-2
所以我創建了一個編輯配置文件頁面，一切都很順利。但後來我想用類似的代碼來更新數據庫中的另一件事。它是一個非常奇怪的bugg，因爲當我添加了許多需要更新的值時，它拒絕工作。無法更新SQL數據庫
<?php
session_start();
if(isset($_POST['submit'])){
$Creator = $_SESSION['username'];
$dbName = $_POST['Name'];
$dbBase_damage = $_POST['Base_damage'];
$dbDamage = $_POST['Damage'];
$dbPellets = $_POST['Pellets'];
$dbAttackspeed = $_POST['Attackspeed'];
$dbRS = $_POST['RS(first)'];
$dbRS2 = $_POST['RS(consecutive)'];
$dbLoaded = $_POST['Loaded'];
$dbReserve = $_POST['Reserve'];
$dbCritical = $_POST['Critical'];
$dbAbility = $_POST['Ability'];
$dbPiece_set = $_POST['Piece_set'];
$dbBonus_set = $_POST['Bonus_set'];
$dbDescription = $_POST['Description'];
if ($dbName && $dbBase_damage && $dbDamage && $dbAttackspeed && $dbLoaded && $dbCritical) {
mysql_connect ("localhost", "root", "") or die ("Could not connect");
mysql_select_db("weapons") or die ("Could not connect to the database");
$wepquery = mysql_query("SELECT * FROM weapon WHERE Creator='$Creator' AND Name='$dbName'") or die ("The query could not be completed, please try again later!");
if (mysql_num_rows($wepquery) != 0) {
mysql_query ("
UPDATE weapon SET Base_Damage='$dbBase_damage', Damage='$dbDamage',
Pellets='$dbPellets', Attackspeed='$dbAttackspeed', RS(first)='$dbRS', RS(consecutive)='$dbRS2',
Loaded='$dbLoaded', Reserve='$dbReserve', Critical='$dbCritical', Ability='$dbAbility', Piece_set='$dbPiece_set',
Bonus_set='$dbBonus_set' Description='$dbDescription' WHERE Creator='$Creator' AND Name='$dbName'
") or die ("Could not update!");
echo "It just werks!";
}else echo ("That username could not be found!");
}else echo ("Something went wrong!");
}
?>
當我使用這段代碼時，出現'Base_Damage'是未定義索引的錯誤。但是當我刪除'Base_Damage'時，它根本無法更新。我已經加倍檢查了數據庫並創建了「武器構想」，沒有任何問題。但由於某種原因，當我減少需要更新的值的數量時，它工作得很好。
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8" />
<meta name="viewport" content="initial-scale=1.0" />
<title>TF2 NUT</title>
<link rel="stylesheet" href="css/style.css" />
</head>
<div id="Container">
<div id="header">
<h1>Welcome to TF2 NUT!</h1>
</div>
<div id="nav">
<?php
include_once "nav.php";
?>
</div>
<div id="Content">
<h2>Update your submissions</h2>
<form action="edit_sub.php" method="POST">
<table>
<tr>
<td>
Choose weapon to edit:
</td>
<td>
<input type="text" name="Name" id="Name" />
</td>
</tr>
<tr>
<td>
Base damage:
</td>
<td>
<input type="text" name="Base_damage" id="Base_damage" />
</td>
</tr>
<tr>
<td>
Damage:
</td>
<td>
<input type="text" name="Damage" id="Damage" />
</td>
</tr>
<tr>
<td>
Pellets:
</td>
<td>
<input type="text" name="Pellets" id="Pellets" />
</td>
</tr>
<tr>
<td>
Attackspeed:
</td>
<td>
<input type="text" name="Attackspeed" id="Attackspeed" />
</td>
</tr>
<tr>
<td>
Reload speed:
</td>
<td>
<input type="text" name="RS(first)" id="RS(first)" />
</td>
</tr>
<tr>
<td>
Reload(consecutive):
</td>
<td>
<input type="text" name="RS(consecutive)" id="RS(consecutive)" />
</td>
</tr>
<tr>
<td>
Loaded:
</td>
<td>
<input type="text" name="Loaded" id="Loaded" />
</td>
</tr>
<tr>
<td>
Reserve:
</td>
<td>
<input type="text" name="Reserve" id="Reserve" />
</td>
</tr>
<tr>
<td>
Critical:
</td>
<td>
<input type="text" name="Critical" id="Critical" />
</td>
</tr>
<tr>
<td>
Ability:
</td>
<td>
<input type="text" name="Ability" id="Ability" />
</td>
</tr>
<tr>
<td>
Piece set:
</td>
<td>
<input type="text" name="Piece_set" id="Piece_set" />
</td>
</tr>
<tr>
<td>
Bonus set:
</td>
<td>
<input type="text" name="Bonus_set" id="Bonus_set" />
</td>
</tr>
</table>
Description:
<br />
<textarea style="resize: none;" rows="8" cols="40" id="Descripton" name="Description"></textarea>
<br />
<input type='submit' name="submit" value='Update' />
</form>
</div>
<div id="aside">
</div>
<div id="footer">
<h6>Responsible for the stuff on this site is Anton Magnusson</h6>
<h6>If I have done something wrong(I hope not ;_;) just contact me: [email protected]</h6>
</div>
</div>
任何人都可以提供幫助嗎？ ：P
首先，你應該停止使用mysql_命令，因爲它們將在不久的將來被取消。而是使用mysqli或PDO。 – Suyash
所以，你從查詢中得到一個錯誤，但是當你編輯它的查詢時，它起作用了？聽起來像查詢要麼有語法錯誤，要麼正在引用一個不存在的數據庫對象。格式化查詢的可讀性並顯示數據庫表的結構。 – David
1）Post表結構2）這可能是一個區分大小寫的問題，將字段名稱括在'-s –