Question

<?php
include("dbinit.php");
$text="helow";
    $eadd = $_SESSION['account'];
    $result = mysqli_query($link,"SELECT * FROM account where Eadd='". $eadd ."'");
        while($row = mysqli_fetch_array($result))
            {
                $name = $row['Name'];
            }
                $ctext = $_POST['ctext'];
                $sql = "INSERT INTO chat (By, Content, Reported) VALUES ('$name','$ctext','No')";
                mysqli_query($link,$sql);
                mysqli_close($link);
                $text=$name . $ctext;
echo $text;
?>

这是我的代码。在我的另一页中，这可以工作但是......当我更改“插入到”中的值时，为什么我不能将它存储到数据库中？

<?php
session_start();
        $link = mysqli_connect("localhost", "xxx", "xxx", "xxx");

        /* check connection */
        if (mysqli_connect_errno()) {
        printf("Connect failed: %s\n", mysqli_connect_error());
        exit();
        }
        mysqli_select_db($link, "xxx");
        header('Content-type: text/html; charset=utf-8');
?>

这是我的dbinit文件

Answer 1

$sql = "INSERT INTO `chat` (`By`, `Content`, `Reported`) VALUES ('".$name."','".$ctext."','No')";

to protect from mysql injections read this.

另外，你改变了什么样的价值不起作用？

Answer 2

BY是保留关键字，你应该使用反引号

     $sql = "INSERT INTO chat (`By`, Content, Repo.........

Answer 3

用双引号将PHP变量包装在$ sql中，不要忘记点。

 $sql = "INSERT INTO chat (By, Content, Reported) VALUES ('".$name."','".$ctext."','No')";

插入数据库不起作用

3 个答案: