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.
Related
I have a database where teams will have multiple entries each with different locations. Each entry will have a team name. So for example, team1 might appear several times but each time the location will be different.
The structure of the DB is (each of these represents a column header):
team_name, first_name, last_name, location, arrival_time
My current working code creates HTML tables grouped by team name but currently only creates one row to show the first location and the time of arrival for the first location. I need this to dynamically create more rows to show all locations and arrival times for each team.
The desired result would look like this -
https://codepen.io/TheBigFolorn/pen/LqJeXr
But current result looks like this -
https://codepen.io/TheBigFolorn/pen/qgMppx
And here is an example of how the DB table might look -
https://codepen.io/TheBigFolorn/pen/daqJze
I've tried breaking up the echo and adding a second while loop before the row that I want to apply the above logic to but it seems to break everything. Any input on how I get this to work without having to use separate queries for each team would be very much appreciated. I'm new to php so please go easy on me :)
<?php
$leaders = "SELECT *, COUNT(location) FROM my_example_table GROUP BY team_name";
$result = mysqli_query($connect, $leaders) or die ("<br>** Error in database table <b>".mysqli_error($connect)."</b> **<br>$sql");
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "
<div class='red-border'>
<h2>". $row["team_name"]. "<br><small>Total locations visited: ". $row["COUNT(location)"]. "</small></h2>
</div>
<div class='data-holder'>
<table>
<tr>
<th>Location</th>
<th>Time of arrival</th>
</tr>
<tr><td>". $row["location"]. "</td> <td>". $row["arrival_time"]. "</td></tr>
</table>
</div>
";
}
} else {
echo "0 results";
}
?>
Your problem is due to the GROUP BY, as you've probably realised. This is necessary in order to get a count per team, but causes the number of rows output to be only 1 per team - that's what grouping does. Fundamentally, running an aggregate query such as a COUNT or SUM is incompatible with also outputting all of the row data at the same time. You either do one or the other.
Now, you could run two queries - one to get the counts, and one to get all the rows. But actually you don't really need to. If you just select all the rows, then the count-per-team is implicit in your data. Since you're going to need to loop through them all anyway to output them in the HTML, you might as well use that process to keep track of how many rows you've got per team as you go along, and create the "Total number of locations" headings in your HTML based on that.
Two things are key to this:
1) Making the query output the data in a useful order:
SELECT * FROM my_example_table Order By team_name, arrival_time;
2) Not immediately echoing HTML to the page as soon as you get to a table row. Instead, put HTML snippets into variables which you can populate at different times in the process (since you won't know the total locations per team until you've looped all the rows for that team), and then string them all together at a later point to get the final output:
$leaders = "SELECT * FROM my_example_table Order By team_name, arrival_time;";
$result = mysqli_query($connect, $leaders) or die ("<br>** Error in database table <b>".mysqli_error($connect)."</b> **<br>$sql");
$currentTeam = "";
$locationCount = 0;
$html = "";
$teamHtmlStart = "";
$teamHtmlEnd = "";
if ($result->num_rows > 0)
{
while($row = $result->fetch_assoc())
{
//run this bit if we've detected a new team
if ($currentTeam != $row["team_name"]) {
//finalise the previous team's html and append it to the main output
if ($currentTeam != "") $html .= $teamHtmlStart.$locationCount.$teamHtmlEnd."</table></div>";
//reset all the team-specific variables
$currentTeam = $row["team_name"];
$teamHtmlStart = "<div class='red-border'><h2>".$currentTeam."<br><small>Total locations visited: ";
$locationCount = 0;
$teamHtmlEnd = "</small></h2>
</div>
<div class='data-holder'>
<table>
<tr>
<th>Location</th>
<th>Time of arrival</th>
</tr>";
}
$teamHtmlEnd .= "<tr><td>". $row["location"]. "</td> <td>". $row["arrival_time"]. "</td></tr>";
$locationCount++;
}
//for the final team (since the loop won't go back to the start):
$html .= $teamHtmlStart.$locationCount.$teamHtmlEnd."</table></div>";
echo $html;
}
else {
echo "0 results";
}
Here's a runnable demo (using some static data in place of the SQL query): http://sandbox.onlinephpfunctions.com/code/2f52c1d7ec242f674eaca5619cc7b9325295c0d4
I have a webpage where it shows the lists of Projects and the monitoring of its progress/finances for every quarter. As shown below:
As you can see, my table is comprises of Project Name and a lists of sub-title's underneath it. And a series of columns per each quarter. Thru PHP I was able to populate the list of sub-titles under the Project Name, which also being fetched from the server side. Here's the code:
$sql = mysqli_query($con," My SELECT Statement ");
$i=0;
while($row = mysqli_fetch_assoc($sql)){
$ptitle = $row['Title'];
$iname = $row['Item'];
if($i%1)
{
?>
<?php } else { ?>
<tr>
<?php } ?>
<td width="25%"><?php echo $ptitle; ?></td>
<td></td>
</tr>
<tr>
<td><?php echo "<ul style='list-style-type: none;'><li>".nl2br($iname)."</li></ul>"; ?></td>
<td contenteditable="true" name="v1"></td>
Note: ptitle = ProjectName and iname = Semi-title underneath the Project's name.
