I have this code:
<?php
$user = $_COOKIE["user"];
$password = $_COOKIE["password"];
$localhost = "localhost";
$userdb = "xxxxx";
$passworddb = "xxxxx";
$database = "xxxxx";
$conn = mysqli_connect($localhost, $userdb, $passworddb, $database);
$vyber = "SELECT PASSWORD FROM Login WHERE User=".$user;
$result = mysqli_query($conn, $vyber);
echo $result;
?>
Cookie are set and if I use $vyber in database so everything is good. But there PHP write nothing. Can anybody tell, what I doing wrong? (Without comand $vyber every thing running perfect)
instead of,
echo $result
try to do that :
while ($row = mysqli_fetch_row($result)){
echo $row[0];
}
It is query error change query to :
$vyber = "SELECT PASSWORD FROM Login WHERE User='$user'";
if did not work use die function to dispaly error message :
mysqli_query($conn, $vyber) or die(mysqli_error($conn));
To fetch records :
while ($row = mysqli_fetch_array($result)){
echo $row[0];
}
Related
I'am setting up a new wampserver and created a new database with a table named 'users'
apache 2.4.23 & PHP 7.1.10 & mySQL 5.7.14
<?php
$server = 'localhost';
$serverUsername = 'root';
$serverPassword = '';
$dbName = 'test';
$connection = mysqli_connect($server,$serverUsername,$serverPassword);
msqli_select_db($dbName);
if(!$connection){
echo 'connection failed to database '.mysqli_connect_error();
}
$sql = "SELECT * FROM users";
$query = mysqli_query($sql);
while($row = mysqli_fetch_array($query)){
print_r($row);
}
?>
my value is really in th data base but nothing appears in the page after running code
Have a look at the comments mentioning the fix
$server = 'localhost';
$serverUsername = 'root';
$serverPassword = '';
$dbName = 'test';
// FIX 1
// You need to mention the database name as the last argument
$connection = mysqli_connect($server,$serverUsername,$serverPassword, $dbName);
if(!$connection){
echo 'connection failed to database '.mysqli_connect_error();
}
$sql = "SELECT * FROM users";
// FIX 2
// The first argument should be your mysqli connection
$query = mysqli_query($connection, $sql);
// Check for errors
if (!$query) {
printf("Error: %s\n", mysqli_error($connection));
exit();
}
while($row = mysqli_fetch_array($query)){
print_r($row);
}
?>
Reference for mysqli_connect: https://secure.php.net/manual/en/function.mysqli-connect.php
Reference for mysqli_query: https://secure.php.net/manual/en/mysqli.query.php
I'm a beginner in learning how to set up database & PHP script and follow example
to do that, then when I run login.php script I can't retrieve data from the database ,
I really feel that is a very simple question for others but I tried to solve it But didn't succeed, so can someone take look on my code then Corrects it?
here is my php script :
init.php :
<?php
$db_name = "webapp";
$mysql_username = "root";
$mysql_password = "";
$server_name = "localhost";
$con=mysqli_connect($server_name, $mysql_username, $mysql_password, $db_name);
if (!$con) {
echo "Connection Error ......." . mysqli_connect_error();
} else {
echo "<h3>Database connection Success .....</h3>";
}
?>
login.php :
<?php
require "init.php";
$user_name = "YASER";
$user_phone = "123456";
$sql_query = "select name from user_info where user_name like'$user_name'and
user_phone like'$user_phone';";
$result = mysqli_query($con,$sql_query);
if (mysqli_num_rows($result)>0)
{
$row = mysqli_fetch_assoc($result);
$name = $row["name"];
echo "<h3> Hello And Wellcome" . $name . "</h3>";
} else {
echo " No Info Is Available .......";
}
?>
1st : First check that query is executing or failing
if(!$result){ echo mysqli_error($con); }
2nd : use = instead of like
$sql_query = "select name from user_info
where user_name='$user_name' and
user_phone='$user_phone'";
3rd : You need to give proper spacing In query
like'$user_name'and
^^^ ^^^
To
like '$user_name' and
You have error in your query.
try this to find error
$result = mysqli_query($con,$sql_query) or die(mysqli_error($con));
Your query should be like as follow...
'SELECT name FROM user_info WHERE user_name LIKE "'.$user_name.'" AND user_phone LIKE "'.$user_phone.'"';
I am connected to the database, a page has lots of content so I'll only share the part that doesnt return a value in php, but it returns a value in MySQL
Here is the code;
$query = "SELECT firstname FROM users WHERE id = '17'";
$query_run = mysql_query($query);
$row = mysql_fetch_row($query_run);
echo $row[0];
Changing the code I shared first to this, solved the problem. Thanks to anyone who tried to help.
$query = "SELECT firstname FROM users WHERE id = '17'";
$query_run = mysqli_query($conn, $query);
$row = mysqli_fetch_row($query_run);
echo $row[0];
And made bit changes to the connect.inc.php which I also shared in comment.
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "notsitesi";
$conn = new mysqli($servername, $username, $password, $dbname);
if($conn->connect_error) {
die("Connection failed");
}
?>
I believe my PHP to be functioning perfectly, therefore I think it's a query error. When I proceed, with form details stored in the session... it happily returns my Posted information but doesn't seem to be pulling anything from my database - there is a row in my database containing the email address I am using. Does anybody see anything blatantly wrong with this PHP?
Thanks for your help.
<?php
session_start();
$servername = "localhost";
$username = "privatedbroot";
$password = "not4ulol";
$dbname = "pdb_inventory";
$status = $_GET["action"];
$_SESSION["Cemail"] = $_POST["CEMAIL"];
$_SESSION["Access"] = md5($_POST["ACCESS"]);
$conn = new mysqli($servername, $username, $password, $dbname);
$sql = "SELECT CEMAIL, ACCESS FROM POPU WHERE `CEMAIL`= ".$_SESSION['Cemail'];
echo $sql;
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
if ($_SESSION["Access"] == $row["ACCESS"]){
echo "password correct!";
} else {
echo "password wrong!";
}
}
}else{
echo "ur email is wrong m8.";
}
?>
Try this:
$cemail = $_SESSION['Cemail'];
$sql = "SELECT CEMAIL, ACCESS FROM POPU WHERE `CEMAIL`= '$cemail'";
Here is what I'm trying to do: When user adds a contact to his list, the number of this contact gets run by with the numbers in the database and it gives feedback if the user is already in the database or not. Right now I always get back "User is in database" even though he isn't. Then again I'm not that well acquainted with php. I changed the code a bit again, now it doesn't work at all, because it doesn't like the part
$number = ($_GET["number"] from $DB_Table);
Full code
<?php
$DB_HostName = "localhost";
$DB_Name = "db";
$DB_User = "user";
$DB_Pass = "pw";
$DB_Table = "contacts";
$number = ($_GET["number"] from $DB_Table);
$fnumber = ($_GET["fnumber"]);
if ($number == $fnumber) {
echo "This user is already in database";
} else {
echo "This user isn't in the database";
}
$con = mysql_connect($DB_HostName,$DB_User,$DB_Pass) or die (mysql_error());
mysql_select_db($DB_Name,$con) or die(mysql_error());
mysql_close($con);
?>
I don't actually see you executing the database query. You could do something like this:
<?php
$DB_HostName = "localhost";
$DB_Name = "db";
$DB_User = "user";
$DB_Pass = "pw";
$DB_Table = "contacts";
$con = mysql_connect($DB_HostName,$DB_User,$DB_Pass) or die (mysql_error());
mysql_select_db($DB_Name,$con) or die(mysql_error());
$fnumber = mysql_real_escape_string($_GET["fnumber"]);
$result = mysql_query("SELECT * FROM $DB_Table WHERE Something = '$fnumber'", $con);
if ($result) {
// Check the number of rows in the result set
if (mysql_num_rows($result) > 0) {
echo "This user is already in database";
}
else echo "This user isn't in the database";
}
mysql_close($con);
?>
This is not valid PHP code: $number = ($_GET["number"] from $DB_Table);
$_GET["number"] represents the value of the "number" parameter that you find in the url of your page.
Example: http://example.com/index.php?number=7 so $_GET["number"] is 7.
In your code, $DB_Table is a just a string ("contact") and "from" does not fit there using php syntax.
mysql_select_db($DB_Name,$con) or die(mysql_error());
is valid PHP but you are not doing anything with what you get from the database. I suggest you at least take a look at this tutorial php mysql select