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假设我已将多个表中的值提取到表单，并希望更改一个或多个输入，即。电话号码或地址。

所以这是我的选择查询：

SELECT c.*, u.username
FROM   client c
JOIN   users u ON u.id = c.credid
WHERE  credid = :id

考虑到上面的链接问题（和答案），我如何为已更改的值进行准备更新查询？我的表是InnoDB。

编辑：我需要将用户名表和用户表的所有其他内容放入用户表。（clients表字段credid是用户表主键id的外键）

<form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>" method="post">
<?php
echo 'Username: <input type="text" name="1" value="' . $getuserinfo['username'] . '" /><br>';
echo 'Client: <input type="text" name="2" value="' . $getuserinfo['company'] . '" /><br>';
echo 'Address: <input type="text" name="3" value="' . $getuserinfo['address1'] . '" /><br>';
echo 'Address 2: <input type="text" name="4" value="' . $getuserinfo['address2'] . '" /><br>';
echo 'ZIP: <input type="text" name="5" value="' . $getuserinfo['zip'] . '" /><br>';
echo 'City: <input type="text" name="6" value="' . $getuserinfo['city'] . '" /><br>';
echo 'Country: <input type="text" name="7" value="' . $getuserinfo['country'] . '" /><br>';
echo 'E-mail: <input type="text" name="8" value="' . $getuserinfo['email'] . '" /><br>';
echo 'Phone number: <input type="text" name="9" value="' . $getuserinfo['phone'] . '" /><br>';
?>

    <input type="submit" name="submit" value="Save " /><br>
</form>

编辑：我想构建一个像这样的SQL。

UPDATE c.(name, address, zip, email, phone, etc.),u.username 
VALUES (:1, :2, :3, :etc) 
FROM client c 
JOIN users u ON u.id = c.credid 
WHERE credid = :id

这附近有吗？或者可能是这样的：

UPDATE users,client
SET users.username = :username,
    client.value1 = :value1,
    client.value2 = :value2,
    etc...

 WHERE client.credid=users.id

Answer 1

您需要做的就是将输入重命名为表格列的名称，例如。

foreach ($getuserinfo as $key => $val) {

    echo ucfist($key).': <input type="text" name="'.$key.'" value="' . $val . '" /><br>';
}

上面的内容不会为您提供准确的标签，但很容易添加if语句来改变它。

在您最初获取信息后，将其添加到会话数组以检查是否有任何更改

$_SESSION = $getuserinfo;

然后在回发表单后（请记住，除了检查$_SESSION中是否已存在索引之外，这不包含任何验证）

if ($_POST['username'] != $_SESSION['user_edit_info']['username']) {

    //run validation and query to update username
}

$posted = $_POST;
unset($posted['username']);

$sql_array = array();

$params = array();

foreach ($posted as $key => $val) {

    //This should prevent extra fields being posted
    if (isset($_SESSION['user_edit_info'][$key]) && $_SESSION['user_edit_info'][$key] != $val) {

        $sql_array[] = "$key=:$key";
        $params[$key] = $val;
    }
}

if (!empty($sql_array)) {

    $params['id'] = $_SESSION['user_id']; //or whatever you set it to in your session

     //I don't know exactly how you're running your PDOs but below should at least show
     //what you need to do
    ('UPDATE client SET '.implode(',', $sql_array).' WHERE credid=:id', $params);
}

//finally you don't need this anymore so just unset it
unset($_SESSION['user_edit_info']);

希望这有帮助！

sql - 仅将表单中更改的值更新为多个表

1 个答案: