I would to like to create a dynamic table that the rows and columns of the table depends on the data in mysql data. Please see below my codes.
<?php
require "connect/db.php";
$query = mysqli_query($mysqli, "SELECT COUNT(level) level, shelf_no, bin_id FROM location_bin WHERE rack_id =1 GROUP BY shelf_no");
while($res=mysqli_fetch_array($query)){
$col = $res['level']; //col
$row = $res['shelf_no']; //rows
echo "<table border='1'>";
$i = 0;
while ($i < $col){
if ($i==$col){
echo "<tr>";
}
echo "<td>".$res['bin_id']."</td>";
$i++;
}
echo "</tr>";
echo "</table>";
}
?>
What I want is to display A-1-01 up to A-1-06 in the first row then A-2-01 to A-2-03 on the second row. Note that the data is dynamic.
Can you try this code . It works. change db.php path as your system path.
require "db.php";
$query = mysqli_query($conn, 'SELECT level, shelf_no, bin_id FROM location_bin ORDER BY shelf_no asc, level asc ');
echo "<table border='1'>";
echo "<tr>";
$i=0;
while($res = mysqli_fetch_assoc($query)) {
$row=$res['shelf_no'];
if ($i < $row){
$i=$row;
echo "</tr><tr>";
}
echo "<td>".$res['bin_id']."</td>";
}
echo "</table>";
?>
Related
i need help.. im trying to echo all the data from my database but from the last to the first.
im a college student and i need ur help.
while (($row = mysqli_fetch_assoc($result)) && ($i < 6)) { // my lecture ask me to echo 5 data only
echo "<tr>";
foreach ($row as $field => $value) {
echo "<td>" . $value . "</td>";
}
echo "</tr>";
$i = $i + 1;
}
that is my code for now, and its only echoing from the first table to the last. my lecture told me i need to echo from the last.
edit :
i think i haven't give all the information.. sorry, this is the first time im here :(.
<?php
$i = 1;
include_once("function/helper.php");
include_once("function/koneksi.php");
$query = mysqli_query($koneksi, "SELECT * FROM transaksi ");
$pemilik = mysqli_fetch_assoc($query);
?>
<?php
$i = 1;
$sql = "SELECT mutasi, waktu_tanggal,tujuan FROM transaksi WHERE user_id='$user_id' ORDER BY waktu_tanggal";
$result = mysqli_query($koneksi, $sql);
if(mysqli_num_rows($result) == 0) {
?>
<h1>Anda belum melakukan transaksi apapun</h1>
<?php
}else {
echo "<br>";
echo "<table border='1'>";
?>
<tr>
<th>Mutasi</th>
<th>Waktu</th>
<th>Keterangan</th>
</tr>
<?php
while (($row = mysqli_fetch_assoc($result)) && ($i < 6)) {
echo "<tr>";
foreach ($row as $field => $value) { line like this: foreach($row as $value) {
echo "<td>" . $value . "</td>";
}
echo "</tr>";
$i = $i + 1;
}
echo "</table>";
}
?>
that is the entire page.
SELECT column1, column2, ...
FROM table_name
ORDER BY column1, column2, ... ASC|DESC;
The sql [ORDER BY] command is used to sort the result set in ascending or descending order.
SELECT column1, column2 FROM table_name ORDER BY column1 DESC;
Try php function mysqli_fetch_all () --- >> you don't need a while-loop.
Then create a new array and populate it with the required number of records.
Like this:
$query = "SELECT TOP 5*FROM `MyTable`";
// Or alternative query: "SELECT*FROM `MyTable LIMIT 5`
$result = mysqli_query($connection, $query);
if($result != 0) {
$newDataArray = mysqli_fetch_all($result, MYSQLI_BOTH);
}
I want to add data from phpmyadmin into a table format, but this code is missing the first row in the table and I don't understand why, I have tried other examples on SO such as
for ($i=0; $i< count($num); $i++)
But this did not work either.
Can someone see the issue?
Thanks
<?php
include('connect.php');
$con=mysqli_connect("localhost","******","*******","********");
$display = mysqli_query($con,"SELECT * FROM markers");
$num = mysqli_num_rows ($display);
$col = mysqli_num_fields ($display);
$row = mysqli_fetch_assoc($display);
$name = $row['name'];
?>
<?php
// start for loop
for ($i=0; $i<$num; $i++){
$row = mysqli_fetch_row($display); // fetching data
//echo results to table
echo "<tr data-name='$row[1]' data-address='$row[2]' data-lat='$row[3]' data-long='$row[4]'>";
for ($j=0; $j <$col; $j++){ // looping through each row of table
$test = $row [$j];
echo "<td test='$test'>" . $row [$j] . "</td>";
}
echo "</tr>";
}
mysqli_close($conn); // closing the connection
?>
Why do it that way?
I'd do a fetch array and loop that.
$result = mysqli_query($con,"SELECT * FROM markers");
while($row = mysqli_fetch_array($result))
{
echo "<tr data-name='$row['dataName']' data-address='$row['Address']' data-lat='$row['LAT']' data-long='$row['LONG']'>"
}
This would loop til you didn't have any more rows.
I am writing an application in which user can enter a database name and I should write all of its contents in table with using PHP.I can do it when I know the name of database with the following code.
$result = mysqli_query($con,"SELECT * FROM course");
echo "<table border='1'>
<tr>
<th>blablabla</th>
<th>blabla</th>
<th>blablabla</th>
<th>bla</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['blablabla'] . "</td>";
echo "<td>" . $row['blabla'] . "</td>";
echo "<td>" . $row['blablabla'] . "</td>";
echo "<td>" . $row['bla'] . "</td>";
echo "</tr>";
}
echo "</table>";
In this example I can show it since I know the name of table is course and it has 4 attributes.But I want to be able to show the result regardless of the name the user entered.So if user wants to view the contents of instructors there should be two columns instead of 4.How can I accomplish this.I get the table name with html.
Table:<input type="text" name="table">
Edit:Denis's answer and GrumpyCroutons' answer are both correct.You can also ask me if you didnt understand something in their solution.
Quickly wrote this up, commented it (This way you can easily learn what's going on, you see), and tested it for you.
<form method="GET">
<input type="text" name="table">
</form>
<?php
//can be done elsewhere, I used this for testing. vv
$config = array(
'SQL-Host' => '',
'SQL-User' => '',
'SQL-Pass' => '',
'SQL-Database' => ''
);
$con = mysqli_connect($config['SQL-Host'], $config['SQL-User'], $config['SQL-Pass'], $config['SQL-Database']) or die("Error " . mysqli_error($con));
//can be done elsewhere, I used this for testing. ^^
if(!isSet($_GET['table'])) { //check if table choser form was submitted.
//In my case, do nothing, but you could display a message saying something like no db chosen etc.
} else {
$table = mysqli_real_escape_string($con, $_GET['table']); //escape it because it's an input, helps prevent sqlinjection.
$sql = "SELECT * FROM " . $table; // SELECT * returns a list of ALL column data
$sql2 = "SHOW COLUMNS FROM " . $table; // SHOW COLUMNS FROM returns a list of columns
$result = mysqli_query($con, $sql);
$Headers = mysqli_query($con, $sql2);
//you could do more checks here to see if anything was returned, and display an error if not or whatever.
echo "<table border='1'>";
echo "<tr>"; //all in one row
$headersList = array(); //create an empty array
while($row = mysqli_fetch_array($Headers)) { //loop through table columns
echo "<td>" . $row['Field'] . "</td>"; // list columns in TD's or TH's.
array_push($headersList, $row['Field']); //Fill array with fields
} //$row = mysqli_fetch_array($Headers)
echo "</tr>";
$amt = count($headersList); // How many headers are there?
while($row = mysqli_fetch_array($result)) {
echo "<tr>"; //each row gets its own tr
for($x = 1; $x <= $amt; $x++) { //nested for loop, based on the $amt variable above, so you don't leave any columns out - should have been <= and not <, my bad
echo "<td>" . $row[$headersList[$x]] . "</td>"; //Fill td's or th's with column data
} //$x = 1; $x < $amt; $x++
echo "</tr>";
} //$row = mysqli_fetch_array($result)
echo "</table>";
}
?>
$tablename = $_POST['table'];
$result = mysqli_query($con,"SELECT * FROM $tablename");
$first = true;
while($row = mysqli_fetch_assoc($result))
{
if ($first)
{
$columns = array_keys($row);
echo "<table border='1'>
<tr>";
foreach ($columns as $c)
{
echo "<th>$c</th>";
}
echo "</tr>";
$first = false;
}
echo "<tr>";
foreach ($row as $v)
{
echo "<td>$v</td>";
}
echo "</tr>";
}
echo "</table>";
<?php
$table_name = do_not_inject($_REQUEST['table_name']);
$result = mysqli_query($con,'SELECT COLUMN_NAME FROM information_schema.COLUMNS WHERE TABLE_NAME='. $table_name);
?>
<table>
<?php
$columns = array();
while ($row = mysql_fetch_assoc($result)){
$columns[]=$row['COLUMN_NAME'];
?>
<tr><th><?php echo $row['COLUMN_NAME']; ?></th></tr>
<?php
}
$result = mysqli_query($con,'SELECT * FROM course'. $table_name);
while($row = mysqli_fetch_assoc($result)){
echo '<tr>';
foreach ($columns as $column){
?>
<td><?php echo $row[$column]; ?></td>
<?php
}
echo '</tr>';
}
?>
</table>
Back with another quick question. I have this code below which echo's out product names from a database. What I want to do is make the echoed out product names a link to another page called product.php, each link needs to have a unique ID, for example
Product Name
How would I go about doing this? Many thanks. I will point out that I am very new to PHP.
<?php
//create an ADO connection and open the database
$conn = new COM("ADODB.Connection");
$conn->open("PROVIDER=Microsoft.Jet.OLEDB.4.0;Data Source=C:\WebData\Northwind.mdb");
//execute an SQL statement and return a recordset
$rs = $conn->execute("SELECT product_name FROM Products");
$num_columns = $rs->Fields->Count();
echo "<table border='1'>";
echo "<tr><th>Name</th></tr>";
while (!$rs->EOF) //looping through the recordset (until End Of File)
{
echo "<tr>";
for ($i=0; $i < $num_columns; $i++) {
echo "<td>" . $rs->Fields($i)->value . "</td>";
}
echo "</tr>";
$rs->MoveNext();
}
echo "</table>";
//close the recordset and the database connection
$rs->close();
$rs = null;
$conn->close();
$conn = null;
?>
Assuming your Products table has a unique ID field called "id", change your select to:
$rs = $conn->execute("SELECT id, product_name FROM Products");
And when you want to create a link, use that field and pass it into the URL. So you'd have product.php?id=<?= $thatIdField; ?>.
Example code:
echo "<table border='1'>";
echo "<tr><th>Name</th></tr>";
while (!$rs->EOF) //looping through the recordset (until End Of File)
{
echo "<tr>";
for ($i=0; $i < $num_columns; $i++) {
echo "<td>" . $rs->Fields($i)->value . "</td>";
}
echo "</tr>";
$rs->MoveNext();
}
echo "</table>";
Hi I'm attempting to display data retrieved from a mysql table horizontally in an html table using php. The code below works well except for the fact that it leaves out the first record (starts at the second record) in my database. I'm sure it has something to do with the counter but I can't seem to figure out how to get it to stop doing this. If anyone can point out my error I'd really appreciate it!
$items = 5;
$query = "SELECT * FROM members ";
$result = mysql_query($query)
or die(mysql_error());
$row = mysql_fetch_array($result);
if (mysql_num_rows($result) > 0) {
echo '<table border="1">';
$i = 0;
while($row = mysql_fetch_array($result)){
$first_name = $row['first_name'];
if ($i==0) {
echo "<tr>\n";
}
echo "\t<td align=\center\">$first_name</td>\n";
$i++;
if ($i == $items) {
echo "</tr>\n";
$i = 0;
}
}//end while loop
if ($i > 0) {
for (;$i < $items; $i++) {
echo "<td> </td>\n";
}
echo '</tr>';
}//end ($i>0) if
echo '</table>';
}else {
echo 'no records found';
}
try and remove the 1st
$row = mysql_fetch_array($result);
you are calling it twice, that's why it skips 1 row in your while loop
try this simpler.
$items = 5;
$query = "SELECT * FROM members ";
$result = mysql_query($query) or die(mysql_error());
if (mysql_num_rows($result) > 0) {
echo '<table border="1">';
while($row = mysql_fetch_array($result)){
$first_name = $row['first_name'];
echo "<tr>";
for ($i=0 ; $i <= $items ;$i++) {
echo "<td align='center'>".$first_name."</td>";
}
}//end while loop
echo "</tr>";
echo '</table>';
}else{ echo 'no records found'; }
I have run into this issue before. Try the do while loop instead. Example
do {
// code
} while($row = mysql_fetch_array($result)); //end while loop
$row = mysql_fetch_array($result);
1) remove this line of code from ur scripts
2) only use while loop code instead.