PHP - Having Problems With Selectbox Entry

Alright,  wasn't quite sure how to summarize this in the title,  but I want to:

Check if a user status is "active" or not based on the UserName input.

I have a table witch holds:

Code: [Select]
VarChar Username      Var CharPassWord     int Active
 Ted                        TedsPW                1
something like the above(assuming it formatted correctly.

In my php script I will want to input a variable for Username to check for:

inputUN in this example would be "Ted".

$UserNameToCheck = $_GET['inputUN'];

Then I want to check for that UserName in the database,  if it exists,  I want pull the value for the "Active" field for just that UserName and echo it.

Thanks for any help.
 

I am using php to upload a file to my server, and at the same time inserting the files name and url into my mysql database.

$sql = "UPDATE uploads SET name = '$name', url='$target_path'";
    $statement = $dbh->prepare($sql); 
    $statement->execute();

This is working, however, when I upload a new file, rather than making a new entry in my database, it just overwrites the first one.

I'm quite new at mysql so was wondering how I would make it add new entrys instead of overwriting the current one?

I have a select box on a form that is populated with the following function:

function getStates(){
  // OPEN CONNECTION TO MYSQL ON LOCALHOST
  $dbhost = 'localhost' ;
  $username = 'root' ;
  $password = '' ;
  $conn = mysqli_connect("$dbhost", "$username", "$password");
  if (!$conn) {
    die('Could not connect: ' . mysqli_error()); 
  }
  mysqli_select_db($conn, "mydb"); 
  $retval = mysqli_query($conn, "SELECT * FROM tblstates") or die("Error: " . mysqli_error($conn));
  echo "<option value=''>Select one...</option>";
  while($row = mysqli_fetch_array($retval, MYSQLI_ASSOC)) {
    $rec = $row['name'];
    echo "<option value = " . $row['name']. ">";
    echo $row['name'];
    echo "</option>";
  }
}

On the form I have several text boxes and the following select box:

<div class="formfieldgroup">
  <label class="label">Birth State/Country: <span class="required">&#149;</span></label>
  <select class="fatherbirthstate" name="varfatherbirthstate" tabindex="9"><?=getStates()?></select>
  <span class="form-error2"><?= $varfatherbirthstateErr ?></span>
</div>

TESTING: I intentionally leave a text box blank so I will get an error and I make sure to make a selection from the select box.

PROBLEM: When the form is submitted, the select box doesn't retain the value selected and is reset to "Select one..."

QUESTION: How do  I go about retaining the selected value when the form errors? 

Help needed

I have a form in which i have 2 selctbox(A & B)
In the Select box A i have the star types of hotels
On the onchange of selctbox A,I need to perform 2 things
1)I need to fill the selectbox B with the hotelnames from database
2)I need to display all the hotel details saved in the database

My code performs the second thing ,but the select box is not filled
Code: [Select]

 <tr><script> function doThis(id)
   {
      var selObj = document.getElementById('star');
      if(selObj.selectedIndex>0)
      {
  var aa=selObj.options[selObj.selectedIndex].value;
  location.href="ab_search.php?star=" + aa;
  //location.href="findcity.php?country="+aa;
  //document.forms["a"].submit();

  ///document.a.submit?star=aa;
        // alert("Selected Value = "+selObj.options[selObj.selectedIndex].value);
     }

    }
</script>
                                      <td width="330"><form action="ab_search.php?star=aa" method="get" id='a' name='a'>
                                          <table width="306" class="f_row">
                                            <tr >
                                              <td  valign="middle" colspan="2"><div class="row"><span class="style17">&nbsp;&nbsp;Abu Dhabi Hotels</span></div></td>
                                            </tr>
                                            <tr>
                                              <td colspan="2"> </td>
                                            </tr>
                                           </table>
                                               

                                           
                                           
     <div style="clear:both"></div>                                 
                                            <table height="202" class="s_row">
                                            <tr align="center">
                                              <td width="144" valign="bottom" colspan="2">
      <label><span class="style4">&nbsp;</span></label>
<label class="style4">Star Type</label>
<br/>
                    <div class="select-box">
                      <table border='0' height="30" align="center"  >
                        <tr>
                          <td align="center" valign="middle"><select id="star" name="star" style='border:none' onChange="doThis(this.value);getCity('findcity.php?country='+this.value)"><!--doThis(this.value);--<option value="0">0</option>-->
                              <option value="1">1</option>
                              <option value="2">2</option>
                              <option value="3">3</option><option value="4">4</option><option value="5">5</option>
                          </select></td>
                        </tr>
                         
                      </table>
                    <p class="style4">&nbsp;</p>   </td>
                                            </tr>
                                            <tr>
<?php
$query="select hotel_name from uae_hotels where star_type='$country' and emirates_name='AbuDhabi'";// and star_type=5";// and star_type=$star";
$result=mysql_query($query);

?>
<div id="citydiv">
<select name="city" onchange="getResult(<?php echo $country;?>,this.value)">
<option>Select Hotel Name</option>
<?php  while($row=mysql_fetch_array($result)) {
?><option value="<?php echo $row['hotel_name']; ?>"><?php echo  $row['hotel_name']; ?></option>
 <?php  } ?>
</select>
</div>
?>
                                              <td colspan="2"><div class="row">
                                                  <label for="destOrHotel_hotel" class="style2"><font color='white'>Hotel Name</font></label>
                                                  <div style="clear:both"></div>
                                              </div>
                                             <div id="citydiv"><select name="city">
<option>Select Hotel Name</option>
        </select></div></td>
                                            </tr>
                                            <tr>
                                              <td height="86" colspan="2" valign="top"><div class="row1">
                                               

                                                <input type='submit' name='submitted' id='submitted' value="Submit" />

                                              </div></td>
                                            </tr>
                                          </table>
                                      </form></td>
                                  </tr>
                                   
                                </table></td>
                            </tr>

For some reason this only allows one SQL to be added...

// SQL Connection
$username="monstert_admin"; $password="admin"; $database="monstert_admin";
$connection = mysql_connect("localhost", $username, $password) 
or die("Connection Failure to Database");
// Select Database
mysql_select_db($database, $connection) or die ($database . "No Database" . $username);

//Select everything from the the table
$MyQuery = "SELECT * FROM photos";
$retrieve = mysql_query($MyQuery) or die(mysql_error());
if(mysql_num_rows($retrieve) != 0):
$row = mysql_fetch_assoc($retrieve);
else:
echo '';
endif;

if(isset($_POST['Submit']) && !$errors)
{
$url = $newname;

   include('img.php');
   $image = new SimpleImage();
   $image->load($url);
   $image->resize(500,315);
   $image->save($newname);

mysql_query("INSERT INTO photos (url) VALUES ('$url')");

echo "File Uploaded Successfully as <i> ";
echo $newname;
echo "</i>";
}

What would the issue be? I only have two columns - ID and url

Thanks in advance!