";
  $result = mysql_query($query)
       or die ("Couldn't execute query.");
        
  $query2 = "SELECT DISTINCT city FROM Proffesional_list ORDER BY city";
  $result2 = mysql_query($query2)
       or die ("Couldn't execute query.");
   
   $anycity ="-Any City-";
   $anyplace ="-Any Discipline-";

  /* create form containing selection list */
  echo "<form action='code/memberdisplaylist.php' name='list1_form' method='POST'>
        <select name='industry'>\n";
echo "<option value='$industry'>$anyplace</option>\n"; 
  while ($row = mysql_fetch_array($result))
  {
     extract($row);
     echo "<option value='$industry'>$industry\n";
  }
  echo "</select>\n";
  ?>
  
//location dropdown menu
<p> </p>
<span class="subheading">Location</span> <br/><?php
  /* create form containing selection list */
  echo "<form action='code/memberdisplaylist.php' name='list1_form' method='POST'>
        <select name='city'>\n";
echo "<option value='$city'>$anycity</option>\n";
  while ($row2 = mysql_fetch_array($result2))
  {
     extract($row2);
     echo "<option value='$city'>$city\n";
  }
  echo "</select>\n";
  ?>
<p> </p>
  <?php
?>

 

Is there anything wrong with my php coding? 

 

(edited to add

tags)