你好亲爱的网络朋友,今天我有一个问题,我不知道如何验证 POST 中的 INPUT 数据,如果有人可以帮助我验证这些数据,我将不胜感激,这就是代码:include("GameEngine/Village.php");$unarray = array(1 => U1, U2, U3, U4, U5, U6, U11, U12, U13, U14, U15, U16, U21, U22, U23, U24, U25, U26, U31, U32, U33, U34, U35, U36, U41, U42, U43, U44, U45, U46, U0);if (isset($_GET['id']) && $_GET['id'] == 1) { $res_qty = $_POST['resqty']; $cost = $res_qty * 30; $allres_qty = $res_qty * 100000; $database->query("UPDATE " . TB_PREFIX . "users SET `gold` = `gold`- " . $cost . " WHERE id =" . $session->uid); $database->query("UPDATE " . TB_PREFIX . "vdata SET `wood` = `wood` + " . $allres_qty . ", `clay` = `clay` + " . $allres_qty . ", `iron` = `iron` + " . $allres_qty . ", `crop` = `crop` + " . $allres_qty . " WHERE wref =" . $village->wid); $_SESSION['info'] = $allres_qty . " Wood, Clay, Iron, Crop, added at the cost of " . $cost . " gold."; header("Location:dorf1.php");}if (isset($_GET['id']) && $_GET['id'] == 2) { $troop_type = $_POST['trooptype']; $troop_qty = $_POST['troopqty']; $cost = $troop_qty * 30; $dt = array(); header("Location:dorf1.php"); for ($x = 1; $x < 11; $x++) { $dt[$x] = 3333; if ($x == $troop_type) $dt[$x] = floor($troop_qty * 3333); } $tribe = $database->getUserField($database->getVillageField($village->wid, "owner"), "tribe", 0); if ($tribe == 1) { $u = ""; } elseif ($tribe == 2) { $u = "1"; } elseif ($tribe == 3) { $u = "2"; } elseif ($tribe == 4) { $u = "3"; } else { $u = "4"; } $database->modifyUnit( $village->wid, array($u . "1", $u . "2", $u . "3", $u . "4", $u . "5", $u . "6", $tribe . "0", "hero"), array($dt[1], $dt[2], $dt[3], $dt[4], $dt[5], $dt[6], 0), array(1, 1, 1, 1, 1, 1) );}您好,我需要验证的是我给我的用户提供的部队和资源的数量,这是在我的游戏中提供部队的脚本,但我需要验证部队和资源,非常感谢!如果有人能帮助我,我将不胜感激!
1 回答
慕容3067478
TA贡献1773条经验 获得超3个赞
这是一般数据验证的示例:
<?php
// define variables and set to empty values
$name = $email = $gender = $comment = $website = "";
if ($_SERVER["REQUEST_METHOD"] == "POST") {
$name = test_input($_POST["name"]);
$email = test_input($_POST["email"]);
$website = test_input($_POST["website"]);
$comment = test_input($_POST["comment"]);
$gender = test_input($_POST["gender"]);
}
function test_input($data) {
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
}
?>
资料来源: https: //www.w3schools.com/php/php_form_validation.asp
主要思想是只接受与您想要作为有效数据接收的内容相匹配的模式。
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