Display Database data - php

I have typed variaties of this code and it doesn't display.
I manage to get another page to display but this one won't display at all.
I am trying to create a page a php file to friendly view Database content, Update , delete and add content if need be.
But first I need to be able to display the content before I can get the rest of the code.
And I did try adding the code for updating =, deleting and adding data from a similar display but to no success.
Your Help please?
No errors shown, its just a blank after the 'th' content tags.
I have tested till which line it compiles until, not sure if that helps.
just fyi.
Sometimes it mentions it does not find database, sometimes but seldom. Don't know why but it is led to the correct source.
<?php
require 'config.php';
$con = mysql_connect( $db_host, $db_username, $db_password, $db_name);
if ($con->connect_error) {
die("Connection failed: " . $con->connect_error);
}//else { echo 'it works'; }
echo '<table>';
echo ' <tr>';
echo ' <th>';
echo ' Comment';
echo ' </th>';
echo ' </tr>';
$myData = "SELECT * FROM demo ";
$res = mysql_query($myData,$con);
// especially not working after this line.
while($row = mysqli_fetch_assoc($res)){
echo ' <tr>';
echo ' <td>';
echo ' <input type=text name=ID value="';
echo $row["ID"];
echo ' "/>';
echo ' </td>';
echo ' <td>';
echo ' <input type=text name=ID value="';
echo $row["Name"];
echo ' "/>';
echo ' </td>';
echo ' <td>';
echo ' <input type=text name=ID value="';
echo $row["Org"];
echo ' "/>';
echo ' </td>';
echo ' <td>';
echo ' <textarea name=Comment value="'.$row["Comment"].'"></textarea>';
echo ' </td>';
echo ' <td>';
echo ' <input type=submit value=Update name=UpBut id=UpBut/>';
echo ' <input type=submit value=Delete name=DelBut id=DelBut/>';
echo ' </td>';
echo ' </tr>';
}
echo '</table>';
$conn->close();

Avoid using the deprecated mysql functions.
Instead of:
$res = mysql_query($myData,$con);
Pass the $con before:
$res = mysqli_query($con, $myData);
This is definitely wrong:
while($row = $res->fetch_assoc()){
Correct way:
while($row = mysqli_fetch_assoc($res)){
Try this:
<?php
require 'config.php';
/*
$db_host = "localhost";
$db_username = "yourusernamehere";
$db_password = "yourpasswordhere";
$db_name = "yourdbnamehere";
*/
$con = new mysqli($db_host, $db_username, $db_password, $db_name);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo '<table>';
echo ' <tr>';
echo ' <th>';
echo ' Comment';
echo ' </th>';
echo ' </tr>';
$myData = "SELECT * FROM demo";
$res = mysqli_query($con, $myData);
while($row = mysqli_fetch_assoc($res)) {
echo '<div id="main">' . '<p>';
echo $row["Comm"]; . '<br>';
echo $row["Name"]; . '<br>';
echo $row["Company"]; . '</p>';
echo "</div>";
}
?>
Notice: Study mysqli or PDO. The proper closing of </table> tag is
up to you as I don't know that else you might have afterwards.

Related

How to Update a Data in MySQL using Radio Button?

Hello Everyone Good Afternoon
Can I ask a question? but before that here is my code.
<html>
<center>
<font size="2" face = "century gothic">
<?php
$con=mysqli_connect("localhost","root","","election2016");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM candidate_info");
echo "<table border='1'>
<tr>
<th>Candidate Name</th>
<th>Position</th>
<th>Vote</th>
<th>Number of Votes</th>
</tr>";
while ($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['CandidateName'] . "</td>";
echo "<td>" . $row['Position'] . "</td>";
echo "<td><input type='radio' name='candidateid'/>";
echo "<td>" . $row['NumberofVotes'] . "</td>";
}
echo "</table>";
mysqli_close($con);
?>
<br>
<br>
<form method = "post" action = "<?php $_PHP_SELF ?>">
<input name = "update" type = "submit" id = "update" value = "Update">
</form>
</center>
</font>
</html>
<?php
if(isset($_POST['update'])) {
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn ) {
die('Could not connect: ' . mysql_error());
}
$candidateid = $row['candidateid'];
$sql = "UPDATE candidate_info SET numberofvotes = '1' WHERE candidateid = $candidateid" ;
mysql_select_db('election2016');
$retval = mysql_query( $sql, $conn );
if(! $retval ) {
die('Could not update data: ' . mysql_error());
}
echo "Updated data successfully\n";
mysql_close($conn);
}
?>
The Output of my Code Here is it will show list of Candidates,Position,Radio Button Number of Votes with a button save.
My error here is that when i select a radio button and click the button update i want to put 1 in numberofvotes field but its not updating. Whats wrong with my code?
Any help would be appreciated.
TY so much
<html>
<center>
<font size="2" face = "century gothic">
<?php
$con=mysqli_connect("localhost","root","","election2016");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM candidate_info");
echo "<table border='1'>
<tr>
<th>Candidate Name</th>
<th>Position</th>
<th>Vote</th>
<th>Number of Votes</th>
</tr>";
while ($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['CandidateName'] . "</td>";
echo "<td>" . $row['Position'] . "</td>";
echo "<td><input type='radio' name='candidateid' value='".$row['candidateid']."' >";
echo "<td>" . $row['NumberofVotes'] . "</td>";
$candidateid=$row['candidateid'];
}
echo "</table>";
mysqli_close($con);
?>
<br>
<br>
<form method = "post" action = "<?php $_PHP_SELF ?>">
<input type="hidden" name="candidateid" value="<?php echo $candidateid;?>">
<input name = "update" type = "submit" id = "update" value = "update">
</form>
</center>
</font>
</html>
<?php
if(isset($_POST['update'])) {
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$candidateid = $_POST['candidateid'];
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn ) {
die('Could not connect: ' . mysql_error());
}
$candidateid = $_POST['candidateid'];
$sql = "UPDATE candidate_info SET numberofvotes = 1 WHERE candidateid = '$candidateid'" ;
mysql_select_db('election2016');
$retval = mysql_query( $sql, $conn );
if(! $retval ) {
die('Could not update data: ' . mysql_error());
}
echo "Updated data successfully\n";
mysql_close($conn);
}
?>

How to insert mysql database in a new file?

I would like to know how can i put the table from mysql in a new file.php.
I want the MySql table to be on the page.
This is my code that inserts data in MySql.
<?php
// Create connection
$con = mysqli_connect("host", "id_", "password", "xxxxxx");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$Task = $_POST['Task'];
$Date = $_POST['Date'];
$Desc = $_POST['Desc'];
$sql = "INSERT INTO tasklist (Task, Date, Description)
VALUES ('$Task', '$Date', '$Desc')";
if (!mysqli_query($con, $sql)) {
die('Error: ' . mysqli_error($con));
}
echo "1 record added";
mysqli_close($con);
?>
<html>
<body>
<form action="addtask.php" method="post">
Task: <input type="text" name="Task">
Date: <input type="text" id="datepicker" name="Date">
Decrption:<textarea type="text" name="Desc"></textarea>
<input type="submit" value="submit">
</form>
</body>
</html>
Try this code:
<?php
$con=mysqli_connect("example.com","peter","abc123","my_db");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM Persons");
echo "<table border='1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['FirstName'] . "</td>";
echo "<td>" . $row['LastName'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
also u can try w3schools sample code :
Display the Result in an HTML Table
The following example selects the same data as the example above, but will display the data in an HTML table:
<?php
$con=mysqli_connect("example.com","peter","abc123","my_db");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM Persons");
echo "<table border='1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['FirstName'] . "</td>";
echo "<td>" . $row['LastName'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
The output of the code above will be:
first code comes from "Jonnny" in this article

Getting data to display in a form

I'm trying to use HTML, PHP and MYSQL to pull data from a database and display it in a form (to later be edited). At this point I'm only trying to pull that data and display it in a form. (I'll worry about updating later). I pull the data but nothing displays in my textboxes:
<?php
$con = mysqli_connect("XXXXX"); //removed for privacy
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$query="select * from VOLUNTEER";
echo '$query';
$result = mysqli_query($con, $query);
echo "<table>";
if ($result)
{
while($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
{
echo '<form method = "post" action="insertvolunteer.php">';
echo '<tr>';
echo '<td>First Name:</td>';
echo '<td>' . '<input type=text name=FirstName' . $row["FirstName"] . '</td>';
echo '<td>' . '<input type=hidden name=VolunteerId' . $row["VolunteerId"] . '</td>';
echo '</tr>';
}
}
echo "</form>";
echo "</table>";
mysqli_close($con);
?>
Text box data needs to be displayed on value as
echo '<td><input type="text" name="FirstName" value="'.$row["FirstName"].'"></td>';
connect.php
<?php
$server = "server";
$user = "user";
$password = "password";
$bd = "yourbd";
$connect = mysql_connect($server, $user, $password);
$connect = mysql_select_db("$bd", $connect);
if (!$connect){
echo mysql_error(); exit;
}
?>
namefile.php
<?php
include('connect.php');
$select = mysql_query("select * from VOLUNTEER");
while ($show = mysql_fetch_assoc($select)):
echo "<table>";
echo "<form method = 'post' action='insertvolunteer.php'>";
echo '<tr>';
echo '<td>First Name:</td>';
echo '<td><input type="text" name="FirstName" value="'.$show["FirstName"].'"></td>';
echo '<td><input type="text" name="FirstName" value="'.$show["VolunteerId"].'"></td>';
echo '</tr>';
echo "</form";
echo "</table>";
endwhile;
?>
When you create an MySQL query, you need to declare this. How?
$var = "SELECT * FROM SOMEWHERE"; wrong
$var = mysql_query("SELECT * FROM SOMEWHERE"); right
'n in echo '<td>' . '<input type=text name=FirstName' . $row["FirstName"] . '</td>';, you need to close the tag. And also have no need of separate <td> of <input>.
Try something like :)
#update I realized that you've got what was wished. Cheers.

Unable to delete rows through php, HTML

I am just unable to understand why the rows are not getting deleted!
please note that i am getting the login values of corrected check boxes in the php page.
from my point of view, most probably error should be in php page where i am using
'DELETE FROM' query.
<?php
session_start();
?>
<html>
<head>
<form id="delete_customers" action="deletecustomers.php" method="post">
<?php
$con = mysql_connect("localhost","root","");
if (!$con) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("car_rental_system", $con);
$result = mysql_query("SELECT * FROM customer");
echo "<table border='1' align='center'>
<tr>
<th>first_name</th>
<th>Last_name</th>
<th>login</th>
</tr>";
while($row = mysql_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['first_name'] . "</td>";
echo "<td>" . $row['login'] . "</td>";
echo "<td>"."<input type='checkbox' name='deletingcustomers[]'
value=$row['login']}"."</td>";
echo "</tr>";
}
echo "</table>";
mysql_close($con);
?>
<p class='submit'>
<button type='submit' name='dscustomer'>Delete selected</button>
</p>
</head>
</html>
//NOW deletecustomers.php
<?php
session_start();
$_SESSION['deletingcustomers'] = $_POST['deletingcustomers'];
$N = count($_SESSION['deletingcustomers']);
$con = mysql_connect("localhost","root","");
if (!$con) die('Could not connect: ' . mysql_error());
mysql_select_db("car_rental_system", $con);
if(empty($_SESSION['deletingcustomers'])) {
echo("No customers selected");
} else {
for ($i=0; $i<$N; $i++) {
$sql1="delete from `customer`
where login='{$_SESSION[deletingcustomers][$i]}'";
if(mysql_query($sql1,$con))
echo 'executed';
}
}
?>
NO! No! Why do people keep using mysql_query().....(head desk)
Please look up PDO. http://php.net/manual/en/book.pdo.php it helps prevent sql injections and gives you a better understanding of how to harness oop's power.
Your $_SESSION[deletingcustomers][$i] needs to be $_SESSION['deletingcustomers'][$i]
Example on its way
$tempVar = $_SESSION['deletingcustomers'][$i];
$dbConnection = new PDO("mysql:host=".$hostName.";dbname=".$dbName, $username, $password);
$sql = "delete from `customer` where login='$tempVar'";
$stmt = $newObj->prepare($sql);
$stmt->execute();
Replace
echo "<td>"."<input type='checkbox' name='deletingcustomers[]' value=$row['login']}"."</td>";
To
echo "<td><input type='checkbox' name='deletingcustomers[]' value='".$row['login']."'</td>";
and try

create a Compare page using checkbox and data from mysql

Im a newbie in PHP and I want to create a simple webpage app for my website, I was able to produce this page base on a tutorial here.
<?php
$con = mysql_connect("localhost","*****","*****");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("*****", $con);
$result = mysql_query("SELECT * FROM products");
echo "<table border='1'>
<tr>
<th>Name</th>
<th>classification</th>
</tr>";
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['classification'] . "</td>";
echo "<td><input type='checkbox' name='{number[]}' value='{$row['prodID']}' /></td>";
echo "</tr>";
}
echo "</table>";
mysql_close($con);
?>
<?php
?>
<html>
<head>
</head>
<form name="form1" method="post" action="result_page.php">
<input type="submit" name="Submit" value="Submit">
</p>
</form>
<body>
</body>
</html>
but my problem is how to create a result_page.php to show the selected entries or data base on the selected checkbox so i can create a comparison page. I have this as being my result_page.php but nothing is showing up. I know Im doing something wrong but I cant find out.
<?php
error_reporting(E_ALL);
$host = 'localhost';
$user = '******';
$pass = '******';
$dbname = '******';
$connection = mysql_connect($host,$user,$pass) or die (mysql_errno().": ".mysql_error()."<BR>");
mysql_select_db($dbname);
$sql = "SELECT * FROM products WHERE prodID IN (";
foreach ($_POST['number'] as $product) $sql .= "'" . $product . "',";
$sql = substr($sql,0,-1) . ")";
$result = mysql_query($sql);
while ($myrow = mysql_fetch_array($result))
{
echo "<table border=1>\n";
echo "<tr><td>Name</td><td>Position</td></tr>\n";
do {
printf("<tr><td>%s %s</td><td>%s</tr>\n", $myrow["1"], $myrow["2"], $myrow["3"]);
} while ($myrow = mysql_fetch_array($result));
echo "</table>\n";
}
?>
A quick glance, the section that generates the output is not correct. You have looped two times for no apperant reason.
while ($myrow = mysql_fetch_array($result)) //<========remove this line
{ //<========remove this line
echo "<table border=1>\n";
echo "<tr><td>Name</td><td>Position</td></tr>\n";
do {
printf("<tr><td>%s %s</td><td>%s</tr>\n", $myrow["1"], $myrow["2"], $myrow["3"]);
} while ($myrow = mysql_fetch_array($result));
echo "</table>\n";
} //<========remove this line
This is done by human parse, but should serves as a starting point.
And to recap tadman, no this is not a good tutorial. And normally you won't need to do printf for the output.

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