How to align php search results displayed? - php

How do I align and put div tags around the displayed data? I am very new to php and hope you guys can help!
Thanks!
James
<?php
$conn = mysql_connect("", "", "");
if (!$conn) {
echo "Unable to connect to DB: " . mysql_error();
exit;
}
{
$search = "%" . $_POST["search"] . "%";
$searchterm = "%" . $_POST["searchterm"] . "%";
}
if (!mysql_select_db("")) {
echo "Unable to select mydbname: " . mysql_error();
exit;
}
$sql = "SELECT name,lastname,email
FROM test_mysql
WHERE name LIKE '$search%' AND lastname LIKE '$searchterm'";
$result = mysql_query($sql);
if (!$result) {
echo "Could not successfully run query ($sql) from DB: " . mysql_error();
exit;
}
if (mysql_num_rows($result) == 0) {
echo "No rows found, nothing to print so am exiting";
exit;
}
while ($row = mysql_fetch_assoc($result)) {
echo $row["name"];
echo $row["lastname"];
echo $row["email"];
}
mysql_free_result($result);
?>
<?php echo $row["name"];?>
<br>
<?php echo $row["lastname"];?>
<br>
<?php echo $row["email"];?>

I think you mean something like this:
while ($row = mysql_fetch_assoc($result)) {
echo '<div class="data">';
echo '<label>' . $row["name"] . '</label>';
echo '<label>' . $row["lastname"] . '</label>';
echo '<label>' . $row["email"] . '</label>';
echo '</div>';
}
Anyway I don't understand the need to have this:
<?php echo $row["name"];?>
<br>
<?php echo $row["lastname"];?>
<br>
<?php echo $row["email"];?>
which will never echo nothing (will always echo the empty string) because when the script reaches this code, $row = false.

Related

How to make my PHP output two values from database

Ive included a block of code below detailling how im placing sections of data from a table in my database in their own divs. However, im new to PHP and cant find out how to also output the "cup_id" from the database into their respective divs here: echo $cup["cup_name"] . "<br />"; Thanks for all the help in advance!
So in short how do i get this to work: echo $cup["cup_id", "cup_name"] . "<br />";
<?php
require_once("action/dbcon.php"); // Get the database connection
$get_cup = "SELECT * FROM cups";
$show_cup = mysqli_query($conn, $get_cup);
if (!$show_cup) {
echo "Could not load cup. " . "(" . mysqli_error($conn) . ")";
}
while ($cup = mysqli_fetch_assoc($show_cup)) {
echo '<div class="cup-info">';
echo $cup["cup_name"] . "<br />";
echo '</div>';
}
?>
Do you want to concatenate strings? use the dot operator:
echo $cup["cup_id"] . $cup["cup_name"];
And if you want to print it in another div, make this:
echo '<div class="cup-info">';
echo $cup["cup_id"] . "<br />";
echo '</div>';
echo '<div class="cup-info">';
echo $cup["cup_name"] . "<br />";
echo '</div>';
if you want the name and id to be in the same raw then you can follow this method
<?php
require_once("action/dbcon.php"); // Get the database connection
$get_cup = "SELECT * FROM cups";
$show_cup = mysqli_query($conn, $get_cup);
if (!$show_cup) {
echo "Could not load cup. " . "(" . mysqli_error($conn) . ")";
}
while ($cup = mysqli_fetch_assoc($show_cup)) {
echo '<div class="cup-info">';
echo $cup["cup_id"] . $cup["cup_name"]."<br />";
echo '</div>';
}
?>
if you want the name and id to be in different raw then you can follow this method
<?php
require_once("action/dbcon.php"); // Get the database connection
$get_cup = "SELECT * FROM cups";
$show_cup = mysqli_query($conn, $get_cup);
if (!$show_cup) {
echo "Could not load cup. " . "(" . mysqli_error($conn) . ")";
}
while ($cup = mysqli_fetch_assoc($show_cup)) {
echo '<div class="cup-info">';
echo $cup["cup_id"]."<br />";
echo '</div>';
echo '<div class="cup-info">';
echo $cup["cup_name"]."<br />";
echo '</div>';
}
?>

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);
?>

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);
?>

Else statement doesn't work in option select

I am trying to implement a dropdown search option. All my search results are working. All the commands that I have assigned to if statements work, but when it does to else it deosn't work.
Here is my code:
if(isset($_REQUEST['submit'])){
$opt = $_POST['opt'];
if($opt==1){//if opt = 1
$sqle = "SELECT * FROM tbl_events WHERE title LIKE '%{$keywords}%'";
$resulte = mysql_query($sqle,$con) or die(mysql_error());
while($row=mysql_fetch_array($resulte)){
echo "<h4>" . $row['title'] . "</h4><br/>";
echo "<p>" . $row['description'] . "<p>";
}
}else if($opt==2){//if opt = 2
$sqls = "SELECT * FROM tbl_games WHERE games_name LIKE '%{$keywords}%'";
$results = mysql_query($sqls,$con)or die(mysql_error());
while($row=mysql_fetch_array($results)){
echo "<h4>" . $row['games_name'] . "</h4><br/>";
echo "<p>" . $row['description'] . "<p>";
}
}else{
echo "Your Searched keyword did not match";
}
}
What to do?
Try this: Take a flag to check if record exists.
$flag = false;
if($opt==1){//if opt = 1
$sqle = "SELECT * FROM tbl_events WHERE title LIKE '%{$keywords}%'";
$resulte = mysql_query($sqle,$con) or die(mysql_error());
if(mysql_num_rows($resulte) > 0) {
$flag = true;
while($row=mysql_fetch_array($resulte)){
echo "<h4>" . $row['title'] . "</h4><br/>";
echo "<p>" . $row['description'] . "<p>";
}
}
}else if($opt==2){//if opt = 2
$sqls = "SELECT * FROM tbl_games WHERE games_name LIKE '%{$keywords}%'";
$results = mysql_query($sqls,$con)or die(mysql_error());
if(mysql_num_rows($resulte) > 0) {
$flag = true;
while($row=mysql_fetch_array($results)){
echo "<h4>" . $row['games_name'] . "</h4><br/>";
echo "<p>" . $row['description'] . "<p>";
}
}
}
if(!$flag){
echo "Your Searched keyword did not match";
}

PHP Show results else nothing to display

I have this script here that shows a list of results. How to do l say "No results" if no results are found. I believe it's the else statement but couldn't quiet get it to work.
<?php
$result = mysql_query("SELECT * FROM emailquotes order by id desc")
or die(mysql_error());
while($row = mysql_fetch_array( $result )) {
echo "<tr height='25px' valign='center'>";
echo '<td valign="middle"><p><img src="../../Images/Icons/table-delete.png"/></p></td>';
echo '<td><p>' . $row['ssp'] . '</p></td>';
echo '<td><p>' . $row['ssp'] . '#someonewhere.com</p></td>';
echo '<td><p>' . $row['surname'] . '</p></td>';
echo '<td><p>Car</p></td>';
echo '<td><p>Show Prices</p></td>';
echo "</tr>";
}
?>
You can try with mysql_num_rows function:
$count = mysql_num_rows($result);
if ($count > 0) {
// loop rows
} else {
// no result
}
Put your while loop in a if-statement and check if there is any results before running the loop. Then you echo "No results" in the else.
It should be like this:
<?php
$result = mysql_query("SELECT * FROM emailquotes order by id desc")
or die(mysql_error());
if( mysql_num_rows($result)) {
while($row = mysql_fetch_array( $result )) {
echo "<tr height='25px' valign='center'>";
echo '<td valign="middle"><p><a href="delete.php?id=' . $row['id'] . '"><img
src="../../Images/Icons/table-delete.png"/></a></p></td>';
echo '<td><p>' . $row['ssp'] . '</p></td>';
echo '<td><p>' . $row['ssp'] . '#someonewhere.com</p></td>';
echo '<td><p>' . $row['surname'] . '</p></td>';
echo '<td><p>Car</p></td>';
echo '<td><p>Show Prices</p></td>';
echo "</tr>";
}
}
else {
echo "No Result";
}
?>

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