Question

我是一个新手，试图为我自愿提供的非营利性做网站。大部分代码都是从Youtube看到的。下面是代码。数据库名称是表所联系的cani。每当我提交的消息都不会给我成功消息时，数据库中不会显示任何数据，并且不会返回到newentry.php。我的脑痛！

<?php

if (isset($_POST['submit'])){

include_once('dbh.inc.php');

$type = mysqli_real_escape_string($conn, $_POST['type']);
$first = mysqli_real_escape_string($conn, $_POST['first']);
$last = mysqli_real_escape_string($conn, $_POST['last']);
$company = mysqli_real_escape_string($conn, $_POST['company']);
$email = mysqli_real_escape_string($conn, $_POST['email']);
$phone = mysqli_real_escape_string($conn, $_POST['phone']);
$add1 = mysqli_real_escape_string($conn, $_POST['add1']);
$add2 = mysqli_real_escape_string($conn, $_POST['add2']);
$city = mysqli_real_escape_string($conn, $_POST['city']);
$state = mysqli_real_escape_string($conn, $_POST['state']);
$zip = mysqli_real_escape_string($conn, $_POST['zip']);

$sql = "INSERT INTO contact (contact_type, contact_first, contact_last,       contact_company, contact_email, contact_phone, contact_add1, contact_add2, contact_city, contact_state, contact_zip) VALUES ('$type', '$first', '$last', '$company', '$email', '$phone', '$add1', '$add2', '$city', '$state', '$zip')";

mysqli_query($conn, $sql);
header("Location: ../newentry.php?Success!");
}else{

header("Location: ../index.html");
exit();
} 
?>

Grr我已部分修复它...现在一切正常，除了数据库中什么都没有显示...正在连接，我确保表名正确。它说成功。但是当我打开phpAdmin并打开表时，它为空。

Answer 1

检查变量$_POST['submit']是否存在

只是找不到我做错的php代码

1 个答案: