mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in C:xampphtdocsatmmanagerindex.php...
0
Im using the following to login it works perfectly but in mamager login its giving following error
mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in C:xampphtdocsatmmanagerindex.php on line 87
<?php
if(isset($_POST['login'])){
$user = mysqli_real_escape_string($con, $_POST['email']);
$pass = mysqli_real_escape_string($con, $_POST['pass']);
$query ="select * from manager where user='$user' and pass='$pass' ";
$res = mysqli_query($con,$query);
if(mysqli_num_rows($res)==1) <--line 87 in my code
{
$query_id ="select * from manager where user='$user'";
$run_id = mysqli_query($con,$query_id);
$run_id2 = mysqli_fetch_array($run_id);
$_SESSION['id'] = $run_id2['id'];
$_SESSION['name'] = $run_id2['name'];
echo "<script>window.open('home.php','_self');</script>";
}
else
{
echo "<script>alert('** Please Enter Correct Information **')</script>";
}
}
?>
php mysqli
asked Nov 21 '18 at 7:00
Mohammed UsmanMohammed Usman
105
add a comment |
0
Im using the following to login it works perfectly but in mamager login its giving following error
mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in C:xampphtdocsatmmanagerindex.php on line 87
<?php
if(isset($_POST['login'])){
$user = mysqli_real_escape_string($con, $_POST['email']);
$pass = mysqli_real_escape_string($con, $_POST['pass']);
$query ="select * from manager where user='$user' and pass='$pass' ";
$res = mysqli_query($con,$query);
if(mysqli_num_rows($res)==1) <--line 87 in my code
{
$query_id ="select * from manager where user='$user'";
$run_id = mysqli_query($con,$query_id);
$run_id2 = mysqli_fetch_array($run_id);
$_SESSION['id'] = $run_id2['id'];
$_SESSION['name'] = $run_id2['name'];
echo "<script>window.open('home.php','_self');</script>";
}
else
{
echo "<script>alert('** Please Enter Correct Information **')</script>";
}
}
?>
php mysqli
asked Nov 21 '18 at 7:00
Mohammed UsmanMohammed Usman
105
clearly$res = mysqli_query($con,$query);is returning a booleanFALSEto$res. the query is failing
– marvinIsSacul
Nov 21 '18 at 7:13
Possible duplicate of mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc... expects parameter 1 to be resource
– Nico Haase
Nov 21 '18 at 7:34
add a comment |
0
0
0
Im using the following to login it works perfectly but in mamager login its giving following error
mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in C:xampphtdocsatmmanagerindex.php on line 87
<?php
if(isset($_POST['login'])){
$user = mysqli_real_escape_string($con, $_POST['email']);
$pass = mysqli_real_escape_string($con, $_POST['pass']);
$query ="select * from manager where user='$user' and pass='$pass' ";
$res = mysqli_query($con,$query);
if(mysqli_num_rows($res)==1) <--line 87 in my code
{
$query_id ="select * from manager where user='$user'";
$run_id = mysqli_query($con,$query_id);
$run_id2 = mysqli_fetch_array($run_id);
$_SESSION['id'] = $run_id2['id'];
$_SESSION['name'] = $run_id2['name'];
echo "<script>window.open('home.php','_self');</script>";
}
else
{
echo "<script>alert('** Please Enter Correct Information **')</script>";
}
}
?>
php mysqli
asked Nov 21 '18 at 7:00
Mohammed UsmanMohammed Usman
105
Im using the following to login it works perfectly but in mamager login its giving following error
mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in C:xampphtdocsatmmanagerindex.php on line 87
<?php
if(isset($_POST['login'])){
$user = mysqli_real_escape_string($con, $_POST['email']);
$pass = mysqli_real_escape_string($con, $_POST['pass']);
$query ="select * from manager where user='$user' and pass='$pass' ";
$res = mysqli_query($con,$query);
if(mysqli_num_rows($res)==1) <--line 87 in my code
{
$query_id ="select * from manager where user='$user'";
$run_id = mysqli_query($con,$query_id);
$run_id2 = mysqli_fetch_array($run_id);
$_SESSION['id'] = $run_id2['id'];
$_SESSION['name'] = $run_id2['name'];
echo "<script>window.open('home.php','_self');</script>";
}
else
{
echo "<script>alert('** Please Enter Correct Information **')</script>";
}
}
?>
php mysqli
php mysqli
asked Nov 21 '18 at 7:00
Mohammed UsmanMohammed Usman
105
asked Nov 21 '18 at 7:00
Mohammed UsmanMohammed Usman
105
asked Nov 21 '18 at 7:00
Mohammed UsmanMohammed Usman
105
asked Nov 21 '18 at 7:00
Mohammed UsmanMohammed Usman
105
asked Nov 21 '18 at 7:00
Mohammed UsmanMohammed Usman
105
105
clearly$res = mysqli_query($con,$query);is returning a booleanFALSEto$res. the query is failing
– marvinIsSacul
Nov 21 '18 at 7:13
Possible duplicate of mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc... expects parameter 1 to be resource
– Nico Haase
Nov 21 '18 at 7:34
add a comment |
clearly$res = mysqli_query($con,$query);is returning a booleanFALSEto$res. the query is failing
– marvinIsSacul
Nov 21 '18 at 7:13
Possible duplicate of mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc... expects parameter 1 to be resource
– Nico Haase
Nov 21 '18 at 7:34
clearly
$res = mysqli_query($con,$query); is returning a boolean FALSE to $res. the query is failing– marvinIsSacul
Nov 21 '18 at 7:13
clearly
$res = mysqli_query($con,$query); is returning a boolean FALSE to $res. the query is failing– marvinIsSacul
Nov 21 '18 at 7:13
Possible duplicate of mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc... expects parameter 1 to be resource
– Nico Haase
Nov 21 '18 at 7:34
Possible duplicate of mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc... expects parameter 1 to be resource
– Nico Haase
Nov 21 '18 at 7:34
add a comment |
1 Answer
1
active
oldest
votes
0
Your code lacks proper error handling - which could help you when debugging. Try doing this...
<?php
if(isset($_POST['login'])){
$user = mysqli_real_escape_string($con, $_POST['email']);
$pass = mysqli_real_escape_string($con, $_POST['pass']);
$query ="select * from manager where user='$user' and pass='$pass' ";
$res = mysqli_query($con,$query);
# $res will be FALSE on query error
if ($res === FALSE){
// you can also check the error message you are getting from mysql
// var_dump(mysqli_error($con));
// you could do a var_dump of $query here to see what exactly you pass as a query to the database.
// var_dump($query);
echo "<script>alert('** Query failed executing **')</script>";
}
elseif(mysqli_num_rows($res)==1) <--line 87 in my code
{
$query_id ="select * from manager where user='$user'";
$run_id = mysqli_query($con,$query_id);
$run_id2 = mysqli_fetch_array($run_id);
$_SESSION['id'] = $run_id2['id'];
$_SESSION['name'] = $run_id2['name'];
echo "<script>window.open('home.php','_self');</script>";
}
else
{
echo "<script>alert('** Please Enter Correct Information **')</script>";
}
}
?>
answered Nov 21 '18 at 7:23
marvinIsSaculmarvinIsSacul
53717
So, if you start debugging, why not check for any error returned by the server?
– Nico Haase
Nov 21 '18 at 7:33
thanks @NicoHaase I had forgotten to add it. I did now
– marvinIsSacul
Nov 21 '18 at 7:44
add a comment |
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1 Answer
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
Your code lacks proper error handling - which could help you when debugging. Try doing this...
<?php
if(isset($_POST['login'])){
$user = mysqli_real_escape_string($con, $_POST['email']);
$pass = mysqli_real_escape_string($con, $_POST['pass']);
$query ="select * from manager where user='$user' and pass='$pass' ";
$res = mysqli_query($con,$query);
# $res will be FALSE on query error
if ($res === FALSE){
// you can also check the error message you are getting from mysql
// var_dump(mysqli_error($con));
// you could do a var_dump of $query here to see what exactly you pass as a query to the database.
// var_dump($query);
echo "<script>alert('** Query failed executing **')</script>";
}
elseif(mysqli_num_rows($res)==1) <--line 87 in my code
{
$query_id ="select * from manager where user='$user'";
$run_id = mysqli_query($con,$query_id);
$run_id2 = mysqli_fetch_array($run_id);
$_SESSION['id'] = $run_id2['id'];
$_SESSION['name'] = $run_id2['name'];
echo "<script>window.open('home.php','_self');</script>";
}
else
{
echo "<script>alert('** Please Enter Correct Information **')</script>";
}
}
?>
answered Nov 21 '18 at 7:23
marvinIsSaculmarvinIsSacul
53717
So, if you start debugging, why not check for any error returned by the server?
– Nico Haase
Nov 21 '18 at 7:33
thanks @NicoHaase I had forgotten to add it. I did now
– marvinIsSacul
Nov 21 '18 at 7:44
add a comment |
0
Your code lacks proper error handling - which could help you when debugging. Try doing this...
<?php
if(isset($_POST['login'])){
$user = mysqli_real_escape_string($con, $_POST['email']);
$pass = mysqli_real_escape_string($con, $_POST['pass']);
$query ="select * from manager where user='$user' and pass='$pass' ";
$res = mysqli_query($con,$query);
# $res will be FALSE on query error
if ($res === FALSE){
// you can also check the error message you are getting from mysql
// var_dump(mysqli_error($con));
// you could do a var_dump of $query here to see what exactly you pass as a query to the database.
// var_dump($query);
echo "<script>alert('** Query failed executing **')</script>";
}
elseif(mysqli_num_rows($res)==1) <--line 87 in my code
{
$query_id ="select * from manager where user='$user'";
$run_id = mysqli_query($con,$query_id);
$run_id2 = mysqli_fetch_array($run_id);
$_SESSION['id'] = $run_id2['id'];
$_SESSION['name'] = $run_id2['name'];
echo "<script>window.open('home.php','_self');</script>";
}
else
{
echo "<script>alert('** Please Enter Correct Information **')</script>";
}
}
?>
answered Nov 21 '18 at 7:23
marvinIsSaculmarvinIsSacul
53717
So, if you start debugging, why not check for any error returned by the server?
– Nico Haase
Nov 21 '18 at 7:33
thanks @NicoHaase I had forgotten to add it. I did now
– marvinIsSacul
Nov 21 '18 at 7:44
add a comment |
0
0
0
Your code lacks proper error handling - which could help you when debugging. Try doing this...
<?php
if(isset($_POST['login'])){
$user = mysqli_real_escape_string($con, $_POST['email']);
$pass = mysqli_real_escape_string($con, $_POST['pass']);
$query ="select * from manager where user='$user' and pass='$pass' ";
$res = mysqli_query($con,$query);
# $res will be FALSE on query error
if ($res === FALSE){
// you can also check the error message you are getting from mysql
// var_dump(mysqli_error($con));
// you could do a var_dump of $query here to see what exactly you pass as a query to the database.
// var_dump($query);
echo "<script>alert('** Query failed executing **')</script>";
}
elseif(mysqli_num_rows($res)==1) <--line 87 in my code
{
$query_id ="select * from manager where user='$user'";
$run_id = mysqli_query($con,$query_id);
$run_id2 = mysqli_fetch_array($run_id);
$_SESSION['id'] = $run_id2['id'];
$_SESSION['name'] = $run_id2['name'];
echo "<script>window.open('home.php','_self');</script>";
}
else
{
echo "<script>alert('** Please Enter Correct Information **')</script>";
}
}
?>
answered Nov 21 '18 at 7:23
marvinIsSaculmarvinIsSacul
53717
Your code lacks proper error handling - which could help you when debugging. Try doing this...
<?php
if(isset($_POST['login'])){
$user = mysqli_real_escape_string($con, $_POST['email']);
$pass = mysqli_real_escape_string($con, $_POST['pass']);
$query ="select * from manager where user='$user' and pass='$pass' ";
$res = mysqli_query($con,$query);
# $res will be FALSE on query error
if ($res === FALSE){
// you can also check the error message you are getting from mysql
// var_dump(mysqli_error($con));
// you could do a var_dump of $query here to see what exactly you pass as a query to the database.
// var_dump($query);
echo "<script>alert('** Query failed executing **')</script>";
}
elseif(mysqli_num_rows($res)==1) <--line 87 in my code
{
$query_id ="select * from manager where user='$user'";
$run_id = mysqli_query($con,$query_id);
$run_id2 = mysqli_fetch_array($run_id);
$_SESSION['id'] = $run_id2['id'];
$_SESSION['name'] = $run_id2['name'];
echo "<script>window.open('home.php','_self');</script>";
}
else
{
echo "<script>alert('** Please Enter Correct Information **')</script>";
}
}
?>
answered Nov 21 '18 at 7:23
marvinIsSaculmarvinIsSacul
53717
edited Nov 21 '18 at 7:44
answered Nov 21 '18 at 7:23
marvinIsSaculmarvinIsSacul
53717
answered Nov 21 '18 at 7:23
marvinIsSaculmarvinIsSacul
53717
answered Nov 21 '18 at 7:23
marvinIsSaculmarvinIsSacul
53717
53717
So, if you start debugging, why not check for any error returned by the server?
– Nico Haase
Nov 21 '18 at 7:33
thanks @NicoHaase I had forgotten to add it. I did now
– marvinIsSacul
Nov 21 '18 at 7:44
add a comment |
So, if you start debugging, why not check for any error returned by the server?
– Nico Haase
Nov 21 '18 at 7:33
thanks @NicoHaase I had forgotten to add it. I did now
– marvinIsSacul
Nov 21 '18 at 7:44
So, if you start debugging, why not check for any error returned by the server?
– Nico Haase
Nov 21 '18 at 7:33
So, if you start debugging, why not check for any error returned by the server?
– Nico Haase
Nov 21 '18 at 7:33
thanks @NicoHaase I had forgotten to add it. I did now
– marvinIsSacul
Nov 21 '18 at 7:44
thanks @NicoHaase I had forgotten to add it. I did now
– marvinIsSacul
Nov 21 '18 at 7:44
add a comment |
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clearly
$res = mysqli_query($con,$query);is returning a booleanFALSEto$res. the query is failing– marvinIsSacul
Nov 21 '18 at 7:13
Possible duplicate of mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc... expects parameter 1 to be resource
– Nico Haase
Nov 21 '18 at 7:34