我不知道问题出在哪里,你能帮助我吗?我无法调试这个。
我试图在任何地方发出回声,看看问题出在哪里,并在if(isset($_POST['SUBMIT'])) {
这是整个php:
<?php
$conn = mysqli_connect($IP, "pavip", "", "c9") or die("Kapcsolatot nem sikerult letrehozni!");
$error = "Here's the problem: ";
$err = false;
$name = $_POST['name'];
$password = $_POST['pass'];
$email = $_POST['email'];
if(isset($_POST['SUBMIT'])) {
echo $name;
if(isset($_POST['email'])){
if($_POST['name'] != "") {
if($_POST['pass'] == $_POST['pass2'] AND strlen($_POST['pass']) < 6) {
if($_POST['email'] != "") {
$name = $_POST['name'];
$password = $_POST['pass'];
$email = $_POST['email'];
$sql = "INSERT INTO user (username, password, email)
VALUES ('$name, '$password','$email)";
}else {
$error .= "No email";
$err = true;
}
}else {
$error .= "Password too short or wrong ujrairas ";
$err = true;
}
}else {
$error .= "No name ";
$err = true;
}
}else {
}
}
if($err == true){ echo '<script type="text/javascript">alert("' . $error . '"); </script>';}
?>
这是我制作的表格:
<form action="index.php" method="Post" class="register">
<input type="text" name="name" placeholder="Name(Can be nick!)">
<input type="pasword" name="pass" placeholder="Password">
<input type="password" name="pass2" placeholder="Password again">
<input type="text" name="email" placeholder="Email address">
<input type="submit" value="Register">
</form>
答案 0 :(得分:0)
请注意您的HTML中没有提到输入名称=“SUBMIT”。
改变你的:
<input type="submit" value="Register">
要
<input type="submit" value="Register" name="SUBMIT" />
希望这有帮助!
答案 1 :(得分:0)
首先,找不到$_POST['SUBMIT']
,因为您的表单没有name
属性。
您需要指定它:
<input name="SUBMIT" type="submit" value="Register"/>
其次,您错过了'$email
的单引号:
$sql = "INSERT INTO user (username, password, email)
VALUES ('$name, '$password','$email')";
第三,您应该使用mysqli_real_escape_string
阻止MySQL注入:
$name = mysqli_real_escape_string($conn, $_POST['name']);
$password = mysqli_real_escape_string($conn, $_POST['pass']);
$email = mysqli_real_escape_string($conn, $_POST['email']);
第四,您可以考虑使用empty()
来检查空字段,例如:
if(!empty($_POST['name'])) {
答案 2 :(得分:-1)
您没有名为&#34;提交&#34;的提交按钮 如果你想以这种方式使用它,你需要添加至少
<input type="submit" name="SUBMIT">
将提交直接路由到不同的控制器功能会更好 请注意,您也可以使用if($ _ POST)afaik