[英]how to pass selected value from dropdown to another dropdown in next page
[英]Pass A Value From Last Page And Auto Selected On The Dropdown List PHP
我正在使用PHP进行项目。 从上一页传递两个参数到Update.php
页面:
$ID = $_GET['PosterId']
$catID = $_GET['CategoryID'];
在Update.php
,一个下拉列表列出了从类别表检索到的所有类别。
$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PosterID的categoryID已传递到此页面,因此默认情况下,我要在下拉列表中显示其categoryname。 如何在<option>
上用PHP编写代码?
谢谢。
尝试这种方式:
$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>";
?>
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.