Display results from DB using Bootstrap - php

I would like to retrieve my results from my DB in this format using Bootstrap.
Below is my PHP code that I'm currently using, the first entry I want the image to be bigger then the rest.
<?php
$article = mysqli_connect("localhost", "root", "", "blog");
// Check connection
if($article === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
// Attempt select query execution
$sql = "SELECT * FROM news";
if($result = mysqli_query($article, $sql)){
if(mysqli_num_rows($result) > 0){
echo "<container>";
echo "<row>";
echo "<th>id</th>";
echo "<th>title</th>";
echo "<th>body</th>";
echo "<th>image</th>";
echo "</div>";
while($row = mysqli_fetch_array($result)){
echo "<row>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['title'] . "</td>";
echo "<td>" . $row['body'] . "</td>";
echo "<td>" . $row['image'] . "</td>";
echo "</row>";
}
echo "</div>";
// Free result set
mysqli_free_result($result);
} else{
echo "No records matching your query were found.";
}
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($article);
}
// Close connection
mysqli_close($article);
?>

I think you should have the condition to check for the first data in your loop to apply a bigger image size.

You can simply add condition:
$resultNum = 1;
while($row = mysqli_fetch_array($result)){
if($resultNum == 1) {
// TODO: show bigger image
} else {
// usual image
echo "<row>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['title'] . "</td>";
echo "<td>" . $row['body'] . "</td>";
echo "<td>" . $row['image'] . "</td>";
echo "</row>";
}
$resultNum++;
}

Just make a counter and then do a condition using the first number of the counter
Your code simplyfied:
$query = mysqli_query($article, "SELECT * FROM news");
$n = 1; //start the counter
if(mysqli_num_rows($query) > 0){ //detect if have rows
foreach ($query as $key => $value) {
if($n == 1){
//print the big image
echo $value["id"];
}else{
//print the little image
echo $value["id"];
}
$n++;
}
}else{
echo "No data founded";
}

Related

images appearing in warning but not in table?

I am working on a website whereby a load of advertisers are stored in the DB and then displayed to the user by there logo. I know storing directly in to the DB for images is not the done thing, however, I am starting out this way, to get the website running and then will refactor to move to a much more suitable approach.
Currently, I have the following PHP code:
<?php
session_start();
require_once "config.php";
// Create connection
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
$sql = "SELECT * FROM advertisers";
if($result = mysqli_query($link, $sql)){
if(mysqli_num_rows($result) > 0){
echo "<table>";
echo "<tr>";
echo "<th>id</th>";
echo "<th>advertiser_Name</th>";
echo "<th>advertiser_URL</th>";
echo "<th>advertiser_Category</th>";
echo "<th>advertiser_logo</th>";
echo "</tr>";
while($row = mysqli_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row['advertiser_id'] . "</td>";
echo "<td>" . $row['advertiser_Name'] . "</td>";
echo "<td>" . $row['advertiser_URL'] . "</td>";
echo "<td>" . $row['advertiser_Category'] . "</td>";
echo "<td>" . $row['<img src="data:image/jpeg;base64,'.base64_encode($row['advertiser_logo']).'"/>'] . "</td>";
echo "</tr>";
}
echo "</table>";
// Free result set
mysqli_free_result($result);
} else{
echo "No records matching your query were found.";
}
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
mysqli_close($link);
?>
However, the images are displayed when called from the DB but they are displayed in the warning message rather than in the table?
<?php
session_start();
require_once "config.php";
// Create connection
if ($link === false)
{
die("ERROR: Could not connect. " . mysqli_connect_error());
}
$sql = "SELECT * FROM advertisers";
if ($result = mysqli_query($link, $sql))
{
if (mysqli_num_rows($result) > 0)
{
echo "<table>";
echo "<tr>";
echo "<th>id</th>";
echo "<th>advertiser_Name</th>";
echo "<th>advertiser_URL</th>";
echo "<th>advertiser_Category</th>";
echo "<th>advertiser_logo</th>";
echo "</tr>";
while ($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['advertiser_id'] . "</td>";
echo "<td>" . $row['advertiser_Name'] . "</td>";
echo "<td>" . $row['advertiser_URL'] . "</td>";
echo "<td>" . $row['advertiser_Category'] . "</td>";
echo "<td><img src='data:image/jpeg;base64," . base64_encode($row['advertiser_logo']) . "'/></td>";
echo "</tr>";
}
echo "</table>";
// Free result set
mysqli_free_result($result);
}
else
{
echo "No records matching your query were found.";
}
}
else
{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
mysqli_close($link);
?>
The fact that is showing the image in the warning is because you're using a tag with the source as an array key which is not correct.
The array keys, so what is inside the square bracket, is the reference to the array position. If you're familiar with C for example is the 0, 1, ecc.. and not the value itself.
Yes as #NigelRen mentioned this row $row['<img src="data:image/jpeg; looks very bad.
I think you should use:
echo "<td><img src='data:image/jpeg;base64," . base64_encode($row['advertiser_logo']) . "'/></td>";

PHP dropdown inside table

I have a table where all the values are selected from the database, but some cells are empty. How do i put a dropdown list inside that empty cell. The values from the dropdown must come from the database
This is my code:
<?php
include("css/style.php");
/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
$link = mysqli_connect("localhost", "root", "Iamthebest1009", "dktp");
// Check connection
if ($link === false) {
die("ERROR: Could not connect. " . mysqli_connect_error());
}
$dropdown_list = '';
$sql = "SELECT * FROM orden";
$result_list = mysqli_query($link, $sql);
if (mysqli_num_rows($result_list) > 0) {
$dropdown_list = '<select>';
while ($row = mysqli_fetch_array($result_list)) {
unset($id, $name);
$id = $row['id'];
$name = $row['id'];
$dropdown_list .= '<option value="' . $id . '">' . $name . '</option>';
}
$dropdown_list .= '</select>';
}
// Attempt select query execution
$sql = "SELECT * FROM Norm LEFT JOIN Cluster ON norm.cluster_id = cluster.id LEFT JOIN Orden ON norm.orden_id = orden.id ORDER BY norm_name";
if ($result = mysqli_query($link, $sql)) {
if (mysqli_num_rows($result) > 0) {
echo '<form method="POST">';
echo "<table>";
echo "<tr>";
echo "<th>Norm id</th>";
echo "<th>Norm</th>";
echo "<th>Omschrijving</th>";
echo "<th>Clusteren</th>";
echo "<th>Ordenen</th>";
echo "</tr>";
while ($row = mysqli_fetch_array($result)) {
if ($row['orden_name']) {
$data_list = $row['id'];
} else {
$data_list = $dropdown_list;
}
echo "<tr>";
echo "<td>" . $row['norm_id'] . "</td>";
echo "<td>" . $row['norm_name'] . "</td>";
echo "<td>" . $row['description'] . "</td>";
echo "<td>" . $row['cluster_name'] . "</td>";
echo "<td>" . $data_list . "</td>";
echo "</tr>";
}
echo "</table>";
echo '<input type="submit" </input><form>';
// Free result set
mysqli_free_result($result);
} else {
echo "No records matching your query were found.";
}
} else {
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
if(isset($_POST['submit']))
{
$sql = "INSERT INTO norm (orden_id) VALUES ('$data_list')";
if(mysqli_query($link, $sql)){
echo "Records added successfully.";
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
}
// Close connection
mysqli_close($link);
?>
you can check this code. when $row['cluster_name'] empty then generate dropdown and first create dropdown then check your data exit or not but not tested
<?php
include("css/style.php");
/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
$link = mysqli_connect("localhost", "root", "Iamthebest1009", "dktp");
// Check connection
if ($link === false) {
die("ERROR: Could not connect. " . mysqli_connect_error());
}
$dropdown_list = '';
$sql = "SELECT * FROM orden";
$result_list = mysqli_query($link, $sql);
if (mysqli_num_rows($result_list) > 0) {
$dropdown_list = '<select>';
while ($row = mysqli_fetch_array($result_list)) {
unset($id, $name);
$id = $row['id'];
$name = $row['orden_name'];
$dropdown_list .= '<option value="' . $id . '">' . $name . '</option>';
}
$dropdown_list .= '</select>';
}
// Attempt select query execution
$sql = "SELECT * FROM Norm LEFT JOIN Cluster ON norm.cluster_id = cluster.id LEFT JOIN Orden ON norm.orden_id = orden.id ORDER BY norm_name";
if ($result = mysqli_query($link, $sql)) {
if (mysqli_num_rows($result) > 0) {
echo "<table>";
echo "<tr>";
echo "<th>Norm id</th>";
echo "<th>Norm</th>";
echo "<th>Omschrijving</th>";
echo "<th>Clusteren</th>";
echo "<th>Ordenen</th>";
echo "</tr>";
while ($row = mysqli_fetch_array($result)) {
if ($row['cluster_name']) {
$data_list = $row['cluster_name'];
} else {
$data_list = $dropdown_list;
}
echo "<tr>";
echo "<td>" . $row['norm_id'] . "</td>";
echo "<td>" . $row['norm_name'] . "</td>";
echo "<td>" . $row['description'] . "</td>";
echo "<td>" . $data_list . "</td>";
echo "<td>" . $row['orden_name'] . "</td>";
echo "</tr>";
}
echo "</table>";
// Free result set
mysqli_free_result($result);
} else {
echo "No records matching your query were found.";
}
} else {
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
// Close connection
mysqli_close($link);
?>

how to check if mysql query return no result(record not found) using php?

i am passing images file names via textarea to php script to find information about each image in mysql db .The problem is i am trying to output those image file names that not found in mysql db and inform the user which image file names not found in mysql. my current code fails to output those missing records in db but it correctly outputs information about those images found in db. could any one tell me what i am doing wrong ?
foreach ($lines as $line) {
$line = rtrim($line);
$result = mysqli_query($con,"SELECT ID,name,imgUrl,imgPURL FROM testdb WHERE imgUrl like '%$line'");
if (!$result) {
die('Invalid query: ' . mysql_error());
}
//echo $result;
if($result == 0)
{
// image not found, do stuff..
echo "Not Found Image:".$line;
}
while($row = mysqli_fetch_array($result))
{
$totalRows++;
echo "<tr>";
echo "<td>" . $row['ID'] ."(".$totalRows. ")</td>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['imgPURL'] . "</td>";
echo "<td>" . $row['imgUrl'] . "</td>"; echo "</tr>";
}
};
echo "</table>";
echo "<br>totalRows:".$totalRows;
You can use mysqli_num_rows() in mysqli
if(mysqli_num_rows($result) > 0){
while($row = mysqli_fetch_array($result))
{
$totalRows++;
echo "<tr>";
echo "<td>" . $row['ID'] ."(".$totalRows. ")</td>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['imgPURL'] . "</td>";
echo "<td>" . $row['imgUrl'] . "</td>";
echo "</tr>";
}
} else {
echo "<tr><td colspan='4'>Not Found Image:".$line.'</td></tr>';
}
You want to use mysqli_num_rows
if(mysqli_num_rows($result)) {
// Do your while loop here
}
Use mysqli_num_rows to compare the number of rows in the result set.

PHP to transfer one users data over to another page

I have an admin area in an ecommerce website whereby the admin can view all users on the allusers.php page. The users are listed in a table with their personal information, however i have a 'view profile' button near each user whereby if you was to click on it, it would take you to another page where you can view that specific users past orders.
the following is the code i have for allusers.php:
<?php
$result = mysql_query("SELECT * FROM customers ")
or die(mysql_error()); ;
if (mysql_num_rows($result) == 0) {
echo 'There Arent Any Orders Yet';
} else {
echo "<table border='0'><table border width=100%><tr><th>First Name</th><th>Surname</th><th>Address</th><th>E-Mail</th><th>Username</th><th>View Profile</th>";
while($info = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $info['name']. "</td>";
echo "<td>" . $info['surname']. "</td>";
echo "<td>" . $info['address1']. $info['address2']. $info['city']. $info['postcode']." </td>";
echo "<td>" . $info['email']. "</td>";
echo "<td>" . $info['username']. "</td>";
echo "<td>" . " <a href='view.php'>View</a> </td>";
}
}
echo "</tr>";
echo "</table>";
?>
the view.php page is as follows:
<?php
$result = mysql_query("SELECT * FROM order WHERE ......dont know what to enter here")
or die(mysql_error()); ;
if (mysql_num_rows($result) == 0) {
echo 'There Arent Any Orders For This Customer Yet';
} else {
echo "<table border='0'><table border width=100%><tr><th>Product</th><th>Quantities</th><th>Date</th>";
while($info = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $info['name']. "</td>";
echo "<td>" . $info['quantity']. "</td>";
echo "<td>" . $info['date']. " </td>";
}
}
echo "</tr>";
echo "</table>";
?>
I have a mysql database with the following fields & tables:
Customers - id, name, surname, address1, address2, city, postcode, email, username, password
Products - serial, name, description, price, picture
Order - id, name, quanitity, price, date, username
Thanks for any help provided
Your code lacks any sort of security mechanisms... This is very bad, especially in an e-commerce setting.
Excusing that, you would pass the username to the view page in the URL.
echo "<td>" . " <a href='view.php?user=" . $info['username'] . "'>View</a> </td>";
In your view page, you would get the parameter from the URL and include it with your query.
if (isset($_GET) && isset($_GET['user'])) {
$user = mysql_real_escape_string($_GET['user']);
} else {
header('Location: allusers.php');
exit(); // boot them back to the previous page.
}
$result = mysql_query("SELECT * FROM order WHERE username = '" . $user . "'")
A simple method could be the follow. Replace this line in alluser.php
echo "<td>" . " <a href='view.php'>View</a></td>";
with this one
echo '<td>View</td>';
and then, in your view.php have
if (isset($_GET['username']) && $_GET['username'] != '')
{
$username = mysql_real_escape_string($_GET['username']);
$result = mysql_query("SELECT * FROM order WHERE username = '$username'");
}
else
{
// No user specified. Do other statements
}
Please note the use of:
The user of the mysql_real_escape_string() function to protect from Sql injection (would be better the use of a prepared statements)
The use of the parameter username in the first page to pass the value of the username to the second page
The use of the $_GET global array to retrieve the parameter
Try this:
allusers.php
<?php
$result = mysql_query("SELECT * FROM customers ")
or die(mysql_error()); ;
if (mysql_num_rows($result) == 0) {
echo 'There Arent Any Orders Yet';
} else {
echo "<table border='0'><table border width=100%><tr><th>First Name</th><th>Surname</th><th>Address</th><th>E-Mail</th><th>Username</th><th>View Profile</th>";
while($info = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $info['name']. "</td>";
echo "<td>" . $info['surname']. "</td>";
echo "<td>" . $info['address1']. $info['address2']. $info['city']. $info['postcode']." </td>";
echo "<td>" . $info['email']. "</td>";
echo "<td>" . $info['username']. "</td>";
echo "<td>" . " <a href='view.php?user={$info['username']}'>View</a> </td>";
}
}
echo "</tr>";
echo "</table>";
?>
view.php
<?php
$user = mysql_real_escape_string($_GET['user']);
$result = mysql_query("SELECT * FROM order WHERE user = '$user'")
or die(mysql_error()); ;
if (mysql_num_rows($result) == 0) {
echo 'There Arent Any Orders For This Customer Yet';
} else {
echo "<table border='0'><table border width=100%><tr><th>Product</th><th>Quantities</th><th>Date</th>";
while($info = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $info['name']. "</td>";
echo "<td>" . $info['quantity']. "</td>";
echo "<td>" . $info['date']. " </td>";
}
}
echo "</tr>";
echo "</table>";
?>

printing [mysql_query] result in a table

I want to print mysql_query result in a table. I know how to do it but I am just confused. I tried this.
<?php
mysql_connect("localhost","root","") or die("Could not Connect.");
mysql_select_db("check") or die("Could not Select DB");
$table = "cc";
$i = 1;
$query = "select * from $table";
$sql = mysql_query($query);
if($sql){
echo "<table border='5'><tr>";
while($i<=2 && $row = mysql_fetch_array($sql)){
echo "<td>" . $row[id] . " : " . $row[name] . "</td>";
++$i;
}
echo "</tr><tr>";
while($i<=4 && $row = mysql_fetch_array($sql)){
echo "<td>" . $row[id] . " : " . $row[name] . "</td>";
++$i;
}
echo "</tr><tr>";
while($i<=6 && $row = mysql_fetch_array($sql)){
echo "<td>" . $row[id] . " : " . $row[name] . "</td>";
++$i;
}
echo "</tr><tr>";
while($i<=8 && $row = mysql_fetch_array($sql)){
echo "<td>" . $row[id] . " : " . $row[name] . "</td>";
++$i;
}
echo "</tr><tr>";
echo "</tr></table>";
}
?>
As you can see it is written again and again with a slight change of 2,4,6,8 in the while loop. It works but the problem is I cant rewrite it again and again as when the website will go live it will have more than 1000 entries. Could You guys help me out by suggesting another way to do this?
""** I need it to be like these dots (dots represent records in the database) **"""
. . . .
. . . .
. . . .
THANKS in Advance.
Ramzy
<?php
mysql_connect("localhost","root","") or die("Could not Connect.");
mysql_select_db("check") or die("Could not Select DB");
$table = "cc";
$query = "select * from $table";
$sql = mysql_query($query);
if($sql){
echo "<table border='5'><tr>";
while($row = mysql_fetch_array($sql)){
echo "<td>" . $row['id'] . " : " . $row['name'] . "</td>";
}
echo "</tr></table>";
}
?>
while($row = mysql_fetch_array($sql)) {
echo "<td>" . $row['id'] . " : " . $row['name'] . "</td>";
}
I don't really see what's the problem here.
By the way you should never call you're array like this $row[id] but you should quote the key instead $row['id']; Because if a constant id exists it will screw up your code and also for performance reason.
Just use
$limit = 1000;//place any value you need here to limit the number of rows displayed
while ($i<=$limit && $row = mysql_fetch_array($sql)){
echo "<td>" . $row['id'] . " : " . $row['name'] . "</td>";
++$i;
}
Also, that limit is unnecessary if all you want is to flush every record to the output. You could just do
while ($row = mysql_fetch_array($sql)){
echo "<td>" . $row['id'] . " : " . $row['name'] . "</td>";
}
And it will stop as soon as there are no more records.
To print all database rows into an HTML-table, use:
echo '<table>';
$i = 0; // $i is just for numbering the output, not really useful
while($row = mysql_fetch_array($sql))
{
echo '<tr><td>' . $i . ' - ' . $row['id'] . ' : ' . $row['name'] . '</td></tr>';
$i++;
}
echo '</tr></table>';
here is a general function I use:
function query_result_to_html_table($res, $table_id = NULL, $table_class = NULL, $display_null = true)
{
$table = array();
while ($tmp = mysql_fetch_assoc($res))
array_push($table, $tmp);
if (!count($table))
return false;
echo "<table cellpadding=\"0\" cellspacing=\"0\" border=\"0\" "
. ($table_id ? "id=\"$table_id\" " : "")
. ($table_class ? "class=\"$table_class\" " : "") . ">";
echo "<tr>";
foreach (array_keys($table[0]) as $field_name) {
echo "<th>$field_name";
}
foreach ($table as $row) {
echo "<tr>";
foreach ($row as $col => $value) {
echo "<td>";
if ($value === NULL)
echo "NULL";
else
echo $value;
}
echo "\n";
}
echo "</table>\n";
return true;
}
I Got The Answer.. I wanted it to be like this. I made this and It Actually Works.
<?php
$i = 1;
mysql_connect("localhost" , "root" , "") or die('Could not Connect.');
mysql_select_db("db") or die('Could not select DB.');
$query = "select * from `check`";
$sql = mysql_query($query) or die(mysql_error());
echo "<table border='5' width='50%'><tr><th>Name</th><th>Gender</th></tr></table><table border='5' width='50%'><tr>";
if($i<3){
echo "<td align='center'>".$row['name']."</td>";
echo "<td align='center'>".$row['gender']."</td>";
++$i;
} else {
echo "<td align='center'>".$row['name']."</td><td align='center'>".$row['gender']."</td>";
echo "</tr>";
$i = 1;
echo "<tr>";
}
}
echo "</table>";
?>
</div>
Thank You Guys For Your Support

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