php - 检索一个值并将其显示在 php 的选择选项框中

Question

我正在尝试从数据库检索存储的值并将其显示在选择选项框中。我的问题是它显示了所选的第一个值。我的代码是

$query = "SELECT * FROM bookwizard.book_inventory where bi_SKU='$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>

感谢您的帮助。

Answer 1

您错过了关闭选项。 if 条件也有错误。

替换您的选择框代码如下:

<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>