简单下拉列表来自MYSQL的PHP​​代码

Question

我正在尝试在桌面上做一个简单的下拉列表，但是，我的代码似乎没有工作，我想知道我连接和检索的方式是否有任何问题？或者它只是我的下拉列表代码是错误的。以下是它的代码，下面的屏幕截图包含我的数据库以及我想放下我的下拉列表的地方。谢谢您的时间。

<?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

试试这段代码

使用mysqli->query

<?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();
?>

Answer 2

我认为绑定查询时的问题，为什么不使用下面的代码。

$result = $mysqli->query("Select sbranch_name from branches");

if ($result->num_rows > 0) {

    echo "<select name='images'>";
    while($row = $result->fetch_assoc()) {
      echo "<option value='" . $row['sbranch_name'] . "'>" . $row['sbranch_name'] . "</option>";
    }
    echo "</select>";
}