php - 如何从数据库中读取数据更改它们并更新数据库

Question

我需要从数据库中读取数据然后用 php 编辑这些数据并用这些数据更新数据库。

我可以从数据库中获取数据但无法将它们写回(我添加了另一个查询以查看值是否已更改)

我已经尝试使用 phpMyAdmin 来查看值是否发生了变化，但它们仍然存在一样

$POWER = "";
if ($pressed == "1") {
    echo "pressed\n";
    $sql = "SELECT `POWER` FROM `VolleyTest` WHERE ID = 1";
    $result = mysqli_query($conn, $sql);
    while ($row = mysqli_fetch_row($result)) {
        $POWER = $row[0];
        echo "pressed power request $row[0]\n";
    }
    if ($POWER == "ON") {
        $sql = 'UPDATE `VolleyTest` SET `POWER`=\"OFF\" WHERE ID = 1';
        $conn->query($sql);
        echo " POWER ON to OFF\n";
    } else if ($POWER == "OFF") {
        $sql = 'UPDATE `VolleyTest` SET `POWER`=\"ON\" WHERE ID = 1';
        $conn->query($sql);
        echo " POWER OFF to ON\n";
    }
}
$sql = "SELECT `POWER` FROM `VolleyTest` WHERE ID = 1";
$result = mysqli_query($conn, $sql);
while ($row = mysqli_fetch_row($result)) {
    $POWER = $row[0];
    echo "$row[0]\n";
}
echo $POWER; //. "  ". $pressed . "  " . $_POST['key'];

Answer 1

您不需要在单引号字符串中转义双引号。

这样做会生成这样的字符串

UPDATE `VolleyTest` SET `POWER`=\"ON\" WHERE ID = 1

所以这可能行得通

$POWER = "";
if ($pressed == "1") {
    echo "pressed\n";
    $sql = "SELECT `POWER` FROM `VolleyTest` WHERE ID = 1";
    $result = mysqli_query($conn, $sql);
    while ($row = mysqli_fetch_row($result)) {
        $POWER = $row[0];
        echo "pressed power request $row[0]\n";
    }
    if ($POWER == "ON") {
        $sql = 'UPDATE `VolleyTest` SET `POWER`="OFF" WHERE ID = 1';
        $conn->query($sql);
        echo " POWER ON to OFF\n";
    } else if ($POWER == "OFF") {
        $sql = 'UPDATE `VolleyTest` SET `POWER`="ON" WHERE ID = 1';
        $conn->query($sql);
        echo " POWER OFF to ON\n";
    }
}
$sql = "SELECT `POWER` FROM `VolleyTest` WHERE ID = 1";
$result = mysqli_query($conn, $sql);
while ($row = mysqli_fetch_row($result)) {
    $POWER = $row[0];
    echo "$row[0]\n";
}
echo $POWER; //. "  ". $pressed . "  " . $_POST['key'];