使用PHP和MYSQL更新表 法里斯网站

尝试将窗体更改为操作

<?php

$connect = mysql_connect("localhost","root","");
$sel_database = mysql_select_db("test");

$name = mysql_real_escape_string( $_POST["name"] );
$id = (int) $_GET['id'];

$query = "UPDATE test SET name='{$name}'";
if($run = mysql_query($query)){
    header("location: index.php");
    exit;
    }else{mysql_error();}


?>

---

首先，请注意SQL注入-您的代码对它非常开放。看

PHP代码
当您要更新从下拉列表中选择的记录时。此外，您已将表单方法设置为POST。因此，您应该尝试以下操作：

<?php

require("inc_connect.php");

?>
<h1 align="center">Farris Website</h1>
<hr width="1000">
<p align="center">
  <table align="center" width="1000" border="3" bordercolor="#0066FF" >
    <tr>
      <td><form name="update" method="post" action="ex_update.php?id=<?php echo urlencode($_GET['id']); ?>">
        <p><strong>Enter Name:</strong>
          <input type="text" name="name"><br />
          <label for="select">ID:</label>
          <select name="id" id="select">
<?php 
  $query = "SELECT * FROM test";
  $run = mysql_query($query);
  while( $r = mysql_fetch_array($run) ){
    # I always use short, single character, variables when in loops.
    # Saves alot of characters and potential confusion.
    echo "          <option value='{$r['id']}'>{$r['id']}</option>\n";
  }
?>
          </select>
        </p>
        <p>
          <input type="submit" name="submit" value="Update!">
        </p>
      </form></td>
      <td><?php include("inc_output.php"); ?></td>
    </tr>
  </table>
</p>

这不是您问题的答案，但您应该继续阅读，特别是如果您打算以root用户身份连接到MySQL（不要这样做）。我刚刚检查了一下，当我在没有WHERE id={$id}的情况下更新时，查询没有问题。但是我想根据id来更新它，好吧，三件事。1） 你能在这里分享完整的SQL语句吗？2） 表中是否存在
id
字段？3） 
mysql\u error（）
显示了什么？实际上，表单中已经有了对“id”的引用，所以将表单操作更新为“ex\u update.php”。然后将php代码更改为$id=（int）$\u POST['id']；
<form name="update" method="post" action="ex_update.php?id=<?php echo urlencode($_GET['id']); ?>">

<?php

$connect = mysql_connect("localhost","root","");
$sel_database = mysql_select_db("test");

$name = mysql_real_escape_string( $_POST["name"] );
$id = (int) $_GET['id'];

$query = "UPDATE test SET name='{$name}' WHERE id = {$id}";
if($run = mysql_query($query)){
  header("location: index.php");
  exit;
}else{
  # In production, don't show raw errors to users - log them to a file and
  # present the user with a generic "There was a problem with the database"
  # error. Or people can start sniffing for vulnerabilities in your site.
  echo mysql_error();
}

?>

<?php

require("inc_connect.php");

?>
<h1 align="center">Farris Website</h1>
<hr width="1000">
<p align="center">
  <table align="center" width="1000" border="3" bordercolor="#0066FF" >
    <tr>
      <td><form name="update" method="post" action="ex_update.php?id=<?php echo urlencode($_GET['id']); ?>">
        <p><strong>Enter Name:</strong>
          <input type="text" name="name"><br />
          <label for="select">ID:</label>
          <select name="id" id="select">
<?php 
  $query = "SELECT * FROM test";
  $run = mysql_query($query);
  while( $r = mysql_fetch_array($run) ){
    # I always use short, single character, variables when in loops.
    # Saves alot of characters and potential confusion.
    echo "          <option value='{$r['id']}'>{$r['id']}</option>\n";
  }
?>
          </select>
        </p>
        <p>
          <input type="submit" name="submit" value="Update!">
        </p>
      </form></td>
      <td><?php include("inc_output.php"); ?></td>
    </tr>
  </table>
</p>

<form name="update" method="post" action="ex_update.php?id=<?php echo urlencode($_POST['id']); ?>">