我目前在将数据插入数据库时遇到了一些麻烦。我在网站上放入一些数据的所有内容都会出现错误:'数据库插入1失败'。此外,当我从代码中删除第一个查询时,我得到下一个错误:'数据库插入2失败'。等等...
我做错了什么?他们的代码中有任何缺陷吗?
我的数据库说明:
我该如何解决这个问题?非常感谢你提前!
亲切的问候,
聚苯乙烯。以下代码,对于那些没有注意到的人;)另外,如果有人需要一些额外的信息,当然可以随意询问。我会尽可能地提供它。
Connect.php:
<?php
$connection = mysqli_connect('localhost', 'root', '');
if (!$connection){
die("Database Connection Failed" . mysql_error());
}
$select_db = mysqli_select_db($connection,'datingsite');
if (!$select_db){
die("Database Selection Failed" . mysql_error());
}
?>
Register.php:
<?php
Include('connect.php');
if (isset($_POST['email']) && isset($_POST['wachtwoord'])){
$lidnummer = NULL;
$email = $_POST['email'];
$wachtwoord = $_POST['wachtwoord'];
$Ingelogd = 0;
$Voornaam = $_POST['Voornaam'];
$Tweedenaam = $_POST['Tweedenaam'];
$Achternaam = $_POST['Achternaam'];
$Woonplaats = $_POST['Woonplaats'];
$Provincie = $_POST['Provincie'];
$Ben = $_POST['Ben'];
$Zoek = $_POST['Zoek'];
$Dag = $_POST['Dag'];
$Maand = $_POST['Maand'];
$Jaar = $_POST['Jaar'];
$Hobby1 = $_POST['Hobby1'];
$Hobby2 = $_POST['Hobby2'];
$Opleiding = $_POST['Opleiding'];
$query = "INSERT INTO 'user' (Lidnummer, Email, Wachtwoord, Ingelogd)
VALUES (
'$lidnummer',
'$email',
'$wachtwoord',
'$Ingelogd')";
$query2 = "INSERT INTO 'naam' (Lidnummer, Voornaam, Tweedenaam, Achternaam)
VALUES (
'$lidnummer',
'$Voornaam',
'$Tweedenaam',
'$Achternaam')";
$query3 = "INSERT INTO 'woon' (Lidnummer, Woonplaats, Provincie)
VALUES (
'$lidnummer',
'$Woonplaats',
'$Provincie')";
$query4 = "INSERT INTO 'sexe' (Lidnummer, Ben, Zoek)
VALUES (
'$lidnummer',
'$Ben',
'$Zoek')";
$query5 = "INSERT INTO 'jaar' (Lidnummer, Dag, Maand, Jaar)
VALUES (
'$lidnummer',
'$Dag',
'$Maand',
'$Jaar')";
$query6 = "INSERT INTO 'hobby' (Lidnummer, Hobby1, Hobby2, Opleiding)
VALUES (
'$lidnummer',
'$Hobby1',
'$Hobby2',
'$Opleiding')";
$res = mysqli_query($connection, $query);
if (!$res){
die("Database Insert 1 Failed" . mysql_error());}
$res2 = mysqli_query($connection, $query2);
if (!$res2){
die("Database Insert 2 Failed" . mysql_error());}
$res3 = mysqli_query($connection, $query3);
if (!$res3){
die("Database Insert 3 Failed" . mysql_error());}
$res4 = mysqli_query($connection, $query4);
if (!$res4){
die("Database Insert 4 Failed" . mysql_error());}
$res5 = mysqli_query($connection, $query5);
if (!$res5){
die("Database Insert 5 Failed" . mysql_error());}
$res6 = mysqli_query($connection, $query6);
if (!$res6){
die("Database Insert 6 Failed" . mysql_error());}
if(($res)&&($res2)&&($res3)&&($res4)&&($res5)&&($res6)){
$msg = "User Created Successfully.";
}
}
?>
<!DOCTYPE html>
<html lang="nl">
<head>
<meta charset="utf-8" />
<title>Registreren op Chives</title>
<link href="../Css/inlog.css" rel="stylesheet"/>
<link href="../Css/styles.css" rel="stylesheet" />
</head>
<body class="back">
<?php
if(isset($msg) & !empty($msg)){
echo $msg;
}
?>
<div id="Inlog-Container" align="center">
<form action="" method="post">
<H1> Registreren </H1>
<H2> Email:</H2>
<input name="email" type="email" class="Input-box" required/>
<H2> Wachtwoord:</H2>
<input name="wachtwoord" type="password" class="Input-box" required/>
<div class="Radiolabelbox">
<fieldset class="" id="" >
<H2>Ik ben een:</H2>
<div class="Radiolabel">
<label>
<input type="radio" name="Ben" class="styled-radio" value="Man" required/>
Man
</label> <br />
<label>
<input type="radio" name="Ben" class="styled-radio" value="Vrouw"/>
Vrouw
</label>
</div>
</fieldset>
<fieldset class="">
<H2 class="">Ik zoek een:</H2>
<div class="Radiolabel">
<label>
<input type="radio" name="Zoek" class="styled-radio" value="Man" required/>
Man
</label>
<br />
<label>
<input type="radio" name="Zoek" class="styled-radio" value="Vrouw"/>
Vrouw
</label>
<br />
<label>
<input type="radio" name="Zoek" class="styled-radio" value="Beide"/>
Beide
</label>
</div>
</fieldset>
</div>
<H2> Woonplaats:</H2>
<input name="Woonplaats" type="text" class="Input-box" required/>
<H2> Provincie:</H2>
<input name="Provincie" type="text" class="Input-box" required/>
<H2> Hobby 1:</H2>
<input name="Hobby1" type="text" class="Input-box" required/>
<H2> Hobby 2:</H2>
<input name="Hobby2" type="text" class="Input-box" required/>
<H2> Voornaam:</H2>
<input name="Voornaam" type="text" class="Input-box" required/>
<H2> Tweede naam:</H2>
<input name="Tweedenaam" type="text" class="Input-box" required/>
<H2> Achternaam:</H2>
<input name="Achternaam" type="text" class="Input-box" required/>
<H2> Geboortedag:</H2>
<input name="Dag" type="Number" class="Input-box" min="0" max="31" required/>
<H2> Geboortemaand:</H2>
<input name="Maand" type="Number" class="Input-box" min="0" max="12" required/>
<H2> Geboortejaar:</H2>
<input name="Jaar" type="Number" class="Input-box" min="1920" max="2000" required/>
<H2> Opleiding:</H2>
<input name="Opleiding" type="Text" class="Input-box" required/>
<input type="submit" value="GA VERDER" class="Roundbutton" id="Login" />
</form>
</div>
</body>
</html>
答案 0 :(得分:3)
首先,您将API与mysql_error()
混合使用 - 对所有API使用mysqli_error($connection)
。
这些不同的API不会相互混合。
然后,您使用错误的标识符来所有您的表名,即引号。
阅读主题http://dev.mysql.com/doc/refman/5.0/en/identifier-qualifiers.html
将or die(mysqli_error($connection))
添加到mysqli_query()
会发出信号。
(如果存在空格,连字符或保留关键字,则需要勾号)
即:
$query = "INSERT INTO `user`
或删除它们
$query = "INSERT INTO user
我还必须注意,您目前的代码向SQL injection开放。使用prepared statements或PDO with prepared statements,他们更安全。