传递从上一页的值，然后在下拉列表php上选择的自动

Question

我正在使用PHP进行项目。从上一个页面传递了两个参数到页面Update.php：

$ID = $_GET['PosterId'] 
$catID = $_GET['CategoryID'];

在Update.php上，下拉列表列出了从类别表检索的所有类别。

$sql = "SELECT PosterID, CategoryID, Category, Title, Price FROM tblposters where PosterID =" . $ID;    
$sqlCategory = "select  CategoryID, CategoryName from tblcategory";
$result = mysqli_query($conn, $sqlCategory);
$resultCategory = mysqli_query($conn, $sqlCategory);
// get category list from table
echo "<select id='Category' name='CategoryList'>";
echo "<option value='-1'>Please Select Posters Group</option>";
while( $row =  $resultCategory -> fetch_assoc( )) {     
    $catRowID = $row['CategoryID'];
    $cat = $row['CategoryName'];
    echo "<option value='$catRowID'> $cat </option>";
}
echo "</select></div>"; 
?>

由于此海报的类别ID已将其传递给此页面，因此我想在下拉列表中默认显示其类别名称。如何在<option>上使用PHP编写代码？

谢谢。

Answer 1

尝试以这种方式：

$sql = "SELECT PosterID, CategoryID, Category, Title, Price FROM tblposters where PosterID =" . $ID;    
$sqlCategory = "select  CategoryID, CategoryName from tblcategory";
$result = mysqli_query($conn, $sqlCategory);
$resultCategory = mysqli_query($conn, $sqlCategory);
//get category list from table
echo "<select id='Category' name='CategoryList'>";
echo "<option value='-1'>Please Select Posters Group</option>";
while( $row =  $resultCategory -> fetch_assoc( )){  
$catRowID = $row['CategoryID'];
$cat = $row['CategoryName'];
if($_GET['CategoryID'] == $row['CategoryID'])
{
echo "<option value='$catRowID' selected> $cat </option>";
}
else
{
echo "<option value='$catRowID'> $cat </option>";
}
}
echo "</select></div>"; 
?>