Now, as you can see, the Project Name column literally "conquer" a single row on the left. Yet, the rows under the column of each quarter, should have its own separately, and must be parallel to the every sub-title underneath the Project Name. (Please refer to the image above for this) the only problem am encountering is... how can I make an editable row from inside a row, without affecting mysqli result? coz basically my table right now is kinda look like this:
Anyone who's more experience on this? I need your help.
PS: ...and oh! You might be wondering why do I include JSON in the title? It is because, I originally use JSON for editing those table rows before I even use mysqli_fetch_array. But when I include the results of the array inside the <table> tag, everything's changed and JSON is no longer working. So as of now, I am force to do it manually, meaning typing each <td contenteditable=true> in all of those rows. Yet, its not the desired output since I need another row within an existing row. Ideas? Anyone?
Figure Two:
Figure Three:
Count the number of lines in $iname, and then use a loop to create that many rows of contenteditable cells. You can also use this in the rowspan attribute of the <td> containing the title and subtitles.
$rows = substr_count($iname, "\n") + 1;
for ($i = 0; $i < $rows; $i++) {
echo "<tr>";
if ($i == 0) { ?>
<td rowspan='<?php echo $rows;?>' width='25%'><?php echo $ptitle . "<br>" . nl2br($iname);?></td>
<?php }
?>
<td contenteditable="true" name="v1"></td><td contenteditable="true" name="v2"></td>...
</tr>
<?php }
DEMO
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");
}
?>
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 $row['number'] instead.
I have a database with a few tables in it. The table i'm working on has 6 columns in it,
school_id, title, location, content, class_date and user_id
The school_id is unique (ranging from 1-20) and there are 4 different user_id (ranging from 1-4).....I'm working on a site which has a table that displays the information about the user and it's correspondence. I have a ReadMore link in the last column of the displayed table, my question is this, when I click the ReadMore link I want it to display the content field in the database based on which event is clicked.
For the login, the user is asked their user_id as their password.
//$T is called from a previous page which is the user_id or the password
$T = $_SESSION["ID"];
$INFO= $dbc->query("SELECT content FROM events where school_id='$T'");
$a = $INFO->fetch();
$b = $a['content'];
print_r($b);
THE ABOVE CODE IS HOW I TRY TO PRINT OUT THE CORRECT CONTENT BUT ONLY THE FIRST INSTANCE IS PRINTED OUT
THE BELOW CODE IS HOW THE TABLE IS DISPALYED ON MY SITE
<?php
$gold= $dbc->query("SELECT * FROM events WHERE user_id='$idcheck' ORDER BY event_dateLIMIT 10");
$x = $gold->fetchAll();
echo "<table border = '1'>";
echo "<tr>
<td> Event </td>
<td> Date of Event </td>
<td> Location </td>
<td> Event Information</td>
</tr>";
foreach ($r as $back) {
$title = $back['title'];
$eventdate = $back['class_date'];
$location = $back['location'];
echo "<tr>";
echo "<td>$title</td>";
echo "<td>$eventdate</td>";
echo "<td>$location</td>";
echo "<td>";
?>
<form action="displaycontent.php" method="post">
<input type="submit" value="Read More">
</form>
<?php
echo "</td>";
echo "</tr>";
}
echo "</table>";
?>
fetch method returns next found row from your statement. You have to iterate over your result to get all rows.
Something like
while ($row = $INFO->fetch()) {
$b[] = $row;
}
Then have a look at print_r($b). It should look like
array(0=>array('content'=>'Some content1'),
1=>array('content'=>'Some content2'), ....)
And your way of outputting the result should work.