作为学习php和sql注入,我想参数化我的安全和安全的网站应用程序的查询。然而,我的不能工作,我试图参数化我的更新并选择我的查询,但我没有达到让程序工作的目标。
当前输出抛出错误:找不到?
到目前为止,这是我的代码,我是否遗漏了一些不起作用的东西?
<?php
//connection
$connection = mysqli_connect("hostserver","username","");
$db = mysqli_select_db($connection, 'dbname');
if (isset($_POST['qrname'])) {
$qrid = $_POST['qrid'];
//Query No. 1
$qrQuery = "SELECT * FROM scratch_cards WHERE code='$qrid' ";
$qrQuery_run = mysqli_query($connection,$qrQuery);
//Query No. 2
$qrQuery2 = "UPDATE scratch_cards SET status = 'U' WHERE code='$qrid' ";
$qrQuery_run2 = mysqli_query($connection,$qrQuery2);
$qrQuery2->bind_param("s", $qrid);
$qrQuery2->execute();
while ($qrRow = mysqli_fetch_array($qrQuery_run)) {
$txtQrvalue = $qrRow['amount'];
$txtQrstatus = $qrRow['status'];
// QUERY TO UPDATE THE VALUE
// BIND AND PARAMETIZE MY QUERY
$qrQuery3 = $db->parepare("UPDATE shopusers SET ewallet = ewallet + " . (0+?) . " WHERE id = '?' ");
$qrQuery3->bind_param("ii", $txtQrvalue, $id);
$qrQuery3->execute();
//END
}发布于 2019-05-10 18:25:06
如果我没看错您的问题和代码,那么您可以使用JOIN将其简化为两个查询,这样就可以去掉SELECT语句。对这两种情况都使用预准备语句。
我还将连接的字符集指定为UTF-8 (应该为PHP和HTML头文件以及数据库表设置该字符集)。
<?php
$connection = mysqli_connect("hostserver","username","");
$db = mysqli_select_db($connection, 'dbname');
$connection->set_charset("utf8");
if (isset($_POST['qrname'])) {
$qrid = $_POST['qrid'];
$sql = "UPDATE scratch_cards SET status = 'U' WHERE code=?";
$stmt = $connection->prepare($sql);
$stmt->bind_param("s", $qrid);
$stmt->execute();
$stmt->close();
$sql = "UPDATE shopusers su
INNER JOIN scratch_cards sc
ON sc.qrid = su.code
SET su.ewallet = su.ewallet + sc.amount,
sc.status = 'U'
WHERE sc.code = ?";
$stmt = $connection->prepare($sql);
$stmt->bind_param("s", $qrid);
$stmt->execute();
$stmt->close();
}发布于 2019-05-10 18:49:41
我们在PDO bind param中有foll语法,我把你的update查询作为一个例子,它工作得很好。尝试搜索命名参数绑定
<?php
$user = 'root';
$pass = 'xxxx';
$DB = 'test';
$host = 'localhost';
$mysqlConnection = new \PDO('mysql:host='.$host.';dbname='.$DB, $user, $pass);
$mysqlConnection->setAttribute(\PDO::ATTR_ERRMODE, \PDO::ERRMODE_EXCEPTION);
$sql = 'update info set fname = fname + :fn where id = 1';
$stmt = $mysqlConnection->prepare($sql);
$stmt->bindValue(':fn', '100');
$stmt->execute();
echo $stmt->rowCount();
?>发布于 2019-05-10 19:49:04
这是您想要使用mysqli绑定参数运行的查询吗?
<?php
ini_set('display_errors', 1);
$user = 'root';
$pass = 'xxxx';
$DB = 'test';
$host = 'localhost';
$sql = 'update info set fname = fname + ? where id = 1';
$conn = new mysqli($host, $user, $pass, $DB);
$stmt = $conn->prepare($sql);
$stmt->bind_param("i", $val);
$val = 100;
$stmt->execute();
printf("%d Row inserted.\n", $stmt->affected_rows);
exit;https://stackoverflow.com/questions/56075383
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