Displaying data in tables depending on group - php

I have a question in relation to displaying PHP tables that should be straight forward but I cannot get my head around it at the moment so any help would be appreciated, basically what I want to do is display a team of players in a table, but display multiple tables of users with their team name display above it.
What I currently have : http://puu.sh/ilUJp/4a6ae5e47b.png
What I am looking to achieve : http://puu.sh/ilUJ8/7756033517.png
<div class="col-lg-6">
<h3>Team Name Goes Here </h3>
<?php
echo "<table class='table table-striped'>";
echo " <thead>
<tr>
<th>Firstname</th>
<th>Lastname</th>
<th>Email</th>
</tr>
</thead>
";
while($row = mysqli_fetch_array($result)) {
$teamName = $row['teamName'];
$fName = $row['firstName'];
$surName = $row['surName'];
echo "
<tbody>
<tr>
<td>$teamName</td>
<td>$fName</td>
<td>$surName</td>
</tr>
</tbody>
";
}
echo "</table>";
?>
</div>
with my query :
$sql = "SELECT t.teamID,t.teamName,u.firstName,u.surName From users as u INNER JOIN team as t where u.teamID = t.teamID ";
I know the idea I need to do but cannot get it done, so any help would be appreciated.


Try this code
<?php $teemid=array();
while($row = mysqli_fetch_array($result)) {
if(!in_array($row['teamID'],$teemid)){
array_push($teemid,$row['teamID']);
if(!empty($teemid)){ ?>
</tbody>
</table>
</div>
<?php }
?>
<div class="col-lg-6">
<h3><?php echo $row['teamName']; ?></h3>
<table class='table table-striped'>
<thead>
<tr>
<th>Firstname</th>
<th>Lastname</th>
<th>Email</th>
</tr>
</thead>
<tbody>
<?php } ?>
<tr>
<td><?php echo $row['teamName']; ?></td>
<td><?php echo $row['firstName']; ?></td>
<td><?php echo $row['surName']; ?></td>
</tr>
<?php } ?>
</tbody>
</table>
</div>
SQL Query Change as below
$sql = "SELECT t.teamID,t.teamName,u.firstName,u.surName From users as u INNER JOIN team as t where u.teamID = t.teamID ORDER BY u.teamID";


You can do this logic
$teams = "get all teams sql query";
while ($row = mysqli_fetch_array($teams)) {
$teamid = $row['teamid'];
$teamname = $row['teamname'];
$teammemberquery = "select all member in the where team = $teamid sql query";
echo "<table>";
while ($r = mysqli_fetch_array($teammemberquery)) {
$teamName = $r['teamName'];
$fName = $r['firstName'];
$surName = $r['surName'];
echo "
<tbody>
<tr>
<td>$teamName</td>
<td>$fName</td>
<td>$surName</td>
</tr>
</tbody>
";
}
echo "</table>";
}


Try as below (Please replace table column name as yours and mysql to mysqli):
<?php
$link = mysql_connect('localhost', 'root', 'root');
$db_selected = mysql_select_db('test', $link);
$sql = "SELECT t.team_id,t.team,u.fname,u.lname,u.email From users as u INNER JOIN team as t where u.team_id = t.team_id order by t.team_id ";
$result = mysql_query($sql);
?>
<html><head><title>team</title></head><body><div class="col-lg-6">
<?php
echo "<table>";
$teamName = "";
$i=0;
while($row = mysql_fetch_array($result))
{
if($teamName == "" || $teamName != $row['team'])
{
if($i!=0)
echo "</table>";
echo "<tr><td colspan='3'><h3>".$row['team']."</h3></td></tr>";
$teamName = $row['team'];
$i=0;
}
$fName = $row['fname'];
$surName = $row['lname'];
$email = $row['email'];
if($i==0)
{
echo "<table class='table table-striped'><tr>
<th>Firstname</th>
<th>Lastname</th>
<th>Email</th>
</tr>";
}
echo "<tr>
<td>$fName</td>
<td>$surName</td>
<td>$email</td>
</tr>";
$i++;
}
echo "</table>";
?>
</div></body></html>

Related

TWO MySQL commands (SELECT to View), but only one is showing output

I have a table and each , I want to select a data from the same table in my database.
For example, first <td> is first name, then the second <td> is phone number.
I got the command, but only the first command is showing output.
This is my php codes to open and connect to the database :
<?php
include("./inc/db_connect.php");
$conn = OpenCon();
?>
This is the php codes for the table including <th> and <td> :
<div class="layer55">
<h3>
<table class="flat-table">
<tbody>
<tr>
<th>
<?php
$sql = "SELECT * FROM sharp_emp WHERE employee_id = 'AA170336'";
if ($result = $conn->query($sql)) {
if ($result->num_rows > 0) {
echo "Name";
}
}
?>
</th>
<th>
<?php
$sql = "SELECT * FROM sharp_emp WHERE employee_id = 'AA170336'";
if ($result = $conn->query($sql)) {
if ($result->num_rows > 0) {
echo "Phone Number";
}
}
?>
</th>
</tr>
<tr>
<td>
<?php
$sql = "SELECT first_name FROM sharp_emp WHERE employee_id = 'AA170336'";
while ($row = $result->fetch_array()) {
echo "" . $row['first_name'] . "";
}
?>
</td>
<td>
<?php
$sql = "SELECT phone FROM sharp_emp WHERE employee_id = 'AA170336'";
while ($row = $result->fetch_array()) {
echo "" . $row['phone'] . "";
}
?>
</td>
</tr>
</tbody>
</table>
</h3>
</div>
This is the php codes for db_connect.php :
<?php
function OpenCon()
{
$dbhost = "localhost";
$dbuser = // Hidden;
$dbpass = // Hidden;
$db = "sharp_db";
$conn = new mysqli($dbhost, $dbuser, $dbpass,$db) or die("Connect failed: %s\n". $conn -> error);
return $conn;
}
function CloseCon($conn)
{
$conn -> close();
}
?>
The expected output :
|----------|----------|
|Name |Phone Number|
|----------|----------|
|John |179898765 |
The current output :
|----------|----------|
|Name |Phone Number|
|----------|----------|
|John |Null (empty) |

You are running the same query multiple times, overwriting the $result variable for no reason, having useless $sql for the later 2 fetch without using them, and fetching a single $result twice by mistake.
So there are multiple concept problem with your code. I think your current code is something equivalant to this:
<div class="layer55">
<h3>
<table class="flat-table">
<tbody>
<tr>
<?php
$sql = "SELECT * FROM sharp_emp WHERE employee_id = 'AA170336'";
if ($result = $conn->query($sql)) {
if ($result->num_rows > 0) {
?>
<th>Name</th>
<th>Phone Number</th>
<?php } else { ?>
<th></th>
<th></th>
<?php } ?>
<?php } ?>
</tr>
<tr>
<?php if ($row = $result->fetch_array()) { ?>
<td><?php echo "" . $row['first_name'] . ""; ?></td>
<td><?php echo "" . $row['phone'] . ""; ?></td>
<?php } else { ?>
<td></td>
<td></td>
<?php } ?>
</tr>
</tbody>
</table>
</h3>
</div>
But frankly, it makes no sense to me to print an empty table when there is no result. So what you need is probably something like this.
<?php
$sql = "SELECT * FROM sharp_emp WHERE employee_id = 'AA170336'";
if (
($result = $conn->query($sql))
&& ($result->num_rows > 0)
&& ($row = $result->fetch_array())
):
?>
<div class="layer55">
<h3>
<table class="flat-table">
<tbody>
<tr>
<th>Name</th>
<th>Phone Number</th>
</tr>
<tr>
<td><?php echo $row['first_name']; ?></td>
<td><?php echo $row['phone']; ?></td>
</tr>
</tbody>
</table>
</h3>
</div>
<?php endif; ?>

Query data from multiple database tables into form table (MYSQL,PHP)

As I'm new to PHP, I want to know that how to put data from different database tables into one table form on the page.
My codes so far as below,
<?php
include('DBconnect.php');
mysql_query("USE onlinerecruitment");
$username =$_SESSION['user'];
$result = mysql_query("SELECT * FROM application_data_file");
$rows = mysql_fetch_array($result, MYSQL_ASSOC);
$pos_id = $rows['Position_ID'];
$resultt = mysql_query("SELECT * FROM position WHERE Position_ID = '".$pos_id."' ");
$resulttt = mysql_query("SELECT * FROM resume_data_file WHERE App_Email = '".$pos_id."' ");
?>
<TABLE border ='1'>
<table style="width:100%">
<tr>
<th>Application ID</th>
<th>Applicant E-mail</th>
<th>Position Selected</th>
<th></th>
<th></th>
<th></th>
</tr>
<?php
while ($row = mysql_fetch_array($result, MYSQL_ASSOC) & $rowss = mysql_fetch_array($resultt, MYSQL_ASSOC)){
echo "<TR>";
echo "<TD>".$row['App_Data_ID']."</TD>";
echo "<TD>".$row['App_Email']."</TD>";
echo "<TD>".$rowss['Position_Name']."</TD>";
echo "<TD><a href='view-app-form.php?app_mail=".$row['App_Email']."'>View Application Data</a></TD>";
echo "<TD><a href='view-resume-form.php?app_mail=".$row['App_Email']."'>View Resume Data</a></TD>";
echo "<TD><a href='view-test-score.php?app_mail=".$row['App_Email']."'>View Testing Score Data</a></TD>";
echo "</TR>";
}
?>
</table>
I will focus the part here.
<TABLE border ='1'>
<table style="width:100%">
<tr>
<th>Application ID</th>
<th>Applicant E-mail</th>
<th>Position Selected</th>
<th></th>
<th></th>
<th></th>
</tr>
<?php
while ($row = mysql_fetch_array($result, MYSQL_ASSOC) & $rowss = mysql_fetch_array($resultt, MYSQL_ASSOC)){
echo "<TR>";
echo "<TD>".$row['App_Data_ID']."</TD>";
echo "<TD>".$row['App_Email']."</TD>";
echo "<TD>".$rowss['Position_Name']."</TD>";
echo "<TD><a href='view-app-form.php?app_mail=".$row['App_Email']."'>View Application Data</a></TD>";
echo "<TD><a href='view-resume-form.php?app_mail=".$row['App_Email']."'>View Resume Data</a></TD>";
echo "<TD><a href='view-test-score.php?app_mail=".$row['App_Email']."'>View Testing Score Data</a></TD>";
echo "</TR>";
}
?>
</table>
But if there is any problem in the section that I didn't focused, I still appreciate your solution.
Thank you in advance.

To do this you would need to use a JOIN in the sql statement.
mysql_query("SELECT resume_data_file.App_Email, position.Position_ID FROM position INNER JOIN resume_data_file ON position.Position_ID = position.Position_ID WHERE position.Position_ID = '".$pos_id."' ");
http://www.w3schools.com/sql/sql_join.asp

Don't display if no records exist in mysql

I have this form to search names in mysql database
<form action="search.php" method="GET">
<input type="text" placeholder="Search" name="name">
<input type="submit" value="Search">
this is the search.php
<?php
name = $_GET['name'];
require_once("connect.php");
$records = $connect->query("SELECT * FROM Userlists WHERE Name = '$name'");
echo
"<table>
<thead>
<tr>
<th>Name</th>
<th>Email</th>
<th>Description</th>
</tr>
</thead>
<tbody>";
if (mysqli_num_rows($records)== 0){
echo "No data available for that name specified";
}
else {
while($row=mysqli_fetch_array($records)) {
$name = $row['Name'];
$email = $row['Email'];
$desc = $row['Desc'];
echo
"<tr>
<td>".$name."</td>
<td>".$email."</td>
<td>".$desc."</td>
</tr>";
}
}
echo
"</tbody>
</table>";
?>
so there is no problems when I search for a name that exists in database it displays correctly, but the problem comes when I search for a name that doesn't exist in database.. I want it to display only
"No data available for that name specified" for the output but I will also see empty table in the output like this ------------> IMAGE..
so how can I get rid of the empty table for the output?

Just pu the if outside the table....
<?php
name = $_GET['name'];
require_once("connect.php");
$records = $connect->query("SELECT * FROM Userlists WHERE Name = '$name'");
if (mysqli_num_rows($records)== 0){
echo "No data available for that name specified";
} else {
echo
"<table>
<thead>
<tr>
<th>Name</th>
<th>Email</th>
<th>Desc</th>
</tr>
</thead>
<tbody>";
while($row=mysqli_fetch_array($records)) {
$name = $row['Name'];
$email = $row['Email'];
$desc = $row['Desc'];
echo
"<tr>
<td>".$name."</td>
<td>".$email."</td>
<td>".$desc."</td>
</tr>";
}
echo
"</tbody>
</table>";
}
?>

change your if clause as below and remember to add exit() or die() function,this will end your php if there is no any data in database, and if there is any it will then start creating table for once and repeatedly fill up the table rows for given rows of data on database.
if (mysqli_num_rows($records)== 0){
echo "No data available for that name specified";
exit();
} else {
echo
"<table>
<thead>
<tr>
<th>Name</th>
<th>Email</th>
<th>Desc</th>
</tr>
</thead>
<tbody>";
while($row=mysqli_fetch_array($records)) {
$name = $row['Name'];
$email = $row['Email'];
$desc = $row['Desc'];
echo
"<tr>
<td>".$name."</td>
<td>".$email."</td>
<td>".$desc."</td>
</tr>";
}
echo
"</tbody>
</table>";
}

Move your echo
"<table>
<thead>
<tr>
<th>Name</th>
<th>Email</th>
<th>Desc</th>
</tr>
</thead>
<tbody>";
into the the if statement. This way it will only display the table when data is available!

$name = $_GET['name'];
require_once("connect.php");
$records = $connect->query("SELECT * FROM Userlists WHERE Name = '$name'");
if (mysqli_num_rows($records)== 0){
echo "No data available for that name specified";
}
else {
echo
"<table>
<thead>
<tr>
<th>Name</th>
<th>Email</th>
<th>Desc</th>
</tr>
</thead>
<tbody>";
while ($row = mysqli_fetch_array($records)) {
$name = $row['Name'];
$email = $row['Email'];
$desc = $row['Desc'];
echo
"<tr>
<td>" . $name . "</td>
<td>" . $email . "</td>
<td>" . $desc . "</td>
</tr>";
}
echo "</tbody></table>";
}
Try this one. But don't forget to escape $_GET['name'] like htmlspecialchars and real_escape_string

Start tag em seen in table

I am coding php and seen this error while I trying to do a HTM5 validator. But the things is there is no in my whole file. Any suggestions what is wrong here?
Error Line 27, Column 4: Start tag em seen in table.
<em><br />
Edit note : I rarely want to add this code here due to the fact that's it is long and have nothing to do with the error (like I said above, there is no tag in the html)
<section>
<table>
<?php
include("./settings.php");
$conn = #mysqli_connect($host, $user, $pwd, $sql_db);
if (!$conn) {
echo "<p>Databse connection error</p>";
} else {
$eoinumber = trim($_POST["num"]);
$lname = trim($_POST["lname"]);
$query = "SELECT * FROM eoi WHERE EOInumber = $eoinumber AND lname = '$lname'";
$result = mysqli_query($conn,$query);
if (!$result) {
echo "<p>There is something wrong with the $query</p>";
} else {
$temp = mysqli_num_rows($result);
if ($temp == 0) {?>
<p>0 enquiry found. Please check again your EOInumber and last name</p>
<?php } else {
?>
<thead>
<tr>
<th>EOI number</th>
<th>Job ID</th>
<th>First Name</th>
<th>Last Name</th>
<th>Date of Birth</th>
<th>Gender</th>
<th>Street</th>
<th>Town</th>
<th>State</th>
<th>Postcode</th>
<th>Mail</th>
<th>Telephone</th>
<th>Skills</th>
<th>Other Skills</th>
</tr>
</thead>
<?php
while ($row = mysqli_fetch_assoc($result)) {
echo "<tbody>";
echo "<tr>";
echo "<td>",$row["EOInumber"],"</td>";
echo "<td>",$row["job_num"],"</td>";
echo "<td>",$row["fname"],"</td>";
echo "<td>",$row["lname"],"</td>";
echo "<td>",$row["bday"],"</td>";
echo "<td>",$row["gender"],"</td>";
echo "<td>",$row["street"],"</td>";
echo "<td>",$row["town"],"</td>";
echo "<td>",$row["state"],"</td>";
echo "<td>",$row["postcode"],"</td>";
echo "<td>",$row["mail"],"</td>";
echo "<td>",$row["tele"],"</td>";
echo "<td>",$row["skill"],"</td>";
echo "<td>",$row["other"],"</td>";
echo "</tr>";
echo "</tbody>";
// echo "<p>Sucess</p>";
}
?>
</table>

Regardless of the presence or otherwise of the <em> tag, you are going to run into problems because you are opening the table tag and then potentially putting a <p> or other elements directly inside it, which is illegal (you can put <thead>, <tbody>, <caption>, <tr>, <tfoot> elements inside a <table>). It would be better if you opened the table after checking whether the queries had run successfully or not, i.e.:
<section>
<?php
include("./settings.php");
$conn = #mysqli_connect($host, $user, $pwd, $sql_db);
if (!$conn) {
echo "<p>Databse connection error</p>";
} else {
$eoinumber = trim($_POST["num"]);
$lname = trim($_POST["lname"]);
$query = "SELECT * FROM eoi WHERE EOInumber = $eoinumber AND lname = '$lname'";
$result = mysqli_query($conn,$query);
if (!$result) {
echo "<p>There is something wrong with the $query</p>";
} else {
$temp = mysqli_num_rows($result);
if ($temp == 0) {
echo "<p>0 enquiry found. Please check again your EOInumber and last name</p>";
} else {
echo "<table>";
... echo the table header...
while ($row = mysqli_fetch_assoc($result)) {
... echo table results...
}
echo "</table>";
}
}
?>

You missed a few }
So this should work:
<section>
<table>
<?php
include("./settings.php");
$conn = #mysqli_connect($host, $user, $pwd, $sql_db);
if (!$conn) {
echo "<p>Databse connection error</p>";
exit;
}
$eoinumber = trim($_POST["num"]);
$lname = trim($_POST["lname"]);
$query = "SELECT * FROM eoi WHERE EOInumber = $eoinumber AND lname = '$lname'";
$result = mysqli_query($conn,$query);
if (!$result) {
echo "<p>There is something wrong with the $query</p>";
exit;
} else {
$temp = mysqli_num_rows($result);
}
if ($temp == 0) { ?>
<p>0 enquiry found. Please check again your EOInumber and last name</p>
<?php } else { ?>
<thead>
<tr>
<th>EOI number</th>
<th>Job ID</th>
<th>First Name</th>
<th>Last Name</th>
<th>Date of Birth</th>
<th>Gender</th>
<th>Street</th>
<th>Town</th>
<th>State</th>
<th>Postcode</th>
<th>Mail</th>
<th>Telephone</th>
<th>Skills</th>
<th>Other Skills</th>
</tr>
</thead>
<?php
while ($row = mysqli_fetch_assoc($result)) {
echo "<tbody>";
echo "<tr>";
echo "<td>",$row["EOInumber"],"</td>";
echo "<td>",$row["job_num"],"</td>";
echo "<td>",$row["fname"],"</td>";
echo "<td>",$row["lname"],"</td>";
echo "<td>",$row["bday"],"</td>";
echo "<td>",$row["gender"],"</td>";
echo "<td>",$row["street"],"</td>";
echo "<td>",$row["town"],"</td>";
echo "<td>",$row["state"],"</td>";
echo "<td>",$row["postcode"],"</td>";
echo "<td>",$row["mail"],"</td>";
echo "<td>",$row["tele"],"</td>";
echo "<td>",$row["skill"],"</td>";
echo "<td>",$row["other"],"</td>";
echo "</tr>";
echo "</tbody>";
// echo "<p>Sucess</p>";
}
}
?>
</table>
<!--...-->

how to give style to the SQL table when checkbox selected?

SO i have form which consist of "Event Name" "Event Description" "Event Date" and checkbox "is important". When i check checkbox value "yes" its important, it sends to the sql value = "1" to table "is_important". Everything is all right, but i give the bootstrap style "bg-danger" for that "is_important" = 1 table and it doesnt show up. What's the problem?
You can see in the code:
<?php
if (isset($_POST['important'])) {
$error = array();
$success = array();
$eventTime = time();
$important = $_POST['important'];
$eventName = trim(mysql_real_escape_string($_POST['EventName']));
$eventDesc = htmlentities(trim(mysql_real_escape_string($_POST['EventDesc'])), ENT_QUOTES);
if (!isset($eventName) || empty($eventName)) {
$error['eventName'] = "Prasome ivesti ivykio varda";
} else if (strlen($eventName) > 32 || strlen($eventName) < 3) {
$error['eventName'] = "Ivykio pavadinimas turi buti tarp 3 ir 32 simboliu";
}
if (!isset($eventDesc) || empty($eventDesc)) {
$error['eventDesc'] = "Prasome ivesti ivykio aprasyma";
}
if (empty($error)) {
$sql = "INSERT INTO notes_list (title, description, timestamp,is_important) VALUES ('$eventName', '$eventDesc','$eventTime','$important')";
$result = mysqli_query($con, $sql);
$success[] = "SEKME !";
} else {
}
}
?>
<table class="table table-striped">
<thead>
<tr>
<th>Event name</th>
<th>Event description</th>
<th>Event date</th>
</tr>
</thead>
<tbody>
<?php
$query = "SELECT * FROM notes_list ORDER BY id DESC LIMIT 10";
$result2 = mysqli_query($con, $query);
print_r($_POST);
if ($result2) {
while ($note = mysqli_fetch_assoc($result2)) {
?>
<tr<?php echo (($note['is_important'] == 1) ? "class='bg-danger'" : ""); ?>>
<td><?php echo $note['title']; ?></td>
<td><?php echo $note['description'] ?></td>
<td><?php echo date('l M jS', $note['timestamp']); ?></td>
</tr>
<?php
}
mysqli_free_result($result2);
}
/* close connection */
mysqli_close($con);
?>
</tbody>
</table>
Full Example in this picture:
https://www.dropbox.com/s/h650h2spy2487dm/chechbox.jpg?dl=0

This:
<tr<?php echo (($note['is_important'] == 1) ? "class='bg-danger'" : ""); ?>>
would render this:
<trclass='bg-danger'>
in case is_important is 1. You need a space there, before the class.

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