dpb42021 2014-04-04 08:34
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检查此电子邮件是否已具有帐户功能不适用于PHP HTML和MySQL

I have following code:

<meta charset="UTF-8">

<?php

include_once 'init/init.funcs.php';

function emaili_pikkus(){
global $email;
if (strlen($email)>45){
    echo 'e-mail ei tohi olla pikem kui 45 tähemärki';
}
else{
    parooli_pikkus();
    }
}

function parooli_pikkus()
{
global $parool;
$pikkus = strlen($parool);
if ($pikkus<6){
    echo "Parool peab olema vähemalt 6 tähemärki pikk";
    }
else {
    varasem_olemasolu();
}

}

function varasem_olemasolu()
{

global $email;

if(!empty($_POST['email']))
{
    $query = mysql_query("SELECT * FROM kasutajad ") or die(mysql_error());
    $array = mysql_fetch_array($query);
    if(in_array($email, $array)) 
    {
        echo "Selle e-mailiga on kasutaja juba registreeritud.";
    }
    else
    {
        paroolide_kattuvus();
    }
}
}

function paroolide_kattuvus() 
{
$parool =  $_POST['parool'];
$parool_uuesti =  $_POST['parooluuesti'];

    if($parool==$parool_uuesti)
    {
        NewUser();
        }
    else{
    echo "Paroolid ei kattu.";
    {}  
}
}

function NewUser()
{
global $sql;
if (mysql_query( $sql))
{
echo "Kasutaja loodud";
}
}
    emaili_pikkus();


?>

And file init.funcs.php which contains following:

<?php
session_start ();

$db = mysql_connect ( 'localhost', 'root', 'aaaa' );

$email = mysql_real_escape_string($_POST['email']);
$eesnimi = mysql_real_escape_string($_POST['eesnimi']);
$perekonnanimi = mysql_real_escape_string($_POST['perekonnanimi']);
$parool =  $_POST['parool'];
$parool_uuesti =  $_POST['parooluuesti'];
$salt = rand(10000,99999);
$hashed_pwd = sha1('$parool'.$salt);
$sql="INSERT INTO kasutajad (e_mail, eesnimi, perenimi, parool, salt ) VALUES ('$email','$eesnimi', '$perekonnanimi', '$parool', '$salt')";

if (! $db) {
    header ( "location: /" );
    die ();
} else {
    mysql_select_db ( 'ta2014' );
}



include_once 'functions/user.funcs.php';

?>

My HTML looks like this:

<!DOCTYPE html>
<meta charset="UTF-8">
<html>
<head>
<title>Registreerimine</title>
</head>
<body>
<strong>Registreerimiseks täida järgnevad väljad: </strong><br>
<br>
<form method="POST" action="registreerimine4.php">
<table>
<tr><td>Sinu Tieto e-maili aadress: </td><td><input type="text" name="email"></td></tr>
<tr><td>Eesnimi: </td><td><input type="text" name="eesnimi"></td></tr>
<tr><td>Perekonnanimi: </td><td><input type="text" name="perekonnanimi"></td></tr>
<tr><td>Parool: </td><td><input type="text" name="parool"></td></tr>
<tr><td>Parool uuesti: </td><td><input type="text" name="parooluuesti"></td></tr>
</table>
<br>
<input type="submit" value="Registreeri" name="Registreeri">
</form>
</body>
</html>

Now when I run my HTML and PHP everything works properly except one function. varasem_olemasolu() does not work. This function is meant for checking if this email address already has an account registred. Everything worked properly when I used following code, but its too long and overly complicated to really use:

<meta charset="UTF-8">

<?php

function emaili_pikkus(){
$con = mysql_connect("localhost","root","aaaa");
$email = mysql_real_escape_string($_POST['email']);
if (strlen($email)>45){
    echo 'e-mail ei tohi olla pikem kui 45 tähemärki';
}
else{
    parooli_pikkus();
    }
}

function parooli_pikkus()
{
$parool =  $_POST['parool'];
$pikkus = strlen($parool);
if ($pikkus<6){
    echo "Parool peab olema vähemalt 6 tähemärki pikk";
    }
else {
    varasem_olemasolu();
}

}

function varasem_olemasolu()
{
$con = mysql_connect("localhost","root","aaaa");
mysql_select_db("ta2014", $con);
$email = mysql_real_escape_string($_POST['email']);
if(!empty($_POST['email']))
{
    $query = mysql_query("SELECT * FROM kasutajad ") or die(mysql_error());
    $array = mysql_fetch_array($query);
    if(in_array($email, $array)) 
    {
        echo "Selle e-mailiga on kasutaja juba registreeritud.";
    }
    else
    {
        paroolide_kattuvus();
    }
}
}

function paroolide_kattuvus() 
{
$parool =  $_POST['parool'];
$parool_uuesti =  $_POST['parooluuesti'];

    if($parool==$parool_uuesti)
    {
        NewUser();
        }
    else{
    echo "Paroolid ei kattu.";
    {}  
}
}

function NewUser()
{
$con = mysql_connect("localhost","root","aaaa");
if (!$con)

  {

  die('Could not connect: ' . mysql_error());

  }

mysql_select_db("ta2014", $con);
$email = mysql_real_escape_string($_POST['email']);
$eesnimi = mysql_real_escape_string($_POST['eesnimi']);
$perekonnanimi = mysql_real_escape_string($_POST['perekonnanimi']);
$parool =  $_POST['parool'];
$parool_uuesti =  $_POST['parooluuesti'];
$salt = rand(1000000,99999999);
$hashed_pwd = sha1('$parool'.$salt);
$sql="INSERT INTO kasutajad (e_mail, eesnimi, perenimi, parool, salt ) VALUES ('$email','$eesnimi', '$perekonnanimi', '$parool', '$salt')";

 if (!mysql_query($sql,$con))

  {

  die('Error: ' . mysql_error());

  }

echo "Kasutaja loodud";
}

    emaili_pikkus();


?>

It is a long question and I would be very thankful if someone answers me.

  • 写回答

1条回答 默认 最新

  • douwen1937 2014-04-04 08:37
    关注

    You do not require to fetch complete table record to search email in it. Just change your query as below & check if it returns row is greater than 0.

    SELECT * FROM `kasutajad` WHERE `e_mail` = $email

    Waring: Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO, or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial.

    Example with PDO:

    $email = $_POST['email'];
$con = new PDO( 'mysql:host=localhost;dbname=DB_NAME;charset=UTF-8', 'DB_USER_NAME', 'DB_USER_PASS' );    
$query = $con->prepare( "SELECT `e_mail` FROM `kasutajad` WHERE `e_mail` = ?" );
$query->bindValue( 1, $email);
$query->execute();    
if( $query->rowCount() > 0 ) { # If rows are found for query
     echo "Email Already exits!";
}
else {
     echo "Email not found!";
}
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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