检索值并在php中的选择选项框中显示它

Question

I am trying to retrieve a stored value from database and show it in a select option box. My problem is it's showing the first value as selected. my code is

$query = "SELECT * FROM bookwizard.book_inventory where bi_SKU='$SKU'";
$result = mysql_query($query) or die(mysql_error());

while($row = mysql_fetch_array($result))

{
    $bid=($row{'bi_bookcondition_id'});
}

<select class='form-control' id='condition' name='condition'>
    <option value='' <?php if (!empty($bid) && $bid == '' ) echo 'selected = 'selected'; ?> </option>      
    <option value='1' <?php if (!empty($bid) && $bid == '1')  echo 'selected = 'selected'; ?>Good</option>
    <option value='2' <?php if (!empty($bid) && $bid == '2')  echo 'selected = 'selected'; ?>Old</option>
    <option value='3' <?php if (!empty($bid) && $bid == '3')  echo 'selected = 'selected'; ?>Bad</option>
</select>

Thanks for helping.

Answer 1

You have missed to close the options. As well as if conditions are having error.

Replace your selectbox code as below :

<select class='form-control' id='condition' name='condition'>
    <option value='' <?php if (!empty($bid) && $bid == '' ) echo "selected = 'selected'"; ?>></option>
    <option value='1' <?php if (!empty($bid) && $bid == '1')  echo "selected = 'selected'"; ?>>Good</option>
    <option value='2' <?php if (!empty($bid) && $bid == '2')  echo "selected = 'selected'"; ?>>Old</option>
    <option value='3' <?php if (!empty($bid) && $bid == '3')  echo "selected = 'selected'"; ?>>Bad</option>
</select>