我想从 SQL 表中创建一个具有两个条件的选择菜单选项 第一个是没有用户 ID 的表 ID 基础,第二个是用户 ID 基础。
include '../conn.php';
$userid=$_SESSION['currentid'];
$sql = "SELECT wname FROM bankw where UserId='$userid' INNER JOIN bankw where id='1' ";
$result = mysqli_query($con,$sql);
echo "<select class='form-control' id='item' name='item'>";
while ($row = mysqli_fetch_array($result)) {
echo "<option value='" . $row['wname'] ."'>" . $row['wname'] ."</option>";
}
echo "</select>";
两者都显示在同一选项菜单中,但我的代码不显示任何选项或仅显示用户 ID 基本选项
最佳答案
只要改变一下,我就得到了我需要的
$userid=$_SESSION['currentid'];
$sql = "SELECT wname FROM bankw where UserId='$userid' or id='1' ";
$result = mysqli_query($con,$sql);
echo "<select class='form-control' id='item' name='item'>";
while ($row = mysqli_fetch_array($result)) {
echo "<option value='" . $row['wname'] ."'>" . $row['wname'] ."</option>";
}
echo "</select>";
谢谢大家
关于php - 如何创建表行作为选择选项,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/57799275/