Question

我有这段代码，允许用户从网址链接重置帐户

<?php
$servername = "localhost";
$username = "  ";
$password = " ";
$dbname = "   ";

$code = $_GET['code'];

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT com_code FROM user WHERE com_code = ".$_GET['code'];
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        echo "<form action='reset.php?code=" . $row["com_code"]. "' method='post'>Enter New Password: <input type='text' name='new_password' placeholder='New Password'><br><input type='submit' value='Submit'></form>";
    }
} else {
    echo "0 results";
}
$conn->close();
?>

<?php
$servername = "localhost";
$username = "  ";
$password = "  ";
$dbname = "    ";

  $pword = $_POST['new_password'];

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 
if (isset($_POST['Submit'])) {
$sql1 = "UPDATE user SET password='$pword', com_code=NULL WHERE com_code = '$code'";
}
if ($conn->query($sql1) === TRUE) {
    echo "Password has been change successfully!";
} else {
    echo "Error updating record: " . $conn->error;
}
?>

我一直收到错误：

警告：mysqli :: query（）：空查询第47行/home/u590953899/public_html/notify/reset.php错误更新记录：

按下提交按钮时，假设更新com_code = $ GET url的数据库
的 BUT
会发生什么是它只重新加载页面，我该如何解决这个问题呢？

指向它的链接是：http://notify.bithumor.co/reset.php?code=123456789

Answer 1

您应该将代码更改为内部isset：

if (isset($_POST['Submit'])) {
    $sql1 = "UPDATE user SET password='$pword', com_code=NULL WHERE com_code = '$code'";

    if ($conn->query($sql1) === TRUE) {
        echo "Password has been change successfully!";
    } else {
        echo "Error updating record: " . $conn->error;
    }
  }

Answer 2

在您的代码中进行以下更改：

if (isset($_POST['Submit'])) {
    $sql1 = "UPDATE user SET password='$pword', com_code IS NULL WHERE com_code = '$code'";

    if ($conn->query($sql1) === TRUE) {
        echo "Password has been change successfully!";
    } else {
        echo "Error updating record: " . $conn->error;
    }
}

Answer 3

我们使用IS NULL来检查mysql中的NULL

if (isset($_POST['Submit'])) {
     $sql1 = "UPDATE user SET password='$pword', com_code IS NULL WHERE com_code = $code";
}

在MYSQL中阅读NULL Values

Answer 4

如果您的提交按钮没有$_POST['Submit']，则永远不会设置

name="submit"。仅拥有type="submit"或value="submit"或id="submit"将无法执行此操作。您需要name属性。

Answer 5

首先检查您的输入类型是否为name="Submit"，如果没有添加它。在那之后回显您的查询，

if (isset($_POST['Submit'])) {

  echo "UPDATE user SET password='$pword', com_code=NULL WHERE com_code = '$code'";

   $sql1 = "UPDATE user SET password='$pword', com_code=NULL WHERE com_code = '$code'";

  if ($conn->query($sql1) === TRUE) {
            echo "Password has been change successfully!";
   } else {
            echo "Error updating record: " . $conn->error;
        }

  }

所有代码，即查询执行和消息都应该在同一个if语句中（if（isset（$ _ POST [＆＃39; Submit＆＃39;]）））。

我希望这适合你。

当POST设置时，UPDATE查询不起作用

5 个答案: