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验证来自 POST 值的输入?

验证来自 POST 值的输入?

PHP
慕码人2483693 2023-04-21 10:13:28
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!
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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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反对 回复 2023-04-21
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