Php 准备功能使用错误?
任何人都知道我为什么会出现这个错误:Php 准备功能使用错误?,php,function,Php,Function,任何人都知道我为什么会出现这个错误: Call to a member function prepare() on a non-object 对于此代码段: public function check_availability($dataID) { $query = "SELECT dump FROM dumps WHERE dump = '".$dataID."'"; if ($stmt = $mysqli->prepare($query)) { $
Call to a member function prepare() on a non-object
对于此代码段:
public function check_availability($dataID) {
$query = "SELECT dump FROM dumps WHERE dump = '".$dataID."'";
if ($stmt = $mysqli->prepare($query)) {
$stmt->execute();
if($stmt->rowCount() > 0) {
return FALSE;
} else {
return TRUE;
}
$stmt->close();
}
}
特别是这一行:
if ($stmt = $mysqli->prepare($query)) {
变量$mysqli不会出错,因为:
$mysqli = new mysqli($config['sql_hostname'], $config['sql_username'], $config['sql_password'], $config['sql_database']);
更新:
我在浏览器中调用此脚本:
define('SECURE', true);
include "storescripts/connect_to_mysql.php";
require 'AvailabilityChecker.php';
$config = array('DB' => $mysqli,
'Table' => 'dumps',
'Row' => 'dump',
'Output' => true,
'Format' => 'JSON');
$availabilityChecker = new AvailabilityChecker($config);
if($availabilityChecker->check_availability($_POST['dataID'])) {
echo "ok";
} else {
echo "not ok";
}
$mysqli是在“connect_to_mysql.php”中定义的,因此?您应该创建
mysqli
的对象,并检查结果
$mysqli = new mysqli("localhost", "username", "password", "dbname");
if(!$mysqli)
{
die("can not access the db");
}
然后你可以使用这个函数
$mysqli->prepare($query)
显然,$mysqli
是错误的。您在哪里定义了$mysqli
?请检查我更新的这个。你能看一看吗?不,你看了很多重复的@user2999787ehm中的一个,我已经创建了一个对象,你没有看到吗?在使用$mysqli之前,你应该使用if($mysqli){}检查它