使用PHP变量填充文本字段

I'm retaining values in form elements after a form submit. I've got it to work fine with a select box using the following:

<select name="BranchManager" class="formfield" id="BranchManager"onchange="document.forms[0].submit();SEinit();"><option value="">-- Select Manager --</option>
         <?php
            $area = $_POST['Area'];
            if ($area); {

               $BMquery = "SELECT DISTINCT Branch_Manager FROM Sales_Execs WHERE AREA = '$area' ".
               "ORDER BY Branch_Manager";

               $BMresult = mysql_query($BMquery);

               while($row = mysql_fetch_array($BMresult))
               {
                  echo "<option value=\"".$row['Branch_Manager']."\">".$row['Branch_Manager']."</option>
  ";
               }
            }

        $branchmanager = $POST['BranchManager'];
        ?>

       <script type="text/javascript">
document.getElementById('BranchManager').value = <?php echo json_encode(trim($_POST['BranchManager']));?>;
</script>

Which works fine (apologies if it isn't the cleanest/most efficient code, I'm doing my best!) The next field is a text field that needs to be populated based off the Branch Managers name above. So I've used :

<input name="BranchNum" type="text" class="formfield" id="BranchNum" size="3" maxlength="3" />
                      <?php

                $bm = $_POST['BranchManager'];

                if ($bm); {

                $BNumquery = "SELECT DISTINCT BRANCH_NUM FROM Sales_Execs WHERE Branch_Manager = '$bm' ";

                $BNumresult = mysql_query($BNumquery);

                }

            $branchnum = $POST['BranchNum'];
            ?>

   <script type="text/javascript">
document.getElementById('BranchNum').value = <?php echo json_encode($BNumresult);?>;
</script>

Which isn't working... where am I going wrong here?

You need to put value="<? echo $variableName; ?>" inside the input field

The reason is because a) you're not echoing, and b) you must echo in a different spot than in a select. You must echo in the value portion of the input.

<?php
    $bm = mysql_real_escape_string($_POST['BranchManager']);
    if ($bm) {
        $BNumquery = "SELECT DISTINCT BRANCH_NUM FROM Sales_Execs WHERE Branch_Manager = '$bm' ";
        $BNumresult = mysql_query($BNumquery);
    }
    $branchnum = $POST['BranchNum'];
?>
<input name="BranchNum" 
    type="text" 
    class="formfield" 
    id="BranchNum" 
    size="3" 
    maxlength="3" 
    value="<?php echo htmlspecialchars($branchnum); ?>" />

As per the comments, they are correct; you should not be using mysql_*. Instead, look at PDO; though this is outside the scope of your question.

why are you having semicolons after if condition checks?

if ($bm);

if ($area);

This will always terminate the statement and whatever is in the curly braces will always get executed irrespective of the value in $bm or $area

You need mydql_fetch functions to retrive data from $result.

if($row = mysql_fetch_array($BNumresult))
    $branchNum = $row[BRANCH_NUM];

why are you using json_encode when your input tag has size = 3?