php - 来自 MYSQL 的简单下拉列表 PHP 代码

Question

我正在尝试在表格上做一个简单的下拉列表，但是，我的代码似乎不起作用，我想知道我的连接和检索方式是否存在任何问题？或者只是我的下拉列表代码是错误的。这是它的代码，下面的屏幕截图包含我的数据库以及我想要放置下拉列表的位置。感谢您抽出时间。

<?php
$mysqli = new mysqli(spf, dbuser, dbpw, db);
$stmt = $mysqli->prepare("Select sbranch_name from branches");
$result = $stmt->execute();
$stmt->bind_result($sbranch_name);

//while ($stmt->fetch())
//{
//    $stmt .="<option>". $row['sbranch_name']. "</option>";
    //echo '<input type="checkbox" name="sbranch_name[]" value="'.$sbranch_name.'". <br>';
//  echo $stmt;
//}
if ($result->num_rows > 0) {
    echo "<select name='sbranch_name'>";
    while($row = $result->fetch_assoc()) {
        echo "<option value='" . $row['sbranch_name'] . "'>" . $row['sbranch_name'] . "</option>";
    }
    echo "</select>";
}   
$stmt->close();
$mysqli->close();
?>

Answer 1

试试这个代码

使用mysqli->查询

<?php
$mysqli = new mysqli(spf, dbuser, dbpw, db);
$sql="Select sbranch_name from branches";

$result = $mysqli->query($sql);
//$stmt->bind_result($sbranch_name);

//while ($stmt->fetch())
//{
//    $stmt .="<option>". $row['sbranch_name']. "</option>";
    //echo '<input type="checkbox" name="sbranch_name[]" value="'.$sbranch_name.'". <br>';
//  echo $stmt;
//}
if ($result->num_rows > 0) {
    echo "<select name='sbranch_name'>";
    while($row = $result->fetch_assoc()) {
        echo "<option value='" . $row['sbranch_name'] . "'>" . $row['sbranch_name'] . "</option>";
    }
    echo "</select>";
}   
//$stmt->close();
$mysqli->close();
?>