Alert box not show when sign up has same username
I am doing my a sign up using php, I want to show up an alert box that show "User already taken". However when I sign up an account that have same information like the last account, it's not show an alert box. It will go back to blank login page like I set it. The data duplicate not write to the mySQL but the alert box not show.
Thanks
<?php
session_start();
header('location:login.php');
$con = mysqli_connect('localhost','root','');
mysqli_select_db($con, 'userregistration');
$name = $_POST['user'];
$pass = $_POST ['password'];
$email = $_POST['mail'];
$phone = $_POST['phone'];
$s = " SELECT * FROM usertable WHERE name = '$name'";
$result = mysqli_query($con, $s);
$num = mysqli_num_rows($result);
if ($num == 1){
$message="User already taken";
echo "<script type='text/javascript'> alert('$message');</script>";
}else {
$reg = "INSERT INTO usertable(name, password, email, phone ) values ('$name', '$pass','$email', '$phone')";/*, '$email','$phone'*/
mysqli_query($con, $reg);
echo"Registration successfull";
}
?>
php html
asked Nov 22 '18 at 3:16
beam291beam291
147
add a comment |
I am doing my a sign up using php, I want to show up an alert box that show "User already taken". However when I sign up an account that have same information like the last account, it's not show an alert box. It will go back to blank login page like I set it. The data duplicate not write to the mySQL but the alert box not show.
Thanks
<?php
session_start();
header('location:login.php');
$con = mysqli_connect('localhost','root','');
mysqli_select_db($con, 'userregistration');
$name = $_POST['user'];
$pass = $_POST ['password'];
$email = $_POST['mail'];
$phone = $_POST['phone'];
$s = " SELECT * FROM usertable WHERE name = '$name'";
$result = mysqli_query($con, $s);
$num = mysqli_num_rows($result);
if ($num == 1){
$message="User already taken";
echo "<script type='text/javascript'> alert('$message');</script>";
}else {
$reg = "INSERT INTO usertable(name, password, email, phone ) values ('$name', '$pass','$email', '$phone')";/*, '$email','$phone'*/
mysqli_query($con, $reg);
echo"Registration successfull";
}
?>
php html
asked Nov 22 '18 at 3:16
beam291beam291
147
1
In case you prevented the page from showing more alerts, try appending your message after yourscripttag and see if the message shows up. Also, your code is vulnerable to SQL injections
– Thum Choon Tat
Nov 22 '18 at 3:37
add a comment |
I am doing my a sign up using php, I want to show up an alert box that show "User already taken". However when I sign up an account that have same information like the last account, it's not show an alert box. It will go back to blank login page like I set it. The data duplicate not write to the mySQL but the alert box not show.
Thanks
<?php
session_start();
header('location:login.php');
$con = mysqli_connect('localhost','root','');
mysqli_select_db($con, 'userregistration');
$name = $_POST['user'];
$pass = $_POST ['password'];
$email = $_POST['mail'];
$phone = $_POST['phone'];
$s = " SELECT * FROM usertable WHERE name = '$name'";
$result = mysqli_query($con, $s);
$num = mysqli_num_rows($result);
if ($num == 1){
$message="User already taken";
echo "<script type='text/javascript'> alert('$message');</script>";
}else {
$reg = "INSERT INTO usertable(name, password, email, phone ) values ('$name', '$pass','$email', '$phone')";/*, '$email','$phone'*/
mysqli_query($con, $reg);
echo"Registration successfull";
}
?>
php html
asked Nov 22 '18 at 3:16
beam291beam291
147
I am doing my a sign up using php, I want to show up an alert box that show "User already taken". However when I sign up an account that have same information like the last account, it's not show an alert box. It will go back to blank login page like I set it. The data duplicate not write to the mySQL but the alert box not show.
