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我有一個表格，我使用PHP來驗證，使用一堆嵌套的if語句。它工作，但問題當然是錯誤消息一次顯示一個。我想讓他們一次顯示。所以我試圖重寫我的代碼而不使用嵌套的條件語句。我花了至少4小時嘗試，沒有任何工作。有人能給我一些建議嗎？這將是非常感謝...此外，我不確定是否有可能編寫一個if語句，如果條件滿足，它不執行代碼。這就是我的代碼現在正在嘗試做的事情，我確信這是一種可怕的做法，但我想不到另一種方式。如何使用PHP驗證表單，並獲取所有錯誤消息一次顯示？
<?php
if($_POST['submit']==1)
{
//$submit = strip_tags(isset($_POST['submit']));
$fname = strip_tags($_POST['fname']);
$lname = strip_tags($_POST['lname']);
$usernamereg = strip_tags($_POST['usernamereg']);
$passwordreg = strip_tags($_POST['passwordreg']);
$email = strip_tags($_POST['email']);
$emailcheck = strip_tags($_POST['emailcheck']);
$date = date("Y-m-d");
$connect = mysql_connect("localhost","user","password") or die ("There is a problem connecting to the database");
mysql_select_db("user_db",$connect) or die("Couldn't find the database.");
$queryusername = mysql_query("SELECT * FROM users WHERE username = '$usernamereg'");
$numrowsusername = mysql_num_rows($queryusername);
$queryemail = mysql_query("SELECT * FROM users WHERE email = '$email'");
$numrowsemail = mysql_num_rows($queryemail);
if($fname&&$lname&&$usernamereg&&$passwordreg&&$email&&$emailcheck)
{
}
else
{
$all_fields = 'All fields must be filled in';
}
if (strlen($usernamereg)>25||strlen($fname)>25||strlen($lname)>25)
{
$length_error = "First name, last name, and username cannot be longer than 25 characters";
}
if($numrowsusername = 0)
{
}
else
{
$username_error ='username already taken';
}
if(strlen($passwordreg)<6)
{
$password_error ="Password must be atleast 6 characters";
}
else
{
//some form of password encryption
}
if($email != $emailcheck)
{
$emails_no_match = "Emails don't match";
}
if($numrowsemail = 0)
{
}
else
{
$email_in_use_error ="This email is already in use";
}
if(isset($all_fields)||isset($length_error)||isset($username_error)||isset($password_error)||isset($emails_no_match)||isset($email_in_use_error)==1)
{
}
else
{
$queryget =mysql_query("INSERT INTO users(date,fname,lname,username,password,email,emailcheck)VALUES('$date','$fname','$lname','$usernamereg','$passwordreg','$email','$emailcheck')");
echo "You have been registered!";
}
}
?>
Here is my form code
<form action ='' method = 'POST' name ='regform'>
<table width = '500px'>
<tr>
<td id ="form_title" align ='left'>Sign Up Here</td>
</tr>
<tr>
<td id="form_subtitle" align ='left'> 100% Free</td>
</tr>
<tr>
<td align ='left'><input type ='text' name = 'fname' placeholder = 'First Name' class = 'firstname' /><input type ='text' name='lname' placeholder = 'Last Name' class = 'lastname' /></td>
</tr>
<tr>
<td></td>
</tr>
<tr>
<td align ='left'><input type ='text' name = 'usernamereg' placeholder = 'Username' class = 'username2' /></td>
</tr>
<tr>
<td><?php
if(!empty($username_error)){
echo $username_error;
}
?></td>
</tr>
<tr>
<td align ='left'><input type ='password' name='passwordreg' placeholder = 'Password' class = 'password2' /></td>
</tr>
<tr>
<td><?php
if(isset($password_error)){
echo $password_error;
}
?></td>
</tr>
<tr>
<td align ='left'><input type ='email' name = 'email' placeholder = 'Email' class = 'email'/></td>
</tr>
<tr>
<td><?php
if(!empty($emails_no_match)){
echo $emails_no_match;
}
?></td>
</tr>
<tr>
<td align ='left'><input type ='email' name = 'emailcheck' placeholder = 'Re-enter Email' class = 'emailcheck'/></td>
</tr>
<tr>
<td><?php
if(!empty($email_in_use_error)){
echo $email_in_use_error;
}
?></td>
</tr>
<tr>
<td><?php
if(!empty($all_fields)){
echo $all_fields;
}
?> </td>
</tr>
<td align='center'><input type = 'submit' name = 'submit' id= 'regbutton' value = 'Register' /></td>
</form>
我添加了我的表單代碼。感謝您的答覆。不要問愚蠢的問題，但如果我在括號之間放置了一個分號，當我的代碼執行時，條件滿足時，代碼是否會跳轉到代碼的下一部分（在我的情況下是另一個if語句）或它會停止執行嗎？ – user3407857
你可以簡單地檢查一下，但是，它會轉到下一個if語句。 – PBR
和@ user3407857？所有這些代碼都在同一個php源代碼中嗎？ – PBR