Question

问候'im试图从$_POST方法形式获取数据。不幸的是，我从服务器（localhost）收到此警告

警告：mysqli_error（）恰好期望1个参数，第9行的C：\ xampp \ htdocs \ justificativo \ listadoasignado.php中给出的参数为0

代码是下一个

<?php
include("conecta.php");
$c1 = $_POST['rut'];
$c2 = $_POST['nivel'];
$c3 = $_POST['unidad'];
$c4 = $_POST['entidad'];
$c5 = $_POST['reposo'];
$query = mysqli_query($con, "INSERT INTO certificado (id_detalle, n_alumno, n_nivel, n_unidad, dia_reposo) VALUES ('$c1', '$c2', '$c3', '$c4', '$c5')")
or die(mysqli_error());
?>

Answer 1

您需要通过$con，因为$con拥有数据库连接

die(mysqli_error($con));

Answer 2

您使用mysqli_error函数错误。它需要一个参数（连接）。 https://www.w3resource.com/php/function-reference/mysqli_error.php

正确使用：

<?php
include("conecta.php");
$c1 = $_POST['rut'];
$c2 = $_POST['nivel'];
$c3 = $_POST['unidad'];
$c4 = $_POST['entidad'];
$c5 = $_POST['reposo'];
$query = mysqli_query($con, "INSERT INTO certificado (id_detalle, n_alumno, n_nivel, n_unidad, dia_reposo) VALUES ('$c1', '$c2', '$c3', '$c4', '$c5')")
or die(mysqli_error($con));
?>

无法找到我的错误。从$ _POST获取数据时

2 个答案: