Origineel gepost door Jeffrey Bouva
PHP Code:mysql_query("SELECT * FROM `database`.`tabel` WHERE gebruikersnaamwaarschijnlijk = '". $var1 ."' OR gebruikersnaamwaarschijnlijk = '". $var2 ."'");
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16-02-2011, 19:45 #16
Roboo
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Re: Problemen met een inlog script
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16-02-2011, 19:55 #17
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Re: Problemen met een inlog script
Ik heb twee variable als database veld naam Origineel gepost door Robin L
de ene heet schoolid
en de andere contactid
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16-02-2011, 20:13 #18
Roboo
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Re: Problemen met een inlog script
Stuur mij eens je hele script + wat je wilt via PB, dan help ik je wel even verder.. Origineel gepost door Jeffrey Bouva
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17-02-2011, 13:08 #19
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Re: Problemen met een inlog script
Beste mensen zelfs met de hulp van robin werkt het nog niet
misschien kunnen jullie me verder helpen
*thanks alvast btw robin want ben al een stuk verder*
Dit is mijn script nuPHP Code:
if ($person == "company"){
//$usertable = 'contactid';
$query = mysql_query("SELECT * FROM company WHERE contactid = '".$username."' AND password = '".$password."'");
// student? Dan zoeken we in de schoolid tabel
}elseif($person == "students"){
//$usertable = 'schoolid';
$query = mysql_query("SELECT * FROM students WHERE schoolid = '".$username."' AND password = '".$password."'");
// Leraar? Dan zoeken we in de schoolid tabel
}elseif($person == "teacher"){
//$usertable = 'schoolid';
$query = mysql_query("SELECT * FROM company WHERE schoolid = '".$username."' AND password = '".$password."'");
}
// Ongeldige input, type is niet company, noch students of teacher
else {
//login error
$_SESSION['logerror'] = "Please select a type";
header("Location: index.php");
}
Maar als ik als student * of welke dan ook wil inloggen krijg ik :
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''' WHERE username = 'Jeffrey'' at line 1
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17-02-2011, 13:52 #20
Hyn
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Re: Problemen met een inlog script
jeffrey.. Ik geloof niet dat je ook maar 1 van de aanpassingen toegevoegd hebt. Je blijft om de tabelnamen nog steeds de ` missen, die moet je ook om de velden zetten.
Stuur maar per pm je msn, desnoods.
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17-02-2011, 15:40 #21
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Re: Problemen met een inlog script
nu met de ` om alle tabelnamen en veld namen.PHP Code:if ($person == "company"){
//$usertable = 'contactid';
$query = mysql_query("SELECT * FROM `company` WHERE `contactid` = '".$username."' AND `password` = '".$password."'");
// student? Dan zoeken we in de schoolid tabel
}elseif($person == "students"){
//$usertable = 'schoolid';
$query = mysql_query("SELECT * FROM `students` WHERE `schoolid` = '".$username."' AND `password` = '".$password."'");
// Leraar? Dan zoeken we in de schoolid tabel
}elseif($person == "teacher"){
//$usertable = 'schoolid';
$query = mysql_query("SELECT * FROM `company` WHERE `schoolid` = '".$username."' AND `password` = '".$password."'");
}
// Ongeldige input, type is niet company, noch students of teacher
else {
//login error
$_SESSION['logerror'] = "Please select a type";
header("Location: index.php");
}
` <-- dit teken is trouwens geen gewone ' enkele quote.
maar dat klopt denk ik ?
ik krijg de volgende melding
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''' WHERE username = 'test'' at line 1
Nu aangepast met ' een enkele quote
Nu krijg ik de volgende meldingPHP Code:if ($person == "company"){
//$usertable = 'contactid';
$query = mysql_query("SELECT * FROM 'company' WHERE ' contactid' = '".$username."' AND 'password' = '".$password."'");
// student? Dan zoeken we in de schoolid tabel
}elseif($person == "students"){
//$usertable = 'schoolid';
$query = mysql_query("SELECT * FROM 'students' WHERE 'schoolid' = '".$username."' AND 'password' = '".$password."'");
// Leraar? Dan zoeken we in de schoolid tabel
}elseif($person == "teacher"){
//$usertable = 'schoolid';
$query = mysql_query("SELECT * FROM 'company' WHERE 'schoolid' = '".$username."' AND 'password' = '".$password."'");
}
// Ongeldige input, type is niet company, noch students of teacher
else {
//login error
$_SESSION['logerror'] = "Please select a type";
header("Location: index.php");
}
en de codeWarning: mysql_num_rows() expects parameter 1 to be resource, boolean given in D:\xampp\htdocs\internship\login_check.php on line 46
Warning: Cannot modify header information - headers already sent by (output started at D:\xampp\htdocs\internship\login_check.php:46) in D:\xampp\htdocs\internship\login_check.php on line 73
PHP Code:$insert = mysql_num_rows($query); // Checks to see if anything is in the db.
if ($insert == 1) { // There is something in the db. The username/password match up.
$qry_rank = mysql_query("SELECT rank FROM '$type' WHERE username = '$username'") or die (mysql_error());
$rst_rank = mysql_fetch_array($qry_rank);
$qry_id = mysql_query("SELECT firstname FROM '$type' WHERE username = '$username'") or die (mysql_error());
$rst_id = mysql_fetch_array($qry_id);
$_SESSION['rank'] = $rst_rank[0];
$_SESSION['logged'] = 1; // Sets the session.
$_SESSION['firstname'] = $qry_id[0];
header("Location: index2.php"); // Goes to main page.
}
elseif(!$username){
$_SESSION['logerror'] = "Please fill in a Username";
header("Location: index.php");}
elseif ($password == $leeg_md5){
$_SESSION['logerror'] = "Please fill in a Password";
header("Location: index.php");}
elseif ($person == "none"){
$_SESSION['logerror'] = "Please select a type";
header("Location: index.php");}
else{
$_SESSION['logerror'] = "Username or Password didn't match";
header("Location: index.php");
}
}
}
$connector->close() // Closes the connection.
?>
Nu heb ik achter de querys allemaalgezet,PHP Code:or die (mysql_error());
En nu krijg ik de volgende melding
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''students' WHERE 'schoolid' = 'test' AND 'password' = '098f6bcd4621d373cade4e832' at line 1Laatst aangepast door Jeffrey Bouva : 17-02-2011 om 15:52
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17-02-2011, 16:00 #22
Roboo
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Re: Problemen met een inlog script
database en kolom allebij met een ` i.p.v. een '
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17-02-2011, 16:21 #23
Particulier
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Re: Problemen met een inlog script
Opgelost door de code als volgt te coderen
PHP Code:<?php
session_start();
// Starts the session
// Includes the db and form info.
require 'includes/dbconnector.php';
// Create an object (instance) of the DbConnector
$connector = new dbconnector();
if ($_SESSION['logged'] == 1) { // User is already logged in.
header("Location: index2.php"); // Goes to main page.
exit(); // Stops the rest of the script.
}else{
// If the form has been submitted.
if(isset($_POST['posted'])){
$username =($_POST['username']);
$password = md5($_POST['password']); // Encrypts the password.
$leeg_md5 = "d41d8cd98f00b204e9800998ecf8427e";
$person = ($_POST['person']);
if ($person == "company"){
//$usertable = 'contactid';
$query = mysql_query("SELECT * FROM `company` WHERE `contactid` = '".$username."' AND `password` = '".$password."'") or die (mysql_error());
$insert = mysql_num_rows($query) or die (mysql_error()); // Checks to see if anything is in the db.
if ($insert == 1) { // There is something in the db. The username/password match up.
$qry_rank = mysql_query("SELECT rank FROM `company` WHERE `contactid` = '".$username."'") or die (mysql_error());
$rst_rank = mysql_fetch_array($qry_rank);
$qry_id = mysql_query("SELECT firstname FROM `company` WHERE `contactid` = '".$username."'") or die (mysql_error());
$rst_id = mysql_fetch_array($qry_id);
$_SESSION['rank'] = $rst_rank[0];
$_SESSION['logged'] = 1; // Sets the session.
$_SESSION['firstname'] = $qry_id[0];
header("Location: index2.php"); // Goes to main page.
