PHP dropdown inside table - php

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);
?>

Related

Display results from DB using Bootstrap

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";
}

Selecting fields using variable PHP

how can I select the field using variable. I want to select the field (Seriennummer) in the base column (Seriennummer, MacAdresse).
I need to select meta_key columns, and meta_value where the variable $result is equal to meta_key.
Thanks to everyone for help.
I always get an error
Catchable fatal error: Object of class stdClass could not be converted to string in \wp-content\plugins\woocommerce\templates\checkout\thankyou.php
$result = wpuef_get_field('c2');
$value=$result->value;
echo $value,"<br>";
//Sql
$link = mysqli_connect("localhost", "root", "", "paymentdb");
// Check connection
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
// Attempt select query execution
$sql = "SELECT Seriennummer, MacAdresse FROM set_top_box WHERE Seriennummer ='".$result."'";
if($result = mysqli_query($link, $sql)){
if(mysqli_num_rows($result) > 0){
echo "<table>";
echo "<tr>";
echo "<th>Seriennummer</th>";
echo "<th>MacAdresse</th>";
echo "</tr>";
while($row = mysqli_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row['Seriennummer'] . "</td>";
echo "<td>" . $row['MacAdresse'] . "</td>";
echo "</tr>";
}
echo "</table>";
// Free result set
mysqli_free_result($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);
database picture
this is the result
$result = wpuef_get_field('c2');
$value=$result->value;
echo $value,"<br>";
//Sql
$link = mysqli_connect("localhost", "root", "", "paymentdb");
// Check connection
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
// Attempt select query execution
$sql = "SELECT Seriennummer, MacAdresse FROM set_top_box WHERE Seriennummer ='".$value=$result->value."'";
if($result = mysqli_query($link, $sql)){
if(mysqli_num_rows($result) > 0){
echo "<table>";
echo "<tr>";
echo "<th>Seriennummer</th>";
echo "<th>MacAdresse</th>";
echo "</tr>";
while($row = mysqli_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row['Seriennummer'] . "</td>";
echo "<td>" . $row['MacAdresse'] . "</td>";
echo "</tr>";
}
echo "</table>";
// Free result set
mysqli_free_result($result);
} else{
echo "No records matching your query were found.";
}
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);

PHP insert multiple dropdown values in database

I have two dropdownlists inside my table and i want to insert those values in the database. But when I press submit nothing happens.
This is what I have right now:
<?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['orden_name']) {
$data_list = $row['orden_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>" . $row['cluster_name'] . "</td>";
echo "<td>" . $data_list . "</td>";
echo "</tr>";
}
echo "</table>";
echo ' <form method="POST"><input type="submit" </input><form>';
// Free result set
mysqli_free_result($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($link);
?>
This is the part for the insert that isn't working:
if(isset($_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);
}
}
How do I fix that?
Try this solution. You had started form tag before select box. In
order to access you need to add dropdown within form tag.
<?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 '<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['orden_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>" . $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);
?>

How I can display all rows in php

I wrote this code to retrieve some rows form database
session_start();
$con = mysqli_connect('localhost', 'root', '');
if(!$con)
{
die("not ok");
}
mysqli_select_db($con,"uoh");
$q = " SELECT * FROM student WHERE id = " . $_SESSION['user_id'] ." and password = " . $_SESSION['user_pass'];
$result = mysqli_query($con , $q ) ;
if($row = mysqli_fetch_array($result))
{
echo "this academic transcripts for " . $row["name"];
echo " and the id is " . $row["id"];
}
$q1 = " SELECT student_record.course,student_record.grade,student_record.term,coe_courses.crd
FROM student_record INNER JOIN coe_courses ON student_record.course_number = coe_courses.course_number
where student_record.id = ".$_SESSION['user_id'] ;
$result = mysqli_query($con , $q1 ) ;
if($row = mysqli_fetch_array($result))
{
echo "<br />";
echo "<table border=\"1\" style=\"width:500\">";
echo "<tr>";
echo "<th>coe_courses</th>";
echo "<th>terms</th>";
echo "<th>Grades</th>";
echo "<th>CRD</th>";
echo "</tr>";
echo "<tr>";
echo "<td>" . $row["course"]. "</td>";
echo "<td>" . $row["term"]. "</td>";
echo "<td>" . $row["grade"]. "</td>";
echo "<td>" . $row["crd"]. "</td>";
echo "</tr>";
echo "</table>";
}
The problem is that only shows the first row while I have three rows in phpMyAdmin.
enter image description here
You need to call fetch_* repeatedly to retrieve all rows from your result set; each time you call it it retrieves the next row in the result set.
In your sample code above, you would replace
if ($row = mysqli_fetch_array($result))
{
with
while ($row = mysqli_fetch_array($result))
{
This will loop until fetch_array tries to read beyond the last record in $result, at which point fetch_array returns false and the loop exits.

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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