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.
Related
Hi there im in the process of making a page to approve and deny names in a database, I currently have it so it finds an pulls the pending requests im looking for now i wanna know how to update the row once i click approve or deny.
the values it needs to update is the NameState column to either ("approved") or ("REJECTED")
there can be any where from 1 to 100 names at a time.
Thanks Ryan
$DB_HOST = "IP";
$DB_USER = "user";
$DB_PASS = "Pass";
$DB_DTBS = "DB";
mysql_connect($DB_HOST, $DB_USER, $DB_PASS) or die(mysql_error()); // Connect to database server(localhost) with username and password.
mysql_select_db($DB_DTBS) or die(mysql_error()); // Select registration database.
$search = mysql_query('SELECT * FROM TABLE WHERE `NameState` = \'(\"PENDING")\'') or die(mysql_error());
$match = mysql_num_rows($search);
if($match > 0){
echo "<table border='1'>
<tr>
<th>ID</th>
<th>Current Name</th>
<th>Requested Name</th>
<th>Approve or Deny</th>
</tr>";
while($row = mysql_fetch_array($search)) {
echo "<tr>";
echo "<td>" . $row['object_id'] . "</td>";
echo "<td>" . $row['Name'] . "</td>";
echo "<td>" . $row['WishName'] . "</td>";
echo '<td> <form method="post" action=<?php echo Approve:<input type="radio" value="Approve" name="name"> Deny:<input type="radio" value="Deny" name="name"><br /> </td>';
echo "</tr>";
}
echo "</table>";
echo '<input type="Submit" value="Submit" name="Submit"> </form>';
} else {
echo("No Names to Approve");
}
?>
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
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.
I created a drop down list that has been populated by the database and now I'm having trouble retrieving the data. Normally, I would know how to retrieve the value of the drop down list if I had to manually name the data, but in this case, I'm not quite sure how I would name it.
Here is my current code:
<h1>Generate Reports</h1>
<form enctype="multipart/form-data" action="http://localhost/yiiFolder/index.php/create" method="post">
<table>
<tr>
<td><strong>Materials</strong></td>
<?php
mysql_connect('host', 'root', 'password');
mysql_select_db ("db");
$sql = "SELECT material_name FROM materials";
$result = mysql_query($sql);
echo "<td><select name='materials'>";
while ($row = mysql_fetch_array($result))
{
echo "<option value='" . $row['material_name'] . "'>" .
$row['material_name'] . "</option>";
}
echo "</select></td></tr> ";
$sql2 = "SELECT location_name From locations";
$result2 = mysql_query($sql2);
?>
<td><strong>Locations</strong></td>
<?php
echo "<td><select name='locations'>";
while ($row2 = mysql_fetch_array($result2))
{
echo "<option value='" . $row2['location_name'] . "'>" .
$row2['location_name'] . "</option>";
}
echo "</select></td></tr>";
?>
<tr>
<td><button name="submit" type=submit>Generate</button></td>
</tr>
</table>
</form>
<?php
$material = $row['material_name'];
$locations = $row2['location_name'];
$generate = $_POST['submit'];
if(isset($generate))
{
echo $material;
echo $locations;
}
?>
You're trying to capture value before the submit button is hit. Also, as Hanky pointed out you're using the wrong names while referring to select data. You should do this instead
if(isset($_POST['submit'])) // this code will run after the button is clicked
{
$material = $_POST['materials']; // and not material_name
$locations = $_POST['locations']; // and not location_name
echo $material;
echo $locations;
}
PS: You're following a very unsecure way of developing a web application. At the very least you need to switch to PDO and always escape the data.
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