在echo上的php中查找打印变量$email失败

HTML

PHP

它打印：

“更新用户设置密码=”555，其中电子邮件：=：

---

但是，当我填写第二个表单并单击提交时，我回显的上一封$email丢失了。

尝试将该电子邮件添加到更新密码表单中。在PhP中，如果页面刷新，变量将不会被保存，因此您必须保存它，然后再次将其发送到下一个进程。请看下面的更新

另外，在第二个表单中添加了隐藏字段，以保存第一个表单中发布的电子邮件

<?php

$db     = pg_connect("host=localhost port=5432 dbname=CS2102 user=xxx password=xxx");
$email = $_POST['find_email'];
$query = "SELECT * FROM user";
$result = pg_query($db, $query) or die("Cannot execute query: $query\n");       // Query template
$row    = pg_fetch_row($result);        // To store the result row
$count = pg_query($db, "SELECT count(*) FROM user");
if($db) echo "db connected" . "num of data " . pg_num_rows($count) ." . $email;

if (isset($_POST['submit'])) {
            echo "<ul><form name='update' action='update.php' method='POST' >
            <li>New Password:</li> 
            <li><input type='text' name='new_password' value='$row[password]' /></li>
        <li><input type='submit' name='new' /></li>
        </form>
    </ul>";
}

$password = $_POST['new_password'];
$email = $_POST['find_email'];
echo $email;
if (isset($_POST['new'])) { // Submit the update SQL command
    echo $password; 
    $sql = "'UPDATE user SET password = '$password'' WHERE email = '$email'";
    echo $sql . ":)";

}

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请格式化您的帖子。@Script47我可以检查一下我是如何做到的吗？当然：您需要清除代码中所有猖獗的内容。我甚至不能在不知道你的实际代码是什么或不是什么的情况下编辑这些乱七八糟的代码。我给了编辑一次机会，但太多的时间来完全修复。哦，这很有意义。那是个好主意！我不知道每次点击按钮页面都会刷新表单：是的，点击提交按钮总是将页面重定向到操作页面。如果有效，请投票并标记为接受答案

<?php

$db     = pg_connect("host=localhost port=5432 dbname=CS2102 user=xxx password=xxx");
$email = $_POST['find_email'];
$query = "SELECT * FROM user";
$result = pg_query($db, $query) or die("Cannot execute query: $query\n");       // Query template
$row    = pg_fetch_row($result);        // To store the result row
$count = pg_query($db, "SELECT count(*) FROM user");
if($db) echo "db connected" . "num of data " . pg_num_rows($count) ." . $email;

if (isset($_POST['submit'])) {
            echo "<ul><form name='update' action='update.php' method='POST' >
            <li>New Password:</li> 
            <li><input type='text' name='new_password' value='$row[password]' /></li>
        <li><input type='submit' name='new' /></li>
        </form>
    </ul>";
}

$password = $_POST['new_password'];
$email = $_POST['find_email'];
echo $email;
if (isset($_POST['new'])) { // Submit the update SQL command
    echo $password; 
    $sql = "'UPDATE user SET password = '$password'' WHERE email = '$email'";
    echo $sql . ":)";

}

if (isset($_POST['submit'])) {
            echo "<ul><form name='update' action='update.php' method='POST' >
            <li>New Password:</li> 
            <input type='hidden' name='find_email' value='$_POST['find_email']' />
            <li><input type='text' name='new_password' value='$row[password]' /></li>
        <li><input type='submit' name='new' /></li>
        </form>
    </ul>";
}