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php - LAST_INSERT_ID() 无法使此代码工作

转载 作者:行者123 更新时间:2023-11-29 06:40:55 27 4
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我看不出我做错了什么,请帮忙。我正在学习 Select Last_Insert_Id,但我已经被困了几天了。

$result2 = $mysqli1->query("SELECT LAST_INSERT_ID()");

while($row = mysqli_fetch_assoc($result2)) {
return $row;
}
$last_id = $row;

$sql = "UPDATE bao_admin_table SET Name='$name' WHERE ID = '$last_id' ";

if ( $mysqli1->query($sql)){

$_SESSION['message'] = 'Registration successful!';
header("location: success.php");

}else{
$_SESSION['message'] = 'Registration failed!';
header("location: error2.php");
}

最佳答案

我假设您可能正在插入录音,那么只有您才能获取最后插入的ID。有两种方法可以实现它,

  1. 插入后获取最后插入的值。这就是你可以做到的

    $sql = "INSERT INTO MyGuests (firstname, lastname, email) VALUES ('John', 'Doe', 'john@example.com')";
    if (mysqli_query($conn, $sql))
    {
    $last_id = mysqli_insert_id($conn);
    echo "New record created successfully. Last inserted ID is: " . $last_id;
    }
    else
    {
    echo "Error: " . $sql . "<br>" . mysqli_error($conn);
    }
  2. 我希望您至少有主键从表中提取唯一记录,因此,这就是您如何通过按降序应用顺序来获取最后一个 id。

`

$sql = "SELECT * FROM MyGuests ORDER BY id DESC";
$result = $conn->query($sql);
if ($result->num_rows > 0)
{// output data of each row
while($row = $result->fetch_assoc())
{
echo "id: " . $row["id"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "<br>";
}
}
else
{
echo "0 results";
}

`

关于php - LAST_INSERT_ID() 无法使此代码工作,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/51647719/

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