Question

我在PHP中使用表单来选择我想要更改的值，并选择将成为新值的值。 SQL查询成功运行，但更新不会发生。

index.php 的表单部分：

<form action="filmpontozas.php" method="POST">
    <select name="cim">
    <option disabled selected value> -- Válassz egy filmet! -- </option>

    <?php

    $sql = "SELECT cim FROM film";
    $res = mysqli_query($conn, $sql) or die ('Hibás utasítás: '.mysqli_error($conn));


    while ( ($current_row = mysqli_fetch_assoc($res))!= null) {
        $cim = $current_row["cim"];
        echo '<option value="'.$cim.'">'.$cim.'</option>';
    }
    echo '</select><select name="filmAZ">';
    echo '<option disabled selected value> -- Válassz egy pontszámot! -- </option>';


    $sql = "SELECT filmAZ, pontszam FROM film";
    $res = mysqli_query($conn, $sql) or die ('Hibás utasítás: '.mysqli_error($conn));


    while ( ($current_row = mysqli_fetch_assoc($res))!= null) {


        echo '<option value="'.$current_row["filmAZ"].'">'.$current_row["pontszam"].'</option>';
    }
    ?>
    </select>

    <input type="submit" name="update" value="Értékel!">

</form>

filmpontozas.php

<?php

$cim = $_POST["cim"];
$filmAZ = $_POST["filmAZ"];

$conn = mysqli_connect('localhost', 'root','') or die(mysqli_error($conn));

mysqli_select_db($conn, 'magyarimdb') or die(mysqli_error($conn));

$sql = 'UPDATE film SET filmAZ = '.$filmAZ." WHERE cim = '".$cim."'";
mysqli_query($conn, $sql) or die (mysqli_error($conn));

header("Location: koszonjuk.html");
?>

Answer 1

更新您的MySQLi查询，如下所示：

$query = $this->mysqli->prepare("UPDATE film SET filmAZ=? WHERE cim=?");
// Check whether the prepare() succeeded
if ($query === false) {
    trigger_error($this->mysqli->error, E_USER_ERROR);
    return;
}
$query->bind_param('si', $filmAZ, $cim);
$status = $query->execute();
// Check whether the execute() succeeded
if ($status === false) {
    trigger_error($query->error, E_USER_ERROR);
}

UPDATE在sql中不起作用

1 个答案: