Question

我想使用PHP将数据从表单更新到数据库。但是我的代码出现了太多问题，例如＆＃34;无法更新数据：未知列＆＃39; Nabeel＆＃39;在＆＃39;字段列表＆＃39;＆＃34;。

请检查代码并给我解决方案。

谢谢，纳比尔。

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
<script type="text/javascript" src="http://code.jquery.com/jquery-1.11.1.min.js"></script>
<script type="text/javascript">
function doSomething() {
    alert('Are you sure you want to delete this file ??');
}
</script>
</head>

<body>



<h2>Update Data Using PHP</h2>
</div>

<?php
$con = mysql_connect("localhost", "root", "");
$db = mysql_select_db("firstphp", $con);
if (isset($_GET['submit'])) {
$id = $_GET['id'];
$username = $_GET['username'];
$passward = $_GET['passward'];
$name = $_GET['name'];

$query = mysql_query("update users set
username='$username', passward='$passward', name='$name' where id='$id'", $con);
}
$query = mysql_query("select * from users", $con);
while ($row = mysql_fetch_array($query)) 
{
    echo "<b><a href='update.php?update={$row['id']}'>{$row['id']}</a></b>";
    echo "<br />";
}
?>
</div><?php
if (isset($_GET['submit'])) {
$submit = $_GET['submit'];
$query1 = mysql_query("select * from users where id=$submit", $con);
while ($row1 = mysql_fetch_array($query1)) {
echo "<form method='get'>";
echo "<hr/>";
echo"<input class='input' type='hidden' name='id' value='{$row1["id"]}' />";
echo "<br />";
echo "<label>" . "Name:" . "</label>" . "<br />";
echo"<input class='input' type='text' name='username' value='{$row1["username"]}' />";
echo "<br />";
echo "<label>" . "Email:" . "</label>" . "<br />";
echo"<input class='input' type='text' name='passward' value='{$row1["passward"]}' />";
echo "<br />";
echo "<label>" . "Name:" . "</label>" . "<br />";
echo"<input class='input' type='text' name='name' value='{$row1["name"]}' />";
echo "<br />";


echo "<input class='submit' type='submit' name='submit' value='submit' />";
echo "</form>";
}
}
if (isset($_GET['submit'])) {
echo '<div class="form" id="form3"><br><br><br><br><br><br>
<Span>Data Updated Successfuly......!!</span></div>';
}

mysql_close($con);
?>
</bod

></body>
</html>

Answer 1

$query1 = mysql_query("select * from users where id=$submit", $con);

更改为

$query1 = mysql_query("select * from users where id='".$submit."'", $con);

Answer 2

试试这个：

$query1 = mysql_query("select * from users where id={$submit}", $con);

大括号将确保您的PHP变量被解释，而不是简单的字符串。

Answer 3

1）您不能使用where id=$submit，因为Id需要是一个整数。 $submit始终具有值submit（因为您将其设置为：

echo "<input class='submit' type='submit' name='submit' value='submit' />";

您想要使用此行中的ID值：

echo"<input class='input' type='hidden' name='id' value='{$row1["id"]}' />";

2）替换

$query1 = mysql_query("select * from users where id=$submit", $con);

带

$query1 = mysql_query("select * from users where id=$id", $con);

并确保id是一个整数。

3）不要使用mysql，而是使用mysqli：The mysql extension is deprecated and will be removed in the future: use mysqli or PDO instead

4）你拼错了password这个词。从＆＃39; passward＆＃39;变更到password

5）你的HTML在最后被破坏了：

</bod

></body>
</html>

删除</bod >

6）无需使用HTML4 doctype声明。取代

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">

与

<!DOCTYPE html>

使用php将数据从表单更新到数据库

3 个答案: