I have a domain where users have a subdomain with its own name.
Example:
User: Vitor
Subdomain: vitor.google.com
The database has the column user table my_users and have the columns birthday, address, password, city and state in table users_infos.
How do I do that in PHP in index.php can get this information direct user to the database?
this is my incorrect code:
// Get Subdomain
$urlExplode = explode('.', $_SERVER['HTTP_HOST']);
if (count($urlExplode) > 2 && $urlExplode[0] !== 'www') {
$subdomain = $urlExplode[0];
echo $subdomain;
}
// Select DB
$sql = "SELECT * FROM users_infos i INNER JOIN my_users u on u.id = i.id where u.users='$users'";
$result = mysql_query($sql);
if($result === FALSE) {
die(mysql_error());
// TODO: better error handling
}
else {
$row = mysql_fetch_array($result);
$userTitleSite = $row['userTitleSite'];
echo $id_textos = $row["id_textos"];
echo "<br />";
echo $id = $row["id"];
echo "<br />";
echo $phone = $row["phone"];
echo "<br />";
echo "<br />";
}
// Says that the subdomain is = user
$subdomain = $user;
Correct answer below:
Using the help you gave me, it looks like I got here ..
follows the correct code I could do for future users who have the same question:
// Get subdomain
$urlExplode = explode('.', $_SERVER['HTTP_HOST']);
if (count($urlExplode) > 2 && $urlExplode[0] !== 'www') {
$subdomain = $urlExplode[0];
echo $subdomain;
}
// Says that the subdomain is = user
$user = $subdomain;
// Select DB
$sql = "SELECT * FROM vms_textos i INNER JOIN vms_users u on u.id = i.id where u.user='$user'";
$result = mysql_query($sql);
if($result === FALSE) {
die(mysql_error());
// TODO: better error handling
}
else {
$row = mysql_fetch_array($result);
$userTitleSite = $row['userTitleSite'];
echo "<br />";
echo $id_textos = $row["id_textos"];
echo "<br />";
echo $user= $row["user"];
echo "<br />";
echo $id = $row["id"];
echo "<br />";
echo $telefone = $row["telefone"];
echo "<br />";
echo "<br />";
}
echo "<br />";
echo "<br />";
Thanks all for help me! :)
$subdomain = $user;
Should go before the $sql, as your code is written $user is empty.
Are you sure the WHERE clause u.id = i.id is correct? In other words, do the ID columns in your my_users and users_infos table match? It is more common to have a foreign key reference on the users_infos table (e.g. my_users_id), in which case the query would be something like:
SELECT *
FROM users_infos i
INNER JOIN my_users u on u.id = i.my_users_id
WHERE u.users='$users'
Related
I am having a huge issue looping through results, These two queries work hand in hand to check if a restaurant is open today. My problem is i have restaurants, id 1-5(more in the future). But the loop seems to only get restaurant id 5. I have read many posts on here and it seems like i am doing the right thing. But i cannot seem to loop to get the other restaurant id's.
I am blocked now, newbie who is very open to any suggestions or advise.
$sel = "SELECT Rest_Details.Resturant_ID,Delivery_Pcode.Pcode,Delivery_Pcode.Restaurant_ID
FROM Rest_Details INNER JOIN Delivery_Pcode
ON Delivery_Pcode.Restaurant_ID=Rest_Details.Resturant_ID
WHERE Delivery_Pcode.Pcode LIKE'$searchP'";
$res = $dbc->query($sel);
if (!$res) {
echo "invalid query '" . mysqli_error($dbc) . "\n";
}
$i=1;
while ($row_res = $res->fetch_array()) {
$rest_ = $row_res['Resturant_ID'];
$i++;
}
date_default_timezone_set("Europe/London");
$daynum = jddayofweek(unixtojd());
$query = "SELECT *
FROM Opening_hrs WHERE
Restaurant_ID = $rest_
AND Day_of_week = $daynum";
$run_qu = $dbc->query($query);
if ($run_qu->num_rows > 0) {
while ($row_qu = $run_qu->fetch_assoc()) {
$message = "open" . $row_qu["Open_time"] . "</br>";
}
} else {
$message = $message . "close" . $row_qu["Closing_time"] . "</br>";
}
You could either output whatever you want to within your loop or build-up an output string because the value of $rest_ will always be the last value in the loop and i don't think that's what you want... Again you are doing the same with $message. And I am willing to bet that this is what you want to do:
<?php
date_default_timezone_set("Europe/London");
$sel = "SELECT Rest_Details.Resturant_ID,Delivery_Pcode.Pcode,Delivery_Pcode.Restaurant_ID
FROM Rest_Details INNER JOIN Delivery_Pcode
ON Delivery_Pcode.Restaurant_ID=Rest_Details.Resturant_ID
WHERE Delivery_Pcode.Pcode LIKE'$searchP'";
$res = $dbc->query($sel);
if (!$res) {
echo "invalid query '" . mysqli_error($dbc) . "\n";
}
$i=1;
while ($row_res = $res->fetch_array()) {
$rest_ = $row_res['Resturant_ID'];
$i++; // <== YOU DON'T NEED THIS VARIABLE....
// GET THE DATES WITHIN THE LOOP...
$daynum = jddayofweek(unixtojd());
$query = "SELECT *
FROM Opening_hrs WHERE
Restaurant_ID = $rest_
AND Day_of_week = $daynum";
$run_qu = $dbc->query($query);
if ($run_qu->num_rows > 0) {
while ($row_qu = $run_qu->fetch_assoc()) {
$message = "open" . $row_qu["Open_time"] . "</br>";
}
} else {
$message = $message . "close" . $row_qu["Closing_time"] . "</br>";
}
}
I think this is what you are trying to do.
// $searchP should be checked to prevent SQL injection.
$sel = "SELECT Rest_Details.Resturant_ID, Delivery_Pcode.Pcode,
Delivery_Pcode.Restaurant_ID
FROM Rest_Details INNER JOIN Delivery_Pcode
ON Delivery_Pcode.Restaurant_ID = Rest_Details.Resturant_IDW
WHERE Delivery_Pcode.Pcode LIKE '$searchP'";
$res = $dbc->query($sel);
if (!$res) {
echo "invalid query '" . mysqli_error($dbc) . "\n";
}
// set these once as they don't change
date_default_timezone_set("Europe/London");
$daynum = jddayofweek(unixtojd());
// $i=1; - not required, never used
// loop over the original results
while ($row_res = $res->fetch_array()) {
$rest_ = $row_res['Resturant_ID'];
//$i++; not used
// check for a match
$query = "SELECT * FROM Opening_hrs
WHERE Restaurant_ID = $rest_
AND Day_of_week = $daynum";
$run_qu = $dbc->query($query);
if ($run_qu->num_rows > 0) {
// at least one match
while ($row_qu = $run_qu->fetch_assoc()) {
$message = "open" . $row_qu["Open_time"] . "<br />";
$message .= "close" . $row_qu["Closing_time"] . "<br />";
}
} else {
// no matches
$message = "No results for <i>$daynum</i>.";
}
}
It should be possible to get the details in a single query, but I would need to see your SQL tables for that (and you did not ask for that too :]).
Also, it is <br> or <br />, not </br>.
I have a number pages namely
change=1.php
change=2.php
change=3.php
They all have similar coding but leaving 1 or 2 variable values.
And I know its a very bad idea! How can I make a link work like below:
change.php?id=1
change.php?id=2
change.php?id=3
http://oi62.tinypic.com/708gfm.jpg
<?php
include 'connection.php';
session_start();
include 'details.php';
/*$pkmn_id = $_SESSION['pkmn_id'];
$poke = $_SESSION['path'];*/
$data = mysql_query(" SELECT * FROM user_pokemon_db WHERE team = 1 AND user_id = '".$id."' ");
while($rows = mysql_fetch_array($data))
{
$rep_id = $rows[0];
$pkmn_id = $rows['pkmn_id'];
$path = mysql_query(" SELECT * FROM pokemons WHERE pk_id = '".$pkmn_id."' ");
$poke = mysql_result($path, 0, "path");
echo $poke;
echo "<br />";
$level = $rows['level'];
echo $level;
echo "<br />";
$exp = $rows['exp'];
echo $exp;
echo "<br />";
echo "<br />";
}
$data = mysql_query(" SELECT * FROM user_pokemon_db WHERE user_id = '".$id."' AND team = 0");
while($rows = mysql_fetch_assoc($data))
{
$db_id = $rows['id'];
$array[] = $db_id;
$level = $rows['level'];
$array1[] = $level;
$exp = $rows['exp'];
$array2[] = $exp;
$pkmn_id = $rows['pkmn_id'];
$data1 = mysql_query(" SELECT * FROM pokemons WHERE pk_id = '".$pkmn_id."' ");
while($rows = mysql_fetch_assoc($data1))
{
$poke = $rows['path'];
$array3[] = $poke;
}
}
$team = 1;
$_SESSION['team'] = $team;
$_SESSION['rep_id'] = $rep_id;
?>
My PHP code.
You probably want to use GET variables, for which you need to combine all the files into one, named change.php. In this file you need the line $foo = $_GET["id"] which will get the value of the variable "id" in the url change.php?id=1.
if (isset($_GET["id"])) {
$foo = $_GET["id"];
//your code here
}else{
echo 'ERROR!!! No id in URL';
}
You can have several variables in the URL like this: change.php?id=1&a=bar&b=toofoo
You can get current script's file name and parse integer.
__FILE__
gives current script's name. Then,
$myStr = preg_replace('/\.php$/', '', __FILE__);
$result = preg_replace('/change=$/', '', $myStr);
echo $result; // it's your id
I have a user table that contain a lots more data, I wonder how can I improve my select code below
if ($result = $db->query("SELECT * FROM user ")) {
while ($row = mysqli_fetch_assoc($result)) {
echo $row["name"] . "<br />";
echo $row["user_id"] . "<br />";
echo $row["photo"] . "<br />";
//.. a lot more column here
}
}
I use this as a array: $user_info = mysql_fetch_assoc(mysql_query("SELECT * FROM `database` WHERE 1));
and to call it i use $user_info['column']
I have a code which fetches data from a mysql table and converts it into pdf document, the code is working fine except it is skipping row 1.
Here is the code from which i have removed the pdf generation process since the problem is in the loop which is fetching data.
Please help.
<?php
session_start();
if(isset($_SESSION['user']))
{
$cr = $_POST['cour'];
$s = $_POST['sem'];
require('fpdf.php');
include('../includes/connection.php');
$sql = "SELECT * FROM `student` WHERE AppliedCourse ='$cr'";
$rs = mysql_query($sql) or die($sql. "<br/>".mysql_error());
if(!mysql_fetch_array($rs))
{
$_SESSION['db_error'] = "<h2><font color = 'RED'>No such course found! Pease select again.</font></h2>";
header('Location: prinrepo.php');
}
else {
for($i = 0;$i <= $row = mysql_fetch_array($rs);$i++)
{
$formno[$i] = $row ['FormNo'];
$rno[$i] = $row ['rollno'];
$snm[$i] = $row ['StudentNm'];
$fnm[$i] = $row ['FathersNm'];
$mnm[$i] = $row ['MothersNm'];
$addr[$i] = $row['Address'];
$pic[$i] = $row['imagenm'];
$comm[$i] = $row['SocialCat'];
echo $formno[$i]."<br />";
echo $rno[$i]."<br />";
echo $snm[$i]."<br />";
echo $fnm[$i]."<br />";
echo $mnm[$i]."<br />";
echo $addr[$i]."<br />";
echo $pic[$i]."<br />";
echo $comm[$i]."<br />";
echo "<br />";
}
}
mysql_close($con);
}
?>
You are fetching the first row outside of your for() loop then you miss it.
After mysql_query() your should use mysql_num_rows() to check if there are any rows in your result and then fetch them in the for loop.
More info here : http://php.net/manual/fr/function.mysql-num-rows.php
Your code would look like this :
$sql = "SELECT * FROM `student` WHERE AppliedCourse ='$cr'";
$rs = mysql_query($sql) or die($sql. "<br/>".mysql_error());
if(0 == mysql_num_rows($rs)) {
$_SESSION['db_error'] = "<h2><font color = 'RED'>No such course found! Pease select again.</font></h2>";
header('Location: prinrepo.php');
} else {
for($i = 0;$i <= $row = mysql_fetch_array($rs);$i++)
{
// Your code
}
}
I keep getting
Warning: mysql_fetch_row() expects parameter 1 to be resource, boolean given in xx on line 6
error what is the problem with this code ? how can I fix it ?
$read = mysql_query("select * from detail");
while($wr = mysql_fetch_array($read)) {
echo $wr['Who'];
echo "<br />";
echo $wr['Time'];
echo "<br />";
echo $wr['What'];
}
edit; I made it like this still giving an error.
$db = new mysqli('localhost', 'root', '', 'panel');
$sql = "select * from detail";
$read = $db->query($sql);
while($wr = mysql_fetch_array($read)) {
echo $wr['Who'];
echo "<br />";
echo $wr['Time'];
echo "<br />";
echo $wr['What'];
}
You are mixing the MySQL and MySQLi extensions!
Use the appropriate functions of either, not both.
You probably having problems with querying database since your query fails:
$read = mysql_query("select * from detail"); // $read is false
// You can try to discover the error.
$read = mysql_query("select * from detail") || die(mysql_error());
Use this code:
$read = mysql_query("select * from detail");
while($wr = mysql_fetch_array($read, MYSQL_ASSOC)) {
echo $wr['Who'];
echo "<br />";
echo $wr['Time'];
echo "<br />";
echo $wr['What'];
}
See for more information: mysql-fetch-array