Question

嘿，谢谢，我正在尝试从表单中获取数据并将其输入数据库。我正在按照如何做到这一点的指南，但指南没有解释任何不起作用的事情........

我刚刚开始使用PHP，所以我的技能非常有限，但是我确实理解了我编写的所有代码，它只是不起作用！ php脚本不会移动到下面的if语句，我无法找出原因：

// Check for existing user with the new id
$sql = "SELECT COUNT(*) FROM sessions.users WHERE userid = '$_POST[newid]'";
$result = mysqli_query($cxn, $sql);
if (!$result) {
error('A database error occurred in processing your '.
'submission.\nIf this error persists, please '.
'contact you@example.com. This is from error 1');

整个代码：

signup.php

<?php //signup.php
        include 'common.php';
        include 'db.php';

        if(!isset($_POST['submitok'])):
        // display user signup form

        ?>


    <!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>
      <title>New User Registration</title>
      <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
    </head>
    <body>
      <h3>New User Registration Form</h3>
      <p><font color="orangered" size="+1"><tt><b>*</b></tt></font> indicates a required field</p>
      <form method="post" action="<?=$_SERVER['PHP_SELF']?>">
        <table border="0" cellpadding="0" cellspacing="5">
          <tr>
            <td align="right">
              <p>User ID</p>
            </td>
            <td>
              <input name="newid" type="text" maxlength="100" size="25" />
             <font color="orangered" size="+1"><tt><b>*</b></tt></font>
           </td>
        </tr>
        <tr>
          <td align="right">
            <p>Full Name</p>
          </td>
          <td>
            <input name="newname" type="text" maxlength="100" size="25" />
            <font color="orangered" size="+1"><tt><b>*</b></tt></font>
          </td>
        </tr>
        <tr>
          <td align="right">
            <p>E-Mail Address</p>
          </td>
          <td>
            <input name="newemail" type="text" maxlength="100" size="25" />
            <font color="orangered" size="+1"><tt><b>*</b></tt></font>
          </td>
        </tr>
        <tr valign="top">
          <td align="right">
            <p>Other Notes</p>
          </td>
          <td>
            <textarea wrap="soft" name="newnotes" rows="5" cols="30"></textarea>
          </td>
        </tr>
        <tr>
          <td align="right" colspan="2">
            <hr noshade="noshade" />
            <input type="reset" value="Reset Form" />
            <input type="submit" name="submitok" value="   OK   " />
          </td>
        </tr>
      </table>
    </form>
    </body>
    </html>
    <?php
        else:
            //process sign up submission
            dbConnect('sessions');

            if ($_POST['newid']=='' or $_POST['newname']==''
          or $_POST['newemail']=='') {
            error('One or more required fields were left blank.\\n'.
                  'Please fill them in and try again.');
        }

         // Check for existing user with the new id
        $sql = "SELECT COUNT(*) FROM sessions.users WHERE userid = '$_POST[newid]'";
        $result = mysqli_query($cxn, $sql);
        if (!$result) {
        error('A database error occurred in processing your '.
        'submission.\nIf this error persists, please '.
        'contact you@example.com. This is from error 1');
    }

        if (@mysqli_result($result,0,0)>0) {
        error('A user already exists with your chosen userid.\n'.
        'Please try another.');
        }

        $newpass = substr(md5(time()),0,6);

        $sql = "INSERT INTO sessions.users SET
        userid = '$_POST[newid]',
        password = PASSWORD('$newpass'),
        fullname = '$_POST[newname]',
        email = '$_POST[newemail]',
        notes = '$_POST[newnotes]'";
        if (!mysqli_query($sql))
        error('A database error occurred in processing your '.
        'submission.\nIf this error persists, please '.
        'contact you@example.com. This is from error 2');

        // Email the new password to the person.
        $message = "G'Day!

        Your personal account for the Project Web Site
        has been created! To log in, proceed to the
        following address:

            http://www.example.com/

        Your personal login ID and password are as
        follows:

            userid: $_POST[newid]
            password: $newpass

        You aren't stuck with this password! Your can
        change it at any time after you have logged in.

        If you have any problems, feel free to contact me at
        <you@example.com>.

        -Your Name
         Your Site Webmaster
        ";

        mail($_POST['newemail'],"Your Password for the Project Website",
            $message, "From:Your Name <you@example.com>");
        ?>

        <!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>
          <title> Registration Complete </title>
          <meta http-equiv="Content-Type"
            content="text/html; charset=iso-8859-1" />
        </head>
        <body>
        <p><strong>User registration successful!</strong></p>
        <p>Your userid and password have been emailed to
           <strong><?=$_POST['newemail']?></strong>, the email address
           you just provided in your registration form. To log in,
           click <a href="index.php">here</a> to return to the login
           page, and enter your new personal userid and password.</p>
        </body>
        </html>
        <?php
        endif;
        ?>

db.php中

<?php // db.php

    $dbhost = 'localhost';
    $dbuser = 'root';
    $dbpass = '';

    function dbConnect($db='') {
        global $dbhost, $dbuser, $dbpass;

        $cxn = mysqli_connect($dbhost,$dbuser,$dbpass);

        mysqli_select_db($cxn,$db)
        or die ("Couldn’t select database.");


        return $cxn;
    }   
    ?>

的common.php

<?php // common.php

    function error($msg) {
    ?>
    <html>
    <head>
      <script language="JavaScript">
      <!--
      alert("<?=$msg?>");
      history.back();
      //-->
      </script>
    </head>
    <body>
    </body>
    </html>
    <?php
    exit;
    }
    ?>

为什么这个if语句总是产生错误的任何帮助都会很大 - 谢谢。

Answer 1

您未在可用范围内分配数据库连接。你从dbConnect()函数返回它，你只是没有对返回的值做任何事情。

dbConnect('sessions');应为$cxn = dbConnect('sessions');

如果您使用mysqli_error()让MySQL告诉您它不喜欢的内容，系统会向您发出问题的原因。

最后，您应该在查询中使用绑定参数，而不是直接注入用户提供的数据。搜索＆＃34; mysqli绑定参数＆＃34;学习如何做到这一点。你现在所拥有的是可以攻击的。

对于INSERT语句，请使用绑定参数：

$sql = "INSERT INTO sessions.users (userid, password, fullname, email, notes) VALUES (?, ?, ?, ?, ?)";

$stmt = mysqli_prepare($sql);
mysqli_stmt_bind_param($stmt, 'issss', $newID, PASSWORD($newpass), $fullName, $email, $notes);

if(!mysqli_stmt_execute($stmt))
{
    // use this for debugging, but do NOT leave it in production code
    die(mysqli_error($cxn));
}

mysqli_stmt_close($stmt);

以上未经测试，因此您可能需要进行小幅调整。它应该让你很好地了解至少要做什么。

另外，请确保PASSWORD()确实是您想要使用的。我感觉不是，但我不想承担。

尝试使用php从表单将数据存储到DB中

1 个答案: