php programmers i need help

willy ze great · Oct 3, 2018

<?php
session_start();
$servername="localhost";
$username="root";
$password="";
$dbname="hms";
$conn=mysqli_connect($servername,$username,$password,$dbname);
if($_SERVER["REQUEST_METHOD"] == "POST"){
//username and password sent from form
$account = mysqli_real_escape_string($conn,$_POST['account']);
$pword = mysqli_real_escape_string($conn,$_POST['pword']);
$sql="select id from hospital where account='$account' and pword='$pword'";
$result= mysqli_query($conn,$sql);
$row= mysqli_fetch_array($result,MYSQLI_ASSOC);
$count= mysqli_num_rows($result);
//if result matched surname and password table row must be 1 row
if ($count==1){
$_SESSION['pharmacist'] = $account;
header ("location: pharmacy.php");
}elseif($count==1) {
$_SESSION['doctor'] = $account;
header ("location: doctor.php");

}elseif($count==1){
$_SESSION['receptionist'] = $account;
header ("location: reception.php");
}elseif($count==1){
$_SESSION['lab Technician'] = $account;
header ("location: lbtechncian.php");
}elseif($count==1){
$_SESSION['manager/admin'] = $account;
header ("location: index.php");
}else{

echo "<script>alert('wrong password , Try again!!!')</script>";
}

}

?>

user wako registred tayari tatizo nataka waingilie aina za account inayotakiwa na password tu. na impeleke kwenye page husika nimejaribu hizo inakubali kwa header ya kwanza tu.. msaada

problem solved nilifanya hivi ikakubali
<?php
session_start();
$servername="localhost";
$username="root";
$password="";
$dbname="hms";
$conn=mysqli_connect($servername,$username,$password,$dbname);
if($_SERVER["REQUEST_METHOD"] == "POST"){
//username and password sent from form
$account = mysqli_real_escape_string($conn,$_POST['account']);
$pword = mysqli_real_escape_string($conn,$_POST['pword']);
$sql="select id from hospital where account='$account' and pword='$pword'";
$result= mysqli_query($conn,$sql);
$row= mysqli_fetch_array($result,MYSQLI_ASSOC);
$count= mysqli_num_rows($result);
//if result matched surname and password table row must be 1 row
if ($count==1){
$_SESSION['account'] = $account;
if($account=='Doctor'){
echo"<script>window.open('doctor.php','_self')</script>";
}elseif($account=='Receptionist'){
echo"<script>window.open('reception.php','_self')</script>";
}elseif($account=='Lab Technician'){
echo"<script>window.open('lbtechnician.php','_self')</script>";
}elseif($account=='Pharmacist'){
echo"<script>window.open('pharmacist.php','_self')</script>";
}else{
echo"<script>window.open('index.php','_self')</script>";

}
}else{

echo "<script>alert('wrong password , Try again!!!')</script>";
}

}

?>

Kaparo · Oct 3, 2018

Ngoja wakija, ila wakichelewa..
Aione Kang kama vipi mfate piemu (PM) huko..

dronedrake · Oct 3, 2018

umetumia 'if....else if...else if...else if...else if.....else if....' ivi unajua maana yake?

ile ya kwanza kabisa itakayokubali, ZINAZOFATA HAZITA execute

solution: toa izo 'else if' ibaki 'if ' pekeake

RobyMi · Oct 3, 2018

willy ze great said:
<?php
session_start();
$servername="localhost";
$username="root";
$password="";
$dbname="hms";
$conn=mysqli_connect($servername,$username,$password,$dbname);
if($_SERVER["REQUEST_METHOD"] == "POST"){
//username and password sent from form
$account = mysqli_real_escape_string($conn,$_POST['account']);
$pword = mysqli_real_escape_string($conn,$_POST['pword']);
$sql="select id from hospital where account='$account' and pword='$pword'";
$result= mysqli_query($conn,$sql);
$row= mysqli_fetch_array($result,MYSQLI_ASSOC);
$count= mysqli_num_rows($result);
//if result matched surname and password table row must be 1 row
if ($count==1){
$_SESSION['pharmacist'] = $account;
header ("location: pharmacy.php");
}elseif($count==1) {
$_SESSION['doctor'] = $account;
header ("location: doctor.php");

}elseif($count==1){
$_SESSION['receptionist'] = $account;
header ("location: reception.php");
}elseif($count==1){
$_SESSION['lab Technician'] = $account;
header ("location: lbtechncian.php");
}elseif($count==1){
$_SESSION['manager/admin'] = $account;
header ("location: index.php");
}else{

echo "<script>alert('wrong password , Try again!!!')</script>";
}

}

?>

user wako registred tayari tatizo nataka waingilie aina za account inayotakiwa na password tu. na impeleke kwenye page husika nimejaribu hizo inakubali kwa header ya kwanza tu.. msaada

<?php
session_start();
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "hms";
$conn = mysqli_connect($servername, $username, $password, $dbname);
if ($_SERVER["REQUEST_METHOD"] == "POST") {
//username and password sent from form
$account = mysqli_real_escape_string($conn, $_POST['account']);
$pword = mysqli_real_escape_string($conn, $_POST['pword']);
$sql = "select id from hospital where account='$account' and pword='$pword'";
$result = mysqli_query($conn, $sql);
$row = mysqli_fetch_array($result, MYSQLI_ASSOC);
$count = mysqli_num_rows($result);
//if result matched surname and password table row must be 1 row

if($count == 1) {

switch ($account) {

case $_SESSION['pharmacist']:
header("location: pharmacy.php");
break;
case $_SESSION['doctor']:
header("location: doctor.php");
break;

case $_SESSION['receptionist']:
header("location: reception.php");
break;

case $_SESSION['lab Technician']:
header("location: lbtechncian.php");
break;

case $_SESSION['manager/admin']:
header("location: index.php");
break;

}
} else {

echo "<script>alert('wrong password , Try again!!!')</script>";
}

}

Apprently JF doesnt have a code formatting tool ..

willy ze great · Oct 3, 2018

Nimetoa ila sipati kitu nachokitaka

dronedrake said:
umetumia 'if....else if...else if...else if...else if.....else if....' ivi unajua maana yake?

ile ya kwanza kabisa itakayokubali, ZINAZOFATA HAZITA execute

solution: toa izo 'else if' ibaki 'if ' pekeake

Muntu Ya Pori · Oct 3, 2018

willy ze great said:
session_start();
$servername="localhost";
$username="root";
$password="";
$dbname="hms";
$conn=mysqli_connect($servername,$username,$password,$dbname);
if($_SERVER["REQUEST_METHOD"] == "POST"){
//username and password sent from form
$account = mysqli_real_escape_string($conn,$_POST['account']);
$pword = mysqli_real_escape_string($conn,$_POST['pword']);
$sql="select id from hospital where account='$account' and pword='$pword'";
$result= mysqli_query($conn,$sql);
$row= mysqli_fetch_array($result,MYSQLI_ASSOC);
$count= mysqli_num_rows($result);
//if result matched surname and password table row must be 1 row
if ($count==1){
$_SESSION['pharmacist'] = $account;
header ("location: pharmacy.php");
}elseif($count==1) {
$_SESSION['doctor'] = $account;
header ("location: doctor.php");

}elseif($count==1){
$_SESSION['receptionist'] = $account;
header ("location: reception.php");
}elseif($count==1){
$_SESSION['lab Technician'] = $account;
header ("location: lbtechncian.php");
}elseif($count==1){
$_SESSION['manager/admin'] = $account;
header ("location: index.php");
}else{

echo "";
}

}

?>

user wako registred tayari tatizo nataka waingilie aina za account inayotakiwa na password tu. na impeleke kwenye page husika nimejaribu hizo inakubali kwa header ya kwanza tu.. msaada

Kwanini kila unapochukua data na kusave kwenye variables unapotumia hiyo built-in function kuzuia sql injection unapass db connection?
Kingine sijaona code inayotofautisha account ili kufanya decision. Kama account ni doctor we umeset session pharmacist then ume sema iwe doctor na unadirect to pharmacist. Header zako haziendani na account ndio Maana branch ya kwanza tu inakua valid kwa account yoyote atakayoingiza.
Fanya hivi. Use if to check row kama ni 1 else invalid. Kama ni 1 fanya switch hizo account, kama account ni doctor, set session to doctor then direct to doctor page na kuendelea.
Au unaweza fanya nested if ikiwa row ni if(row === 1) {
if($account ==='doctor') {
$_SESSION['doctor'] = $account;
Header(......);
}
if(........) {
}
} else {
Echo ' invalid username or password'
}
Hapo itafanya kazi vizuri tu.

ISLETS · Oct 4, 2018

mi hapa napita tu kuwambia poleni na kazi, na endeleeni na moyo huo huo

Beeb · Oct 4, 2018

Code bana

willy ze great · Oct 4, 2018

Muntu Ya Pori said:
Kwanini kila unapochukua data na kusave kwenye variables unapotumia hiyo built-in function kuzuia sql injection unapass db connection?
Kingine sijaona code inayotofautisha account ili kufanya decision. Kama account ni doctor we umeset session pharmacist then ume sema iwe doctor na unadirect to pharmacist. Header zako haziendani na account ndio Maana branch ya kwanza tu inakua valid kwa account yoyote atakayoingiza.
Fanya hivi. Use if to check row kama ni 1 else invalid. Kama ni 1 fanya switch hizo account, kama account ni doctor, set session to doctor then direct to doctor page na kuendelea.
Au unaweza fanya nested if ikiwa row ni if(row === 1) {
if($account ==='doctor') {
$_SESSION['doctor'] = $account;
Header(......);
}
if(........) {
}
} else {
Echo ' invalid username or password'
}
Hapo itafanya kazi vizuri tu.

nilifnaya hivi
<?php
session_start();
$servername="localhost";
$username="root";
$password="";
$dbname="hms";
$conn=mysqli_connect($servername,$username,$password,$dbname);
if($_SERVER["REQUEST_METHOD"] == "POST"){
//username and password sent from form
$account = mysqli_real_escape_string($conn,$_POST['account']);
$pword = mysqli_real_escape_string($conn,$_POST['pword']);
$sql="select id from hospital where account='$account' and pword='$pword'";
$result= mysqli_query($conn,$sql);
$row= mysqli_fetch_array($result,MYSQLI_ASSOC);
$count= mysqli_num_rows($result);
//if result matched surname and password table row must be 1 row
if ($count==1){
$_SESSION['account'] = $account;
if($account=='Doctor'){
echo"<script>window.open('doctor.php','_self')</script>";
}elseif($account=='Receptionist'){
echo"<script>window.open('reception.php','_self')</script>";
}elseif($account=='Lab Technician'){
echo"<script>window.open('lbtechnician.php','_self')</script>";
}elseif($account=='Pharmacist'){
echo"<script>window.open('pharmacist.php','_self')</script>";
}else{
echo"<script>window.open('index.php','_self')</script>";

}
}else{

echo "<script>alert('wrong password , Try again!!!')</script>";
}

}

?>

Mlolongo · Mar 8, 2021

Yna2 said:
Code bana

Unapenda code? Ulisoma code chuo?

mathsjery · Mar 9, 2021

Vipanga special

sayoo · Mar 9, 2021

mathsjery said:
Vipanga special

Lugha nyepesi hiyo , hiyo ndio lugha nyepesi kuliko zote kwenye upande Wa programming mkuu, usiogope wala nn

Mlolongo · Mar 9, 2021

sayoo said:
Lugha nyepesi hiyo , hiyo ndio lugha nyepesi kuliko zote kwenye upande Wa programming mkuu, usiogope wala nn

Unachekesha mkuu. Unaonekana mgeni humu! Ungemjua unayempa huu ushauri!!

sayoo · Mar 9, 2021

Mlolongo said:
Unachekesha mkuu. Unaonekana mgeni humu! Ungemjua unayempa huu ushauri!!

Mkuu kwan hum kuna haja Ya kujunana, ili iweje sasa

Mlolongo · Mar 9, 2021

sayoo said:
Mkuu kwan hum kuna haja Ya kujunana, ili iweje sasa

Acha ushamba. Wewe unaonekana mgeni humu. Period.

sayoo · Mar 9, 2021

Mlolongo said:
Acha ushamba. Wewe unaonekana mgeni humu. Period.

Yaani wewe umejiunga Jf , juzi Tuu Hapo 2019 kisha unaniambia mm mgeni , jiangalie dogo

sayoo · Mar 10, 2021

Mlolongo said:
Unampumulia? Aisee basi wewe mjanja kwa kumgonga dingi.

umejiona mjanja kumbe bwabwa tu [emoji1787][emoji1787][emoji1787][emoji1787]

php programmers i need help

willy ze great

JF-Expert Member

Kaparo

JF-Expert Member

dronedrake

JF-Expert Member

RobyMi

JF-Expert Member

willy ze great

JF-Expert Member

Muntu Ya Pori

JF-Expert Member

ISLETS

JF-Expert Member

Beeb

JF-Expert Member

willy ze great

JF-Expert Member

Mlolongo

JF-Expert Member

mathsjery

JF-Expert Member

sayoo

JF-Expert Member

Mlolongo

JF-Expert Member

sayoo

JF-Expert Member

Mlolongo

JF-Expert Member

sayoo

JF-Expert Member

sayoo

JF-Expert Member

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