$query = "SELECT username,userid FROM user WHERE username = 'admin' ";
$result=$conn->store_result();
$result1=$conn->store_result();
Can I store same result in two diffrent variable? $result and $result1?
Use this
$query = "SELECT username,userid FROM user WHERE username = 'admin' ";
$result=$conn->store_result();
$result1=$result;
Related
I would like to ask how to select all from table based on username? I mean for example my user 1 insert his/her data and send to localhost and in status page will display their own data only. Below is my status page php.
<?php
require_once('dbConnect.php');
//Creating sql query
$sql = "SELECT * FROM Table";
//getting result
$r = mysqli_query($con,$sql);
//creating a blank array
$result = array();
//looping through all the records fetched
while($row = mysqli_fetch_array($r)){
//Pushing name and id in the blank array created
array_push($result,array(
"id"=>$row['id'],
"username"=>$row['username'],
"name"=>$row['name']
));
}
//Displaying the array in json format
echo json_encode(array('result'=>$result));
mysqli_close($con);
?>
"SELECT * FROM Table WHERE username = 'john'"
This is assuming your table is actually named 'Table' as you have it above. Otherwise replace it with the actual table name.
If you have a variable for user name do this:
"SELECT * FROM Table WHERE username = '" . $username . "'";
You can use WHERE clause in the query like.
"SELECT * FROM Table WHERE username = 'lun L'"
But this is not a good way to fetch data by user name. You should select users on ID base and ID should be unique. Because there are so many users who might have the same name.
"SELECT * FROM Table WHERE youUniqueColName = '$uniqueVal'"
If unique column is id then use id='$id'
SELECT * FROM tablename WHERE username = 'username'
Use this if you want a fixed SQL Query in your variable, or use;
SELECT * FROM tablename WHERE username = '$username'
if you have an input that asks a username to display a specific username..
Basically, I have been having some trouble with sending a request to a MySQL server and receiving the data back and checking if a user is an Admin or just a User.
Admin = 1
User = 0
<?php
$checkAdminQuery = "SELECT * FROM `users` WHERE `admin`";
$checkAdmin = $checkAdminQuery
mysqli_query = $checkAdmin;
if ($checkAdmin == 1) {
echo '<h1>Working!</h1>';
}else {
echo '<h1>Not working!</h1>';
}
?>
Sorry that this may not be as much info needed, I am currently new to Stack Overflow.
Firstly, your SQL query is wrong
SELECT * FROM `users` WHERE `admin`
It's missing the rest of the WHERE clause
SELECT * FROM `users` WHERE `admin` = 1
Then you're going to need fetch the result from the query results. You're not even running the query
$resultSet = mysqli_query($checkAdminQuery)
Then from there, you'll want to extract the value.
while($row = mysqli_fetch_assoc($resultSet))
{
//do stuff
}
These are the initial problems I see, I'll continue to analyze and find more if needed.
See the documentation here
http://php.net/manual/en/book.mysqli.php
You need to have something like user id if you want to check someone in database. For example if you have user id stored in session
<?php
// 1. start session
session_start();
// 2. connect to db
$link = mysqli_connect('host', 'user', 'pass', 'database');
// 3. get user
$checkAdminQuery = mysqli_query($link, "SELECT * FROM `users` WHERE `id_user` = " . $_SESSION['id_user'] );
// 4. fetch from result
$result = mysqli_fetch_assoc($checkAdminQuery);
// 5. if column in database is called admin test it like this
if ($result['admin'] == 1) {
echo '<h1>Is admin!</h1>';
}else {
echo '<h1>Not working!</h1>';
}
?>
// get all admin users (assumes database already connected)
$rtn = array();
$checkAdminQuery = "SELECT * FROM `users` WHERE `admin`=1";
$result = mysqli_query($dbcon,$checkAdminQuery) or die(mysqli_error($dbconn));
while($row = mysqli_fetch_array($result)){
$rtn[] = $row;
}
$checkAdminQuery = "SELECT * FROM `users` WHERE `admin`"; !!!!
where what ? you need to specify where job = 'admin' or where name ='admin'
you need to specify the column name where you are adding the admin string
I don't understand this because I'm just getting into query's and php.
I'm trying to get the user's ID from the database and set that equal to a different users friendreq column.
Don't worry about me not escaping properly, this is only a test so I can practice! Thank you! (Although I'm not sure what escaping is, I'm going to do my research!)
$usernameID = "SELECT Id FROM Users WHERE Username = '$username'";
$sql = "UPDATE Users SET FriendReq = $usernameID WHERE Username = '$usernamebeingreq'";
$result = mysqli_multi_query($con, $usernameID, $sql);
if(!$result)
{
echo 'Failed';
}
else
{
echo 'Friend added!';
}
According to the PHP reference of mysqli_multi_query your two queries need to be concatenated with a semicolon. You're passing each query as its own parameter.
Use the following instead:
$result = mysqli_multi_query($con, $usernameID . "; " . $sql);
This will concatenate your two queries, so that it's the following:
SELECT Id FROM Users WHERE Username = '$username'; UPDATE Users SET FriendReq = $usernameID WHERE Username = '$usernamebeingreq'
<?php
session_start();
$username = "root";
$password = "password";
$database = "meipolytechnic";
mysql_connect('localhost', $username,$password);
#mysql_select_db($database) or die(mysql_error());
$username=$_SESSION['MM_Username'];
$query = "SELECT rollno FROM users where username = '".$username."'";
$result = mysql_query($query) or die(mysql_error());
$num = mysql_num_rows($result);
mysql_close();
$rows = array();
while($r = mysql_fetch_row($result))
{
$rows[] = $r[0];
}
echo ($rows['rollno']);
?>
i want to retrieve only the logged in users roll no from users table in database
when i run this code
and log in as foo
i get the following stuff
Unknown column 'foo' in 'where clause'
There should be session_start() at the top of the page
query need to change as
$query = "SELECT rollno FROM users where username = '".$_SESSION['MM_Username']."' ";
EDIT
Please try something before posting a question here. Please google or go through www.w3school.com for clearing this kind of issues. Make a good knowledge about arrays and mysql connection. And mysql_query function won't work latest PHP version.
Please try following code.
$result = mysql_query($query) or die(mysql_error());
$rows = array();
while($r = mysql_fetch_row($result))
{
$rows[] = $r[0];
}
print_r($rows);
To use an array inside a string you need to put a curly bracket before it and after it
so
$query = "SELECT rollno FROM users where username = {$_SESSION['MM_Username']}";
or
$query = "SELECT rollno FROM users where username = ".$_SESSION['MM_Username'];
First of all start session using start_session()
then change your query:
$query = "SELECT rollno FROM users where username = ".$_SESSION['MM_Username'];
then change:
$num = mysql_num_rows($result); instead of $num = mysql_numrows($result);
Try this query
$query = "SELECT rollno FROM users where username = ".$_SESSION[MM_Username]." ";
And start session on same page.
You can use
$username=$_SESSION[MM_Username];
$query = "SELECT rollno FROM users where username = '".$username."'";
and start the session by using session_start()
and you have used two closing tags omit one make it like below
echo ($rows['rollno']);
?>
This type of error occur when query goes false
May be becouse You have not start session, becouse if you dont have session_start(), then nothing will come in session variable... just try as
$query = "SELECT rollno FROM users where username = '".$_SESSION[MM_Username]."'";
may this help you
You must need to use session_start() before using $_SESSION variable in code.
So put below code at start,
session_start();
Then do some modification in query like,
$query = "SELECT rollno FROM users where username = '".$_SESSION['MM_Username']."'";
first start your session
session_start();
and Change your query like this...
$query = "SELECT rollno FROM users where username = '".$_SESSION['MM_Username']."'";
For the query below, I would like to set a variable called $totalcount equal to the count of all loginids in table called login. How can I do this?
Thanks in advance,
John
$sqlStrcount = "SELECT loginid FROM login";
$sqlQueryStr = "SELECT loginid FROM login";
$sqlQuery = mysql_query($sqlQueryStr);
$totalCount = mysql_num_rows($sqlQuery);
If you only need to count your records in login use this instead
for performance reasons.
$sqlQueryStr = "SELECT COUNT(loginid) as totalCount FROM login";
$sqlQuery = mysql_query($sqlQueryStr);
$row = mysql_fetch_assoc($sqlQuery);
$totalCount = $row['totalCount'];