Thanks
<?php
session_start();
header('location:login.php');
$con = mysqli_connect('localhost','root','');
mysqli_select_db($con, 'userregistration');
$name = $_POST['user'];
$pass = $_POST ['password'];
$email = $_POST['mail'];
$phone = $_POST['phone'];
$s = " SELECT * FROM usertable WHERE name = '$name'";
$result = mysqli_query($con, $s);
$num = mysqli_num_rows($result);
if ($num == 1){
$message="User already taken";
echo "<script type='text/javascript'> alert('$message');</script>";
}else {
$reg = "INSERT INTO usertable(name, password, email, phone ) values ('$name', '$pass','$email', '$phone')";/*, '$email','$phone'*/
mysqli_query($con, $reg);
echo"Registration successfull";
}
?>
php html
php html
asked Nov 22 '18 at 3:16
beam291beam291
147
asked Nov 22 '18 at 3:16
beam291beam291
147
asked Nov 22 '18 at 3:16
beam291beam291
147
asked Nov 22 '18 at 3:16
beam291beam291
147
asked Nov 22 '18 at 3:16
beam291beam291
147
147
1
In case you prevented the page from showing more alerts, try appending your message after yourscripttag and see if the message shows up. Also, your code is vulnerable to SQL injections
– Thum Choon Tat
Nov 22 '18 at 3:37
add a comment |
1
In case you prevented the page from showing more alerts, try appending your message after yourscripttag and see if the message shows up. Also, your code is vulnerable to SQL injections
– Thum Choon Tat
Nov 22 '18 at 3:37
1
1
In case you prevented the page from showing more alerts, try appending your message after your
script tag and see if the message shows up. Also, your code is vulnerable to SQL injections– Thum Choon Tat
Nov 22 '18 at 3:37
In case you prevented the page from showing more alerts, try appending your message after your
script tag and see if the message shows up. Also, your code is vulnerable to SQL injections– Thum Choon Tat
Nov 22 '18 at 3:37
add a comment |
1 Answer
1
active
oldest
votes
You may need to check for successful connection/query first:
if($result=mysqli_query($con,$s)) {
if(mysqli_num_rows($result)==1) {
... user already taken code ...
} else {
... insert new user code ...
}
}
See w3schools ref: https://www.w3schools.com/php/func_mysqli_num_rows.asp
This may also help you debug where the problem is occurring.
answered Nov 22 '18 at 3:31
SparebrainSparebrain
106110
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You may need to check for successful connection/query first:
if($result=mysqli_query($con,$s)) {
if(mysqli_num_rows($result)==1) {
... user already taken code ...
} else {
... insert new user code ...
}
}
See w3schools ref: https://www.w3schools.com/php/func_mysqli_num_rows.asp
This may also help you debug where the problem is occurring.
answered Nov 22 '18 at 3:31
SparebrainSparebrain
106110
add a comment |
You may need to check for successful connection/query first:
if($result=mysqli_query($con,$s)) {
if(mysqli_num_rows($result)==1) {
... user already taken code ...
} else {
... insert new user code ...
}
}
See w3schools ref: https://www.w3schools.com/php/func_mysqli_num_rows.asp
This may also help you debug where the problem is occurring.
answered Nov 22 '18 at 3:31
SparebrainSparebrain
106110
add a comment |
You may need to check for successful connection/query first:
if($result=mysqli_query($con,$s)) {
if(mysqli_num_rows($result)==1) {
... user already taken code ...
} else {
... insert new user code ...
}
}
See w3schools ref: https://www.w3schools.com/php/func_mysqli_num_rows.asp
This may also help you debug where the problem is occurring.
answered Nov 22 '18 at 3:31
SparebrainSparebrain
106110
You may need to check for successful connection/query first:
if($result=mysqli_query($con,$s)) {
if(mysqli_num_rows($result)==1) {
... user already taken code ...
} else {
... insert new user code ...
}
}
See w3schools ref: https://www.w3schools.com/php/func_mysqli_num_rows.asp
This may also help you debug where the problem is occurring.
answered Nov 22 '18 at 3:31
SparebrainSparebrain
106110
answered Nov 22 '18 at 3:31
SparebrainSparebrain
106110
answered Nov 22 '18 at 3:31
SparebrainSparebrain
106110
answered Nov 22 '18 at 3:31
SparebrainSparebrain
106110
106110
add a comment |
add a comment |
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1
In case you prevented the page from showing more alerts, try appending your message after your
scripttag and see if the message shows up. Also, your code is vulnerable to SQL injections– Thum Choon Tat
Nov 22 '18 at 3:37