PHP表单验证不需要';行不通
我试图用php验证制作一个html表单,但如果提交表单时至少有一个填充的文本字段,则会显示电子邮件不匹配(自制的代码部分不应显示)PHP表单验证不需要';行不通,php,html,forms,Php,Html,Forms,我试图用php验证制作一个html表单,但如果提交表单时至少有一个填充的文本字段,则会显示电子邮件不匹配(自制的代码部分不应显示) <?php error_reporting(0); if(isset($_POST["submit"])){ //File Verification if(empty($_POST['username']) && empty($_POST['password1']) && empty($_POST['passw
<?php
error_reporting(0);
if(isset($_POST["submit"])){
//File Verification
if(empty($_POST['username']) && empty($_POST['password1']) && empty($_POST['password2']) && empty($_POST['email1']) && empty($_POST['email2']) && empty($_POST['bday'])){
echo"Kom op, vul alles in";
echo "<meta http-equiv='refresh' content='5;URL=register.php' />";
exit();
}
else{
$email1 = $_POST['email1'];
$email2 = $_POST['email2'];
$pass1 = $_POST['password1'];
$pass2 = $_POST['password2'];
if(email1 == email2){
if(pass1 == pass2){
}
else{
echo "<meta http-equiv='refresh' content='5;URL=register.php' />";
echo"Je wachtwoorden komen niet overeen";
exit();
}
}
else{
echo "<meta http-equiv='refresh' content='5;URL=register.php' />";
echo "Je email gegevens komen niet overeen";
exit();
}
}
}
else{
$form = <<<EOT
<form method="post" action="register.php">
Gebruikersnaam: <input type="text" name="username" placeholder="type hier je gebruikers naam"/><br /><br />
wachtwoord: <input type="password" name="password1" placeholder="type hier je wachtwoord"/><br /><br />
wachtwoord opnieuw: <input type="password" name="password2" placeholder="type je wachtwoord opnieuw in"/><br /><br />
email: <input type="text" name="email1" placeholder="type hier je email"/><br /><br />
email opnieuw: <input type="text" name="email2" placeholder="type hier je email opnieuw"/><br /><br />
Geboorte datum: <input type="date" name="bday"/ placeholder="type je geboorte datum hier"><br /><br />
<input type="submit" name="submit"/>
</form>
EOT;
echo $form;
}
?>
尝试更改:
if(email1 == email2){
if(pass1 == pass2){
致:
您输入了一个打字错误pass1==pass2
。
此外,您可能希望更改为:
if(empty($_POST['username']) &&...
到
这不是你的代码所做的吗?你在两个电子邮件值之间做了一个相等检查,如果它们不匹配,你就会得到else条件。编辑:刚刚注意到你缺少$symbols。请参阅@zzzareckeror reporting的答案应该向你抛出未定义的常量xxx通知,但你决定将其关闭错误报告(0)
if($email1 == $email2){
if($pass1 == $pass2){
if(empty($_POST['username']) &&...
if(empty($_POST['username']) OR...