CODEWITHSUNDEEP
phpmysql
Jacob1
Jacob1

Reputation: 9

retieve data from mysql and display in form

I have a radio box and 2 drop down menus which when submitted save to mysql. The radio box is either yes or no and the 2 dropdowns where created in html. It is currently fully working and saves all the data.

What I now wish to do is when a user logs back in, it will show what they have previously selected (if they have).

PHP SCRIPT:

<?php
session_start(); 
require_once("config.php"); 
if(!isset($_SESSION['username'])){
   header('Location: login.php'); 
   exit; 
}else{
    $sql = "SELECT attendance1 FROM user WHERE username = '".mysql_real_escape_string($_SESSION['username'])."'";
    $res = mysql_query($sql);
    $row = mysql_fetch_array($res);
    if(($row[0] == "Yes") || ($row[0] == "No")){
        header("Location: errorsubmit.html");
        exit;
    }
}
if(isset($_POST['submit'])){         
     $sql = "UPDATE user SET attendance1 = '" . mysql_real_escape_string($_POST['attendance1']) . "' WHERE username = '" .  mysql_real_escape_string($_SESSION['username']) . "'";
mysql_query($sql) or die("Error in SQL: " . mysql_error());  

   $sql = "UPDATE user SET colour1= '" . mysql_real_escape_string($_POST['colour1']) . "' WHERE username = '" . mysql_real_escape_string($_SESSION['username']) . "'";
mysql_query($sql) or die("Error in SQL: " . mysql_error());  

   $sql = "UPDATE user SET shade1= '" . mysql_real_escape_string($_POST['shade1']) . "' WHERE username = '" . mysql_real_escape_string($_SESSION['username']) . "'";
mysql_query($sql) or die("Error in SQL: " . mysql_error());  

   header("Location: thanks.html", true, 303);
}
?>

FORM:

<form>
         <input name="attendance1" type="radio" id="Yes" value="Yes" checked="checked"/>Yes
         <br />
<input name="attendance1" type="radio" id="No" value="No" />No
       </h3></td>
       <td>
              <select name="colour1" id="colour1" >
           <option selected="selected">Please Select</option>
           <option>Red</option>
           <option>White</option>
           <option>Green</option>
         </select>
       </td>
       <td><h3>
                 <select name="shade1" id="shade1" >
           <option selected="selected">Please Select</option>
           <option>light</option>
           <option>heavy</option>
         </select>
       <td>&nbsp;</td>
       <td><label>
         <input type="submit" name="submit" id="button" value="Submit" />
       </label></td>
     </tr>
   </table>

Upvotes: 1

Views: 2304

Answers (2)

user319198
user319198

Reputation:

Try with below:

you need to fetch values from database and match them with select box values to show them selected. 

<select name="shade1" id="shade1" >
  <option>Please Select</option>
  <option value="light" <?php if($val=='light') echo 'selected'; ?>>light</option>
  <option value="heavy"<?php if($val=='heavy') echo 'selected'; ?>>heavy</option>
</select>

Here $val is variable having value retrieved from database.

while adding you should have :

<option value="light" >light</option>
<option value="heavy">heavy</option>

Upvotes: 2

Marc
Marc

Reputation: 257

You just have to check if the value in the database has the value of the option field, and if so you echo a "selected='true'" to your option tag. Like

<option <?php if($row["column_name"] == "light") echo "selected=\"true\""; ?>>light</option>

Upvotes: 1

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