Help! I'm trying to get a login script to work.
I get this error message
MySQL Login Error: You have an error in your SQL syntax near
''jvsd0001_cust omers` WHERE cust_name='test user' AND
cust_pass='test 123'' at line 1
I'm using a database called test
here is mysql table:
mysql> select * from jvsd0001_custom ers;
+---------+-----------+---------------+
| cust_id | cust_name | cust_password |
+---------+-----------+---------------+
| 1 | testuser | test123 |
+---------+-----------+---------------+
1 row in set (0.00 sec)
[jvsd0001@hal] pico login.php
UW PICO(tm) 4.2 File:
login.php Modified
<?php
$username = $_POST['user'];
$password = $_POST['pass'];
if (!$_POST['pass'] && !$_POST['user']) {
?>
<html><b>Memb er Login</b>
<br><form method="POST">U sername:
<br><input type="text" name="user" value="">
<br>Password:
<br><input type="text" name="pass" value="">
<br><input type="submit" name="submit" value="Login">
<?php
} else {
mysql_connect ("localhost" , "abdullah") or die ('My SQL Error: ' .
mysql_error());
mysql_select_db ("test");
$stuff = mysql_query("SE LECT * FROM 'jvsd0001_custo mers` WHERE
username='".$cu st_name."' AND password='".$cu st_pass."'") or
die("MySQL
Login Error: ".mysql_error() );
if (mysql_num_rows ($stuff) > 0) {
echo("Logged in");
} else {
echo("Login Incorrect. Please Try Again!");
}
}
?>
What's wrong???
I can't get this script to work either.
<?php
if(!isset($HTTP _POST_VARS['cust_name'])&&!isset($HTTP _POST_VARS['cust_pass']))
{
//Visitor needs to enter a name and password
?>
<h1>Please Log In</h1>
This page is secret.
<form method="post" action="secretd b.php">
<table border="1">
<tr>
<th> Username </th>
<td> <input type="text" name="cust_name "> </td>
</tr>
<tr>
<th> Password </th>
<td> <input type="password" name="cust_pass "> </td>
</tr>
<tr>
<td colspan="2" align="center">
<input type="submit" value="Log In">
</td>
</tr>
</table>
</form>
<?php
}
else
{
// connect to mysql
$mysql = mysql_connect( 'localhost', 'abdullah');
if(!$mysql)
{
echo 'Cannot connect to database.';
exit;
}
// select the appropriate database
$mysql = mysql_select_db ( 'test' );
if(!$mysql)
{
echo 'Cannot select database.';
exit;
}
// query the database to see if there is a record which matches
$query = "select count(*) from jvsd0001_custom ers where
cust_name = '$cust_name' and
cust_pass = '$cust_pass'";
$result = mysql_query( $query );
if(!$result)
{
echo 'Cannot run query.';
exit;
}
$count = mysql_result( $result, 0, 0 );
if ( $count > 0 )
{
// visitor's name and password combination are correct
echo '<h1>Here it is!</h1>';
echo 'I bet you are glad you can see this secret page.';
}
else
{
// visitor's name and password combination are not correct
echo '<h1>Go Away!</h1>';
echo 'You are not authorized to view this resource.';
}
}
?>
Can abody tell me what I'm doing wrong here, please!
I get this error message
MySQL Login Error: You have an error in your SQL syntax near
''jvsd0001_cust omers` WHERE cust_name='test user' AND
cust_pass='test 123'' at line 1
I'm using a database called test
here is mysql table:
mysql> select * from jvsd0001_custom ers;
+---------+-----------+---------------+
| cust_id | cust_name | cust_password |
+---------+-----------+---------------+
| 1 | testuser | test123 |
+---------+-----------+---------------+
1 row in set (0.00 sec)
[jvsd0001@hal] pico login.php
UW PICO(tm) 4.2 File:
login.php Modified
<?php
$username = $_POST['user'];
$password = $_POST['pass'];
if (!$_POST['pass'] && !$_POST['user']) {
?>
<html><b>Memb er Login</b>
<br><form method="POST">U sername:
<br><input type="text" name="user" value="">
<br>Password:
<br><input type="text" name="pass" value="">
<br><input type="submit" name="submit" value="Login">
<?php
} else {
mysql_connect ("localhost" , "abdullah") or die ('My SQL Error: ' .
mysql_error());
mysql_select_db ("test");
$stuff = mysql_query("SE LECT * FROM 'jvsd0001_custo mers` WHERE
username='".$cu st_name."' AND password='".$cu st_pass."'") or
die("MySQL
Login Error: ".mysql_error() );
if (mysql_num_rows ($stuff) > 0) {
echo("Logged in");
} else {
echo("Login Incorrect. Please Try Again!");
}
}
?>
What's wrong???
I can't get this script to work either.
<?php
if(!isset($HTTP _POST_VARS['cust_name'])&&!isset($HTTP _POST_VARS['cust_pass']))
{
//Visitor needs to enter a name and password
?>
<h1>Please Log In</h1>
This page is secret.
<form method="post" action="secretd b.php">
<table border="1">
<tr>
<th> Username </th>
<td> <input type="text" name="cust_name "> </td>
</tr>
<tr>
<th> Password </th>
<td> <input type="password" name="cust_pass "> </td>
</tr>
<tr>
<td colspan="2" align="center">
<input type="submit" value="Log In">
</td>
</tr>
</table>
</form>
<?php
}
else
{
// connect to mysql
$mysql = mysql_connect( 'localhost', 'abdullah');
if(!$mysql)
{
echo 'Cannot connect to database.';
exit;
}
// select the appropriate database
$mysql = mysql_select_db ( 'test' );
if(!$mysql)
{
echo 'Cannot select database.';
exit;
}
// query the database to see if there is a record which matches
$query = "select count(*) from jvsd0001_custom ers where
cust_name = '$cust_name' and
cust_pass = '$cust_pass'";
$result = mysql_query( $query );
if(!$result)
{
echo 'Cannot run query.';
exit;
}
$count = mysql_result( $result, 0, 0 );
if ( $count > 0 )
{
// visitor's name and password combination are correct
echo '<h1>Here it is!</h1>';
echo 'I bet you are glad you can see this secret page.';
}
else
{
// visitor's name and password combination are not correct
echo '<h1>Go Away!</h1>';
echo 'You are not authorized to view this resource.';
}
}
?>
Can abody tell me what I'm doing wrong here, please!
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