PHP 中的简单添加

Question

好的，所以我想知道是否有一种简单的方法可以在我的代码中制作Addonetothis，以便在运行此代码时添加当前的INT + 1？

<?php
$server = "localhost";
$user = "**********";
$pass = "**********";
$dbname = "**********";
//Creating connection for mysqli
$conn = new mysqli($server, $user, $pass, $dbname);
//Checking connection
if ($conn->connect_error) {
    die("Connection failed:" . $conn->connect_error);
}
$article_id = $_GET['id'];
if ( ! is_numeric($article_id))
    die("Looks like you are lost!  <a href='#'>Back to Home</a> ");
$sql = "UPDATE Example SET addonetothis='+ 1' WHERE `ID` =$article_id";
if ($conn->query($sql) === TRUE) {
    header("Refresh:1; url=example.php?id=$article_id");
    echo "Thank you!";
} else {
    echo "Error" . $sql . "<br/>" . $conn->error;
}
$conn->close();

Answer 1

所以你会看到...

$sql = "UPDATE Example SET Addonetothis='+ 1' WHERE `ID` =$article_id";

如果 Addonetothis 在数据库表中定义为整数，因为它是字符串，则会爆炸。

您正在寻找的是类似...

$sql = "UPDATE Example SET AddOneToThis = AddOneToThis + 1 WHERE `ID` =$article_id";