Currently i'm using php and mysql to fetch downlines of a particular userid. Is there any way i can just use mysql to do that. My code is......
function downlineSearch($mid, $s)
{
global $nums; global $str;
$q = $cn->query("SELECT userid, CONCAT(lastname,' ',firstname,' ',middlename) AS name FROM clientsdata WHERE (recid LIKE '%$s%' OR lastname LIKE '%$s%' OR firstname LIKE '%$s%' OR middlename LIKE '%$s%') AND sponsor = $mid");
while($r = $q->fetch_array(MYSQL_ASSOC))
{
$nums += 1;
$str .= "<tr><td valign='middle'>NRN".$r['userid']."CCN</td><td valign='middle'>".$r['name']."</td></tr>";
downlineSearch($r['userid'], $s);
}
}
$str = '';
downlineSearch($mid, 'janet');
echo '<table id="tblSearch" width="100%">';
echo "<tr><td colspan='4' align='center' style='padding-top:10px;'><h3> $nums records found for [ <span style='color:red'>".$_POST['s']."</span> ]</h3></td></tr>";
echo $str;
echo '</table>';`
PLEASE ANY HELP WOULD BE HIGHLY APPRECIATED. THANKS
My guess it that you need to set the value of $cn, probably something like that:
$cn = new mysqli;
Does that help?
What is the error message you are getting?
Do you know in which line the error happens?
Do you have ability to debug line-by-line and view variables? If not, add several "echo" statements, to view in browser what's going on. Or, use error_log(), to display info in PHP error log, if you have access to it.
(I'm sorry about asking for clarification in an "answer", but the system does not let me add comments to other people's question - yet. As soon as you add some clarification, I will revise this "answer" to be actual answer to your question.)
Related
I'm trying to display data from database and it is important to me that this output is placed on different sides of website. I used php to connect to database, and ajax jquery to refresh data because every 20second values change.
I tried to
echo <div styles='position: absolute; top: 0px' class='text'>{$row['id']}</div>
in a foreach loop but when I do this all 6 of my id's are stacked on top each other.
Making <div> outside loop was unsuccessful too. I guess my problem is in reading data from database because I read all at once but I don't know any other way to do this except wrtiting 6 connection files to gather only the one value that I want to display and then styling it, but I feel like there is smarter way of doing this.
This is my code. Just want to say this is my first contact with php.
<?php
$hostname = "someinfo";
$username = "someinfo";
$password = "someinfo";
$db = "someinfo";
$dbconnect = mysqli_connect($hostname,$username, $password,$db) or die("cant");
if ($dbconnect->connect_error) {
die("Database connection failed: " . $dbconnect->connect_error);
}
$sensor_names = array();
$query2 = mysqli_query($dbconnect,"show tables");
while($row2 = mysqli_fetch_array($query2)){
if($row2[0] == 'sensors' or $row2[0] == 'measurments'){
break;
}
else{
array_push($sensor_names,$row2[0]);
}
}
$query = mysqli_query($dbconnect, "select s.id, s.sensor_name, max(dev.id), dev.temprature, dev.date from sensors s, `{$sensor_names[0]}` dev where s.id=dev.sensor_id gro
up by s.id, s.sensor_name order by s.id asc");
while($row = mysqli_fetch_array($query)){ //i konw this is ugly but this is working placeholder
foreach($sensor_names as $sn){
$query = mysqli_query($dbconnect, "select s.id, s.sensor_name, dev.temprature, dev.date from sensors s, `{$sn}` dev where s.id=dev.sensor_id order by dev.id desc limit 1");
$row = mysqli_fetch_array($query);
echo "
{$row['id']}
{$row['sensor_name']}
{$row['temprature']}
{$row['date']}
<br>";
}
}
?>
This is off-the-cuff from a guy who hasn't touched PHP in a long while, so watch for major bugs. But the basic idea is like this: build the code in a variable, and when done, echo out the entire variable. Makes it easier to add the structure/formatting you want. Note that you can also stick in a style tag along with that code and blurp out the style along with the "table" (Personally, I wouldn't use a table for styling, this is just for demo).
Note: I didn't style the output so that it puts the data on either side of the page - I left that for you to do. It's basic HTML - divs, styles, maybe css grid or flexbox. The point is to create your CSS/HTML/PHP mashup in a string variable and output the entire thing when done.
$out = '<style>.cell_id{font-weight:bold;}</style>';
$out .= '<table><tr><th>Label 1</th><th>Label 2</th><th>Etc</th></tr>'; //<=== ADDED!
while($row = mysqli_fetch_array($query)){
foreach($sensor_names as $sn){
$query = mysqli_query($dbconnect, etc. etc. etc.);
$row = mysqli_fetch_array($query);
$out .= "
<tr>
<td class='cell_id'>{$row['id']}</td>
<td>{$row['sensor_name']}</td>
<td>{$row['temprature']}</td>
<td>{$row['date']}</td>
</tr>";
}
}
echo $out;
Ok I think I got it. Cssyphus's answer got me thinking and I wrote something like that array_push($data, $row) and $data is two dimentional array that hold all data I need and now I can style it easily.
I'm learning this new langage PHP in order to develop modules from this software : Dolibarr
It's the first time I'm using PHP and I don't overcome to display query result in my view.
I would like to know if I wrote something wrong in my script because I don't understand all up to now. I would like to display the number of users in my software. I have to query my llx_user table and display the result in my array.
This is the part of my code :
/*
* View
*/
//Display number of users
$sql = "SELECT COUNT(u.rowid) as total";
$sql.= " FROM ".MAIN_DB_PREFIX."user as u";
$result = $db->query($sql);
print '<table class="noborder" width="100%">';
print '<tr class="liste_titre"><th colspan="2">'.$langs->trans("Statistics").'</th></tr>';
if (! empty($conf->user->enabled))
{
$statUsers = '<tr class="oddeven">';
$statUsers.= '<td>'.$langs->trans("Number of Users").'</td><td align="right">'.round($result).'</td>';
$statUsers.= "</tr>";
}
$total=0;
if ($entity == '0')
{
print $statUsers;
$total=round($result);
}
print '<tr class="liste_total"><td>'.$langs->trans("Total").'</td><td align="right">';
print $total;
print '</td></tr>';
print '</table>';
print '</div></div></div>';
llxFooter();
$db->close();
As I said, it's the first time I'm handling php file and I began to learn php 3 hours ago.
This is what I got :
If I comment like this :
$total=0;
//if ($entity == '0')
//{
print $statUsers;
$total=round($result);
//}
I'm getting this :
But I have 2 users in my table :
Thank you if you could help me
You're doing a good job for that you just started with PHP. Anyway, there's a little mistake in your code.
You actually query the database, but you don't fetch the result.
You have to do the following after your query:
$row = $result->fetch_row();
print $row[0]; // $row[0] will contain the value you're looking for
Also it seems that your $entity is not equal to 0. I don't see you initializing this variable anywhere, are you sure you have defined it? May you want to show us some mor of your code..
I am trying to make a leaderboard and sort my data by kills, but when I try to make it so it only grabs name, kill, death it doesnt grab anything but when I have it grab it all it works. Anyone know why? Code is below please assist.
<?php
$query = $koneksi->prepare("SELECT * from `player`");
$query->execute();
if($query->rowCount() == 0)
I am grabbing my mysql data here, if I change the * to the data I need no data is displayed.
echo "<tr><td colspan='6'><small>There's no player on ban list</small></td></tr>";
}
while($data = $query->fetch())
{
echo "<tr><td>".$data['name']."</td>";
echo "<td>".$data['kill']."</td>";
echo "<td>".$data['death']."</td>";
$kd = $data['kill'] / $data['death'];
echo "<td>".$kd."</td></tr>";
}
?>
Is it something to do with this or is something wrong? I am really confused.
Here you have to use bind_result() and in that you have to pass the number of parameters which is equal to your number of field from your player table.
Because here you are fetching data using select * query.
I am a neophyte programmer in php and I am always seeking for the solution of this problem. If anyone has an idea pls. post your answer and I am thankful for your great ideas to solve this stuff.
In my database table I have data like this:
In my php page I want to present in this way using html table.
Could anyone help me doing this stuff? Thank you very much…
I think you are searching for the functionality of GROUP_CONCAT in mysql (docs). You would get something like:
SELECT name, GROUP_CONCAT( week ), GROUP_CONCAT( there )
FROM presence
GROUP BY name;
which will return the following results you can parse using explode() in php (docs):
Andrew 4th,1st,3rd,2nd Present,Present,Present,Present
John 1st,4th,3rd,2nd Absent,Present,Present,Present
Mark 2nd,3rd,1st,4th Present,Present,Present,Present
Micheal 2nd,3rd,4th,1st Absent,Absent,Absent,Present
On a side note: If you haven't settled on a database scheme yet, it might be better to use an int for the week number column, as it is more reliable when sorting and easier to manipulate.
Sqlfiddle: http://sqlfiddle.com/#!2/fc785/1
Try this:
$SQL = "select NAME, WEEK, STATUS from tblattendance order by NAME, SUBSTRING(WEEK,1,LENGTH(WEEK) - 2) ASC";
$data = $db->query($SQL);
$last_name = "";
echo '<table><tr><th>NAME</th><th>1st WEEK</th><th>2nd WEEK</th><th>3rd WEEK</th><th>4th WEEK</th>';
while($row = $data->fetch_assoc()){
if($last_name != $row["NAME"]){
$last_name = $row["NAME"];
echo '</tr>';
echo '<tr>';
echo '<td>'.$row["NAME"].'</td>';
}
echo '<td>'.$row["STATUS"].'</td>';
}
echo '</tr></table>';
Try this :
$sql="select distinct name from tblattendance;";
$res=$mysqli->query($sql);
if($res){
while($row=$res->fetch_assoc()){
$name=$row['name'];
$sql="select week, status from tblattendance where name='$name';";
$res1=$mysqli->query($sql);
}
}
When I click on any link it opens all movies in my database. I want only that movie which begins with that letter and I don't know where I've made a mistake. Here is my code:
$azRange = range('A', 'Z');
foreach ($azRange as $letter){
echo ''.$letter.' | ';
}
if(isset($_GET["task"]) && $_GET["task"] == "view"){
$naslov = $_GET['naslov'];
$query = "SELECT filmovi.naslov, filmovi.godina, filmovi.trajanje, filmovi.slika
FROM filmovi
ORDER BY naslov";
$result = mysql_query($query)
or die ('SQL Greska: '.mysql_error());
if($result){
while($filmovi = mysql_fetch_array($result)){
echo '<center><b>';
echo '<td><img src="img/'.$filmovi["slika"].'" border="0" width="100" /></td>';
echo '</br>';
echo '<td>'.$filmovi["naslov"].'</td>';
echo '<td> ('.$filmovi["godina"].')</td>';
echo '<br>';
echo '<td>Trajanje: '.$filmovi["trajanje"].' min</td>';
echo '</b></center>';
echo '</tr>';
}
You are not passing the letter to the database query at any point.
$query =
"SELECT filmovi.naslov, filmovi.godina, filmovi.trajanje, filmovi.slika
FROM filmovi
WHERE naslov LIKE '$naslov%'
ORDER BY naslov";
Your query
$query = "SELECT filmovi.naslov, filmovi.godina, filmovi.trajanje, filmovi.slika
FROM filmovi
ORDER BY naslov";
is fetching all the movies from the database. There is no filtering here. Add some where conditions to this query and you'll get the expected result.
Changing to this query might help:
SELECT filmovi.naslov, filmovi.godina, filmovi.trajanje, filmovi.slika
FROM filmovi
WHERE `naslov` LIKE '{$naslov}%'
ORDER BY naslov
Since others have already answered your question (missing WHERE clause), I just want to mention that the <center> HTML tag is deprecated, and you should use CSS instead.
The mysql driver for PHP is also outdated, so instead of using:
mysql_query($query);
you should use
mysqli_query($link, $query);
for better security, OOP support, prepared statements, and transactions.
You can read about it here
Even if you are a beginner and you don't care about what those features mean, you should try and get into the habit of using mysqli anyway, so that when the day comes that you learn to appreciate it, you don't have to go back and update all of your code.