}
elseif(!$username){
$_SESSION['logerror'] = "Please fill in a Username";
header("Location: index.php");}
elseif ($password == $leeg_md5){
$_SESSION['logerror'] = "Please fill in a Password";
header("Location: index.php");}
elseif ($person == "none"){
$_SESSION['logerror'] = "Please select a type";
header("Location: index.php");}
else{
$_SESSION['logerror'] = "Username or Password didn't match";
header("Location: index.php");
}
// student? Dan zoeken we in de schoolid tabel
}elseif($person == "students"){
//$usertable = 'schoolid';
$query = mysql_query("SELECT * FROM `students` WHERE `schoolid` = '".$username."' AND `password` = '".$password."'") or die (mysql_error());
$insert = mysql_num_rows($query) or die (mysql_error()); // Checks to see if anything is in the db.
if ($insert == 1) { // There is something in the db. The username/password match up.
$qry_rank = mysql_query("SELECT rank FROM `students` WHERE `schoolid` = '".$username."'") or die (mysql_error());
$rst_rank = mysql_fetch_array($qry_rank);
$qry_id = mysql_query("SELECT firstname FROM `students` WHERE `schoolid` = '".$username."'") or die (mysql_error());
$rst_id = mysql_fetch_array($qry_id);
$_SESSION['rank'] = $rst_rank[0];
$_SESSION['logged'] = 1; // Sets the session.
$_SESSION['firstname'] = $qry_id[0];
header("Location: index2.php"); // Goes to main page.
}
elseif(!$username){
$_SESSION['logerror'] = "Please fill in a Username";
header("Location: index.php");}
elseif ($password == $leeg_md5){
$_SESSION['logerror'] = "Please fill in a Password";
header("Location: index.php");}
elseif ($person == "none"){
$_SESSION['logerror'] = "Please select a type";
header("Location: index.php");}
else{
$_SESSION['logerror'] = "Username or Password didn't match";
header("Location: index.php");
}
// Leraar? Dan zoeken we in de schoolid tabel
}elseif($person == "teacher"){
//$usertable = 'schoolid';
$query = mysql_query("SELECT * FROM `teacher` WHERE `schoolid` = '".$username."' AND `password` = '".$password."'") or die (mysql_error());
$insert = mysql_num_rows($query) or die (mysql_error()); // Checks to see if anything is in the db.
if ($insert == 1) { // There is something in the db. The username/password match up.
$qry_rank = mysql_query("SELECT rank FROM `students` WHERE `schoolid` = '".$username."'") or die (mysql_error());
$rst_rank = mysql_fetch_array($qry_rank);
$qry_id = mysql_query("SELECT firstname FROM `students` WHERE `schoolid` = '".$username."'") or die (mysql_error());
$rst_id = mysql_fetch_array($qry_id);
$_SESSION['rank'] = $rst_rank[0];
$_SESSION['logged'] = 1; // Sets the session.
$_SESSION['firstname'] = $qry_id[0];
header("Location: index2.php"); // Goes to main page.
}
elseif(!$username){
$_SESSION['logerror'] = "Please fill in a Username";
header("Location: index.php");}
elseif ($password == $leeg_md5){
$_SESSION['logerror'] = "Please fill in a Password";
header("Location: index.php");}
elseif ($person == "none"){
$_SESSION['logerror'] = "Please select a type";
header("Location: index.php");}
else{
$_SESSION['logerror'] = "Username or Password didn't match";
header("Location: index.php");
}
}
// Ongeldige input, type is niet company, noch students of teacher
else {
//login error
$_SESSION['logerror'] = "Please select a type";
header("Location: index.php");
}
}
}
$connector->close() // Closes the connection.
?>
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18-02-2011, 08:29 #24
Hyn
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Re: Problemen met een inlog script
Serieus jeffrey; er word meermaals gezegd dat je velden ook met `` moet omvatten. Je post meermaals code waarin dit al gebeurd is, 2 posts terug meld je dat je nog steeds een error krijgt in de code waar username = 'test' staat.. Deze code is voorheen niet meegegeven, maar daarin was je wel vergeten om om username de ` te zetten. Juist username moet die tildes eromheen hebben, omdat het ook door SQL zelf gebruikt word.
