sorry, I'm still very novice with server-side (back-end) development, thank you for helping in advance.
Recently, I've been developing a web page which you can choose which table you want to display from the mySQL server, then it will create an editable table in HTML using the mySQL table data.
So far, I'm only able to fetch which table to get from mySQL and display it using in the form of an array, which doesn't really look that great.
I wonder if you can display the array in a form of an editable table without knowing the how many columns, column names as it is different for every different table.
Later on, I have to upload the cell that is updated back to mySQL database and I've no idea how to do that as well.
Sorry for the trouble, this website had been a great help for me and the community is great. Thanks a lot!
Image of what I have so far:
Code I have so far:
SelectTable:
<div class="table-responsive">
<b>List of Tables</b>
<table class="table table-condensed selection_table">
<tbody>
<tr>
<?php
if ($tableResult = mysqli_query($conn,"show tables")){
while($table = mysqli_fetch_array($tableResult)) {
echo("<tr> <td>". "<a class = 'list_tables' href = ?clickedTable=$table[0]>". $table[0] . "</a>" ."</td> </tr>");
}
}else{
die("<b>"."No Table in Database!"."</b>");
}
if (isset($_GET['clickedTable'])){
$selectedTable = $_GET["clickedTable"];
}
?>
</tr>
</tbody>
</table>
</div>
Fetch array from selected table:
<?php
if (isset($_GET['clickedTable'])){
echo("<b> Current Table is: </b> ".$selectedTable. "<br/>");
$query = "SELECT * FROM $selectedTable";
if ($result = mysqli_query($conn , $query)) {
while ($row = mysqli_fetch_array($result)){
print_r($row);
}
}
}else{
echo("Please select a table");
}
?>
Related
At the moment I got incomplete what I need, how can I continue to display the SELECT results in their respective tables?
As you can see in the example above - there is a image at the end of topic - the first dynamically generated table was populated with your right data. You can see by the title of the table that refers to the last column Event.
But in the other tables I can not generate the loop.
I'm in the right way ?
What I want to do:
I have two tables in MySQL one with names Events and other Tickets. They have in common a column named event with the same data. In HTML tables generated according to the number of rows in the Events table, I wanted to select the information from the Tickets table and put in their respective HTML tables, where the title of each is a column named event in the Events table.
<?php
include 'conection.php';
$tickets = $con->prepare("SELECT tickets.chair, tickets.name, tickets.event, events.event FROM tickets INNER JOIN events ON tickets.event = events.event");
$tickets ->execute();
$events = $con->prepare("SELECT event FROM events");
$events ->execute();
?>
<!doctype html>
<html>
<body>
<?php
foreach($events as $evt){
echo " <div class='box[]'>
<table border='1px'>
<thread>
<th> ".$evt['event']." </th>
<tr>
<th>chair</th>
<th>name</th>
<th>event</th>
</tr>
</thread> ";
while ($ingr = $tickets->fetch(PDO::FETCH_ASSOC)){
if(($tck['event']) === ($evt['evento'] )){
echo " <tbody> ";
echo " <tr> ";
echo "<td>" .$ingr['chair']. "</td>";
echo "<td>" .$ingr['name']. "</td>";
echo "<td>" .$ingr['event']. "</td>";
}}}
echo " </tr> ";
echo " </tbody> ";
echo " </table> ";
echo "</div>";
?>
</body>
</html>
Exemple
I cant display image yet, but this is what iam done for now.
Figure of the code executed
First, some cleaning up.
The <thead> tag is a "table head", not a "thread" :)
Using a <th> outside of the <tr> will not work; move the event name into a header (<h2> or something) above your table.
You are using the field name 'evento' when you selected the field 'event'.
To answer your main question: you are seeing only one set of tickets because you run through the whole set of tickets when you search for the right ones for your first event. This way, $tickets has been exhausted by the time you want to get the tickets for the second event. You could try putting the ticket information in a variable after selecting (using $tickets->fetchAll), then using that variable in your loop instead of the $tickets resultset.
I'm building an exam management website and one of the pages I'm working on is for adding students to a course. I have a dropdown menu for the student number (which fetches values from a table), however I'd like to make it so that when the teacher selects the student number from the dropdown menu, that student's name and major appear on a table below. I have pretty much all the code for it however I can't seem to make it work. The way it is right now it shows the head of the table but it doesn't show any lines.
The errors are always in the lines where I declare $sql1 and $sql2 and vary according to how I define the condition in the statement.
Code for my dropdown menu : (works fine)
<label class="control-label" for="number">Student Number</label>
<?php
$sql = "SELECT number FROM students";
$result = $conn->query($sql);
echo "<select class=".'"form-control"'.' id="number" name="number" for="number">';
while ($row = $result->fetch_assoc()) {
echo '<option value="' . $row['number'] . '">' . $row['number'] . "</option>";
}
echo "</select>";
?>
Code for my table : (shows only head of table, which is the best I got after moving around the code and getting conversion errors and such)
The errors are always in the lines where I declare $sql1 and $sql2 and vary according to how I define the condition in the statement.
<table class="table">
<thead>
<tr>
<th>Name</th>
<th>Major</th>
</tr>
</thead>
<tbody>
<?php
$sql1 = "SELECT name FROM students WHERE number='$row'";
$result1 = $conn->query($sql1);
$value = $result1->fetch_object();
$sql2 = "SELECT major FROM students where number='$row'";
$result2 = $conn->query($sql2);
$value1 = $result2->fetch_object();
echo "<tr>
<td>".$value."</td>
<td>".$value1."</td>
</tr>";
?>
</tbody>
</table>
Thank you for all your help!!
Before I can formulate a complete answer, I must advise you that there are a few logical errors in your code.
How does your page "know" that a user selected an option from the select? You should perhaps intercept the event and respond to that using an asynchronoys mechanism, e.g. via AJAX.
Anyhow, there's no need to run two queries when you can make it with just one:
SELECT name, major FROM students WHERE number = ...
Once you have described how you mean to address issue #1 we can continue discussing the complete solution.
Well, I think there will be no $row in the the second snippet.
It seems that you didn't pass your $row from 1st snippet to 2nd snippet.
You can read this:
PHP Pass variable to next page
You can use session, cookie, get and post.
Or can just simply use "include", then the variables you defined can be used in the second page.
<?php
include "page1.php";
?>
<table class="table">
<thead>
<tr>
<th>Name</th>
<th>Major</th>
</tr>
</thead>
<tbody>
<?php
$number = $row['number'];
$sql1 = "SELECT name, major FROM students WHERE number='$number'";
$result1 = $conn->query($sql1);
$value = $result1->fetch_object();
echo "<tr>
<td>".$value['name']."</td>
<td>".$value['major']."</td>
</tr>";
?>
</tbody>
</table>
According to godzillante's answer below, the mysql query should be like this:
Anyhow, there's no need to run two queries when you can make it with just one:
SELECT name, major FROM students WHERE number = ...
I notice that you use $row as the key of your second query.
But in the first snippet, the data you fetch is "$row" (it is an array, see PHP - fetch_assoc)
You should use $row['number'] instead.
Ive got a code running smoothly which shows the list of people who are level 8 as shown below
I want when a person click on the username of the people it redirects them to their profile, actually i've got no clue or ideas of how could that be done. So i need a little bit of help to get some points on this.
Here is the code to show the above output
<?php
$lvl8 = 0;
$content = "";
$query = $koneksi->prepare("SELECT `user`, `level`, `LastOnlineDate` FROM `playerdata` WHERE `banned`=0 AND `level`=8");
$query->execute();
while($data = $query->fetch())
{
$lvl8++;
$content .= "<tr><td>".$lvl8."</td>";
$content .= "<td>".$data['user']."</td>";
$content .= "<td>".$data['LastOnlineDate']."</td></tr>";
}
?>
<table class="table table-bordered">
<thead>
<tr>
<td colspan='6'><h4><small>Level 8 - Trusted Admin (Total <?php echo $lvl8 ?>)</small></h4></td>
</tr>
<td><h5>Number</h5></td>
<td><h5>Username</h5></td>
<td><h5>Last Login</h5></td>
</thead>
<?php
if($query->rowCount() == 0)
{
echo "<tr><td colspan='6'><small>No rows found</small></td></tr>";
}
echo $content;
?>
</table>
first thing you should do is, have an a tag for name, like this
$content .= "<td><a href='link to a new file?id=userid from database'>".$data['user']."</a></td>";
In the new page, you can capture the id of the user, run a query to fetch the details of obtained id and then show the details obtained from database.
Create a http query or post form for your action.
like this in your foreach loop.
GET
<?php echo $username ?>
in your ending script for example a controller in an mvc architecture:
$id = $_GET['user_id'];
// do database stuff and view
Do your query using $id and fetch the result then display your view.
Basically you iterate over user id's to create links that carry the id of the clicked user in order to fetch the selected id.
I need to display my database records in a table as rows but it displays as different tables. I am using codeigniter. please help me out.
this is my code http://kl1p.com/Takn
http://ellislab.com/codeigniter/user-guide/database/active_record.html
Codeigniter has not a data grid view if that's yoru question, the best way (and the ony one i know) is :
$query = $this->db->get('mytable');
echo "<table>
<th>
<td>Title</td>
</th>
<tbody>
<tr>";
foreach ($query->result() as $row){
echo "<td>".$row->title."</td>";
}
echo " </tr>
</tbody>
</table>";
I can suggest you something like this : http://www.jeasyui.com/tutorial/datagrid/datagrid24.php and you can remove some propety as you wish. I think this is a good idea. I'm not professionnal but I'm trying to help. Hope you like it!
Im not sure if a similar question has been asked before, but here goes anyway. (I did do a search and found nothing relating to my question).
I am developing a website in which videos are played using the HTML5 video player. I have a connection to my database, a "watch" page that pulls all the correct data using a variable linked to the id (watch.php?v=1). I would like to have an index page where the most recent videos are pulled. They are ordered by the column "id" and everything works when I try and pull one result from the query. How would I go about getting multiple values? Here is my php code (server details hidden):
<?php
$mysqli = new mysqli("HIDDEN", "HIDDEN", "HIDDEN", "HIDDEN");
$sql = "
SELECT id, title, imgsrc, description
FROM videos
ORDER BY id DESC
LIMIT 2
";
$result = $mysqli->query($sql);
$video = mysqli_fetch_array($result, MYSQLI_ASSOC);
mysqli_close($mysqli);
?>
And here is my HTML code for the table.
<table>
<tr>
<td><h2><? echo $video['title']; ?></h2></td>
</tr>
</table>
That isn't the full code, but once I know the procedure I can apply it where needed!
I'm quite new to php and mysql, I can connect to databases but that's about it so a full walkthrough about what does what would be great!
Many Thanks,
James Wassall
You can iterate in for or while loop by calling mysqli_fetch_array($result, MYSQLI_ASSOC):
<table>
<?php
$mysqli = new mysqli("HIDDEN", "HIDDEN", "HIDDEN", "HIDDEN");
$sql = "
SELECT id, title, imgsrc, description
FROM videos
ORDER BY id DESC
LIMIT 2
";
$result = $mysqli->query($sql);
while($video = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
?>
<tr>
<td><h2><? echo $video['title']; ?></h2></td>
</tr>
<?php
}
mysqli_close($mysqli);
?>
</table>
Note that you should consider checking error statements, null controls etc.
try this
while($video = $result->fetch_array(MYSQLI_ASSOC))
{
?>
<tr>
<td><h2><?php echo $video['title']; ?></h2></td>
</tr>
<?php
}
?>
Each time you call mysqli_fetch_array, only one row is fetched.
You need to do something like
while ($video = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
// output $video['title']
}
(according to: http://www.php.net/manual/en/mysqli-result.fetch-array.php)
Try this,
<table>
<?php while($row = $result->fetch_array(MYSQLI_ASSOC)){ ?>
<tr>
<td><h2><? echo $row['title']; ?></h2></td>
<td><h2><? echo $row['description']; ?></h2></td>
</tr>
<?php } ?>
</table>