Question

我正在使用变量记录有关网站用户的信息，以观察他们在我网站上的行为。我的代码如下所示：

<?php
$pages_visited = "www.example.com, www.example.com/example.php, www.example.com/demo.php";
$con = mysqli_connect("localhost", "user", "password", "pages_visited");
    if (mysqli_connect_errno()) {
        echo "An error occurred when connecting to the database. Sorry.";
    }
mysqli_query($con, "INSERT INTO visit_data ('Pages')
VALUES ('$pages_visited')");
mysqli_close($con);

但是，当我访问我的网站时，没有任何反应，并且表visit_data中没有添加新行。为什么会这样？

Answer 1

尝试mysqli预处理语句（错误与不使用预处理语句无关，但我认为最好使用，因为不推荐使用mysql函数）

$db = new mysqli('localhost', 'root', '', 'database');
if ($db->connect_errno) {
echo "Failed to connect to MySQL: (" . $db->connect_errno . ") " . $db->connect_error;
}

$pages_visited = "www.example.com, www.example.com/example.php, www.example.com/demo.php";

$stmt = $db->prepare("insert into `visit_data` (Pages) values(?)");
$stmt->bind_param('s', $pages_visited);
$stmt->execute();

Answer 2

您需要更新您的查询：

INSERT INTO visit_data (`Pages`) VALUES('".$pages_visited."')

实际上你正在使用单引号（＆＃39; Pages＆＃39;），检查波纹管代码。我已在本地验证此代码正在运行

    <?php
$pages_visited = "www.example.com, www.example.com/example.php, www.example.com/demo.php";
$con = mysqli_connect("localhost", "user", "password", "pages_visited");
    if (mysqli_connect_errno()) {
        echo "An error occurred when connecting to the database. Sorry.";
    }

$try = mysqli_query($con, "INSERT INTO visit_data (`Pages`) VALUES('".$pages_visited."')");

if($try === false){
    echo 'error - ';
    echo mysqli_error($con);
}   else{
        echo 'all good';
    }

mysqli_close($con);

Answer 3

使用字符串时，你必须使用'。$。text_var。“'。确保您使用的表列是正确的，并且除非您在创建表时使用默认值，否则不要遗漏任何不接受空值的列。

这应该有效

<?php

$pages_visited = "www.example.com, www.example.com/example.php, www.example.com/demo.php";
$con = mysqli_connect("localhost", "user", "password", "pages_visited");

if ( mysqli_connect_errno() )
{
    echo "An error occurred when connecting to the database. Sorry.";
}

mysqli_query( $con, "INSERT INTO visit_data ( Pages ) VALUES ( '" . $pages_visited . "' )" );

mysqli_close($con);

尝试在PHP中插入MySQL时出现问题

3 个答案: