我試圖通過表單更新我的數據庫。SQL更新代碼不改變數據庫數據
部分代碼正在工作,因爲它從表中檢索數據並將其顯示在窗體中,但sql更新代碼未在後端更改值。
的代碼片段如下,任何幫助都將讚賞:
<html>
<head>
<body>
<?php
$con = mysql_connect("localhost","user","pass");
if(!$con){
die("Cannot Connect to database:" . mysql_error());
}
mysql_select_db("intranet",$con);
$sql = "SELECT * FROM progress_sheet";
$myData = mysql_query($sql,$con);
if(isset($_POST['update'])){
$UpdateQuery = "UPDATE progress_sheet SET jobdescription='$_POST[jobdescription]' WHERE id='$_POST[hidden]'";
mysql_query($UpdateQuery, $con);
};
echo "<table border=1>
<tr>
<th>Job Description</th>
</tr>";
while($record = mysql_fetch_array($myData)){
echo "<form action=save.php method=post>";
echo "<tr>";
echo "<td>" . "<input type=text name=jobdescription value=" . $record['jobdescription'] . " </td>";
echo "<td>" . "<input type=hidden name=hidden value=" . $record['hidden'] . " </td>";
echo "<td>" . "<input type=submit name=update value=update" . " </td>";
echo "</form>";
}
echo "</table>";
?>
</body>
</head>
</html>
*
試圖看到有一個錯誤還檢查發佈數據mysql_query($ updatequery,$ con)或死亡(mysql_error()) – Sedz 2015-02-09 09:56:08
請注意,這可能會讓你陷入一個很大的麻煩,由於SQL注入.. 。 – DonCallisto 2015-02-09 09:57:32
您嘗試與非集PARAMS – donald123 2015-02-09 09:59:24