Question

<?php 

    $data = array(

    0 => 'Natural Chlid 1',
    1 => 'Natural Chlid 2',
    2 => 'Natural Chlid 3'
    ); 
    $link = mysqli_connect('localhost', 'root', '', 'mutli_page_form');

    if (mysqli_connect_errno()) {
        printf("Connect failed: %s\n", mysqli_connect_error());
        exit();
    }

    $serialized = mysqli_real_escape_string($link, serialize($data));

    $result = mysqli_query($link, "INSERT INTO wills_children ('will', 'children') VALUES (123, '$serialized')");

    if (!$result) {
        printf("Error message: %s\n", mysqli_error($link));
    }

?>

这似乎没有将数据发送到数据库，我得到的错误是：

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''will', 'children') VALUES (123, 'a:3:{i:0;s:15:\"Natural Chlid 1\";i:1;s:15:\"N' at line 1

Answer 1

使用：

if (!$result) {
    printf("Error message: %s\n", mysqli_error($link));
}

查看错误消息。

编辑：

试试这个：

$result = mysqli_query($link, "INSERT INTO wills_children (will, children) VALUES (123, '$serialized')");

Answer 2

mysqli_error期望至少有一个链接到数据库的参数。

传递该参数并解决问题

mysqli_error($link);

不要在列名称周围加上引号，而是可以使用反引号来避免保留字错误。

(`will`, `children`)

MYSQLI没有将数据放入数据库

2 个答案: