Question

我正在尝试创建一个函数，使用mysqli（对我来说是全新的）来检查mysql中的重复条目，但我无法理解这个错误。我有3个php文件：db.php - db连接functions.php和我的函数（需要db.php）和submit.php - 接受输入的文件。

db.php中：

<?php
define("HOST", "localhost"); // The host you want to connect to.
define("USER", "user"); // The database username.
define("PASSWORD", "pass"); // The database password. 
define("DATABASE", "db_list"); // The database name.

$mysqli = new mysqli(HOST, USER, PASSWORD, DATABASE);
?>

的functions.php：

require 'db.php';
function check($str, $mysqli) {

$checkdupe = $mysqli->prepare("SELECT name FROM list WHERE str = ?"); //line giving me error
$checkdupe->bind_param("s", $str);
$checkdupe->execute();
$checkdupe->store_result();

    if ($checkdupe->num_rows > 0) {
        //dupe found
        return true;
    } else {
            //no dupe
            return false;
    }
}

submit.php

require 'functions.php';

if (isset($_POST['name'])) {
    if (check($_POST['name'], $mysqli) == true) { //added $mysqli parameter
        echo "success!";
    } else {
        echo "fail;
    }
} else {
  echo 'invalid post';
}
?>

Answer 1

你的功能定义是

function check($str, $mysqli) {

但你只用第一个参数调用它。因此，第二个参数为null。

check($_POST['name'])

所以使用，

check($_POST['name'], $mysqli)

此外，您不会检查db.php中的连接确实成功。在文件末尾添加：

if ($mysqli->connect_error) {
    die('Connect Error (' . $mysqli->connect_errno . ') '
            . $mysqli->connect_error);
}

Answer 2

您的函数check需要2个参数。您只为其分配1个属性：

if (check($_POST['name']) == true)

应该是

if (check($_POST['name'], $mysqli) == true)

Answer 3

db.php中：

<?php
error_reporting(E_ALL);
ini_set('display_errors',1);

define("HOST", "localhost"); // The host you want to connect to.
define("USER", "user"); // The database username.
define("PASSWORD", "pass"); // The database password. 
define("DATABASE", "db_list"); // The database name.

$mysqli = new mysqli(HOST, USER, PASSWORD, DATABASE);
var_dump('db:',is_object($mysqli));

的functions.php：

<?php
require 'db.php'
var_dump('functions:',is_object($mysqli));
function check($str, $mysqli) {
    var_dump('check:',is_object($mysqli));

    $checkdupe = $mysqli->prepare("SELECT name FROM list WHERE str = ?");
    $checkdupe->bind_param("s", $str);
    $checkdupe->execute();
    $checkdupe->store_result();
    return $checkdupe->num_rows;
}

submit.php

<?php
require 'functions.php';
var_dump('submit:',is_object($mysqli));


if (isset($_POST['name'])) {
    if (check($_POST['name'], $mysqli)) {
        echo "success!";
    } else {
        echo "fail";
    }
} else {
  echo 'invalid post';
}

Answer 4

尝试：

global $mysqli;
check($mysqli, ...);

在非对象上调用成员函数prepare（）的致命错误

4 个答案: