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おかげさまでありがとうございます。フォームからデータを取得してデータベースに入力しようとしています。私はこれを行う方法についてのガイドに従っていますが、ガイドには何も作用しないものがありません........PHPを使用してフォームからDBにデータを保存しよう
私はまだPHPで始まったばかりなので、私のスキルはかなり限られています私は書いたすべてのコードを理解していますが、うまくいきません！ PHPスクリプトは、以下のif文を越えて移動しませんし、私は理解することはできません理由：
// 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 [email protected] This is from error 1');
全体コード：
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 [email protected] 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 [email protected] 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: $newpass
You aren't stuck with this password! Your can
change it at any time after you have logged in.
If you have any problems, feel free to contact me at
<[email protected]>.
-Your Name
Your Site Webmaster
";
mail($_POST['newemail'],"Your Password for the Project Website",
$message, "From:Your Name <[email protected]>");
?>
<!DOCTYPE html PUBLIC "-//W3C/DTD XHTML 1.0 Transitional//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title> Registration Complete </title>
<meta http-equiv="Content-Type"
content="text/html; charset=iso-8859-1" />
</head>
<body>
<p><strong>User registration successful!</strong></p>
<p>Your userid and password have been emailed to
<strong><?=$_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;
?>
をdb.php
<?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 $cxn;
}
?>
common.php
<?php // common.php
function error($msg) {
?>
<html>
<head>
<script language="JavaScript">
<!--
alert("<?=$msg?>");
history.back();
//-->
</script>
</head>
<body>
</body>
</html>
<?php
exit;
}
?>
このif文でエラーが発生する理由については、お手数ですがありがとうございます。
あなたがエラーを取得している場合、あなたは私の言葉遣いが少しオフだった申し訳ありませんが、PHPスクリプトは、過去を動かす習慣 –
あなたの質問にエラーが含まれている必要があります //新しいID $ sqlを持つ既存のユーザーをチェック= "SELECT COUNT（*）from sessions.usersどこuserid = '$ _POST [newid]'"; $ result = mysqli_query（$ cxn、$ sql）; if（！$ result）{ error（ '' 'の処理中にデータベースエラーが発生しました。\ nこのエラーが解決しない場合は、' '（[email protected]に連絡してください。 ; – DCWD