I want to retrieve the input value entered by the user and send it to the database. But for some reason it inserts a blank space in the table, instead of the value! Does someone know what's wrong?
<form method="post" name="name" action="pt2.php" >
<?php
//$mysql->commit();
echo "<h3>";
echo "Please enter the name for each seat:<br><p> </p>";
echo "";
foreach($_POST['seats'] AS $seat) {
$rowId = substr($seat, 0, 1);
$columnId = substr($seat, 1);
echo $rowId . $columnId . '<input type="hidden" name="seats[]" value="' . $seat . '"><input name="' . $seat . 'name" type="text"/></br>';
}
?>
<input type="submit" name="submit" value="Submit Form"><br>
</form>
pt2.php
<?php
// Connect to MySQL
mysql_connect("localhost", "root", "root") or die("Connection Failed");
mysql_select_db("tickets")or die("Connection Failed");
$namei = $_POST[$seat . 'name'];
foreach ($_POST['seats'] as $seat){
echo $seat;
echo $namei;
$query = "INSERT INTO seatnames (seatname) VALUES ('$namei')";
mysql_query($query) or die(mysql_error());
}
?>
Related
I'm trying to get the selected value from a dropdown list.
The list has name 'select_employee', when I press the button with name 'save' I hope to get the value. I'm using a POST to to get the value.
I get the error 'Undefined index: select_employee'.
<div class="form-group">
<h2>Enter Certified Course Details</h2>
<?php
// start of connect db
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "TrainingDB";
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//end of connect db
?>
<form id="form" action="" method="post">
<?php
// Drop down list employee
$sql = "SELECT * FROM employees";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
echo "Select Employee<br>";
echo "<select name='select_employee' id='select_employee'>";
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<option value='" . $row['id'] . "'>" . $row['fname'] . " " . $row['sname'] . " </option>";
}
echo "</select>";
echo "<br>";
}
?>
</form>
<?php
if(isset($_POST['save'])){
echo "save<br>";
$employee=$_POST['select_employee']; // error here
echo "Selected Employee" . $employee . "<br>";
}
?>
<form Employee="/employee_page.php" method="post">
<button type="submit" class="btn btn-primary" name="save" value="save">Save</button>
</form>
</div>
You are using two different form so make sure to use only one form.
I have a database table with fields Name, EmailAddress, Qualification I want to perform the search using name, emailaddress, qualification and need to display the user details in my web page can anybody tell how can I do it?
<?php
$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//$name=$_POST['name'];
//$email=$_POST['email'];
//$qualification=$_POST['qualify'];
$sql = "SELECT * FROM form WHERE Name ='kumar' OR EmailAddress = 'kumar#gmail.com' OR Qualification = 'BE' ";
$result=$conn->query($sql);
while($row = $result->fetch_assoc())
{
echo 'Name: '.$row['Name'];
echo '<br /> EmailAddress: ' .$row['EmailAddress'];
echo '<br /> Qualification: '.$row['Qualification'];
echo '<br /> DOB: '.$row['DOB'];
}
mysql_close($con);
$con = mysql_connect ("localhost", "root", "");
mysql_select_db ("myDB", $con);
if (!$con) { die ("Could not connect: " . mysql_error()); }
$sql = mysql_query("SELECT * FROM search WHERE name LIKE '%arun%' OR EmailAddress LIKE '%arun%' OR Qualification LIKE '%arun%' ");
$con->query($sql);
if(count($sql)>0 || $sql !=NULL){
while ($row = mysql_fetch_array($sql, MYSQL_ASSOC))
{
echo 'Name: '.$row['name'];
echo '<br /> Email: ' .$row['email'];
echo '<br /> Address: '.$row['address'];
}
}
else{
echo 'your error here';
}
mysql_close($con);
Use PDO, don´t use mysql or even mysqli. Its not even supported anymore in the latest PHP versions.
$host = 'localhost';
$dbname = 'mydb';
$username = 'root';
$password = 'password';
try {
// create the connection
$conn = new PDO('mysql:host=' . $host . ';dbname=' . $dbname . ';charset=utf8', 'root', 'password');
// set the errmode to exception, set this to ERRMODE_SILENT if you want to hide database errors
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
} catch (PDOException $e) {
echo $e->getMessage(); // catch any connection errors
$conn = false;
}
$select_data = $conn->prepare("SELECT * FROM search WHERE name LIKE :name OR EmailAddress LIKE :email OR Qualification LIKE :qualification ");
$select_data->bindValue(':name', $name); // bind the values to the paramaters
$select_data->bindValue(':email', $email);
$select_data->bindValue(':qualification', $qualification);
$select_data->execute();
if($select_data->rowcount() > 0){ // rowcount returns the amount of results
// atleast 1 result
$results = $select_data->fetchAll(PDO::FETCH_ASSOC); // fetch the results into an array
foreach($results as $row){
echo 'Name: ' . $row['name'];
echo '<br /> Email: ' . $row['email'];
echo '<br /> Address: ' . $row['address'];
}
}
There are other ways to do this with PDO but this is how I tend to do it.
Please try the below code:
<form id="form" action="" method="POST" >
<fieldset>
<label>First Name </label>
<input type="text" placeholder="Name" value="<?php if (isset($_REQUEST['name'])) echo $_REQUEST['name']; ?>" class="form-control required" id="name" name="name">
<label>EmailAddress</label>
<input type="text" placeholder="EmailAddress" value="<?php if (isset($_REQUEST['EmailAddress'])) echo $_REQUEST['EmailAddress']; ?>" class="form-control required" id="EmailAddress" name="EmailAddress">
<label>Qualification </label>
<input type="text" class="form-control" placeholder="Qualification" value="<?php if (isset($_REQUEST['qualification'])) echo $_REQUEST['qualification']; ?>" id="Qualification" name="Qualification">
<br><input value="Search" name="Search" style="width:100%" type="submit" class="btn btn-success">
</fieldset>
</form>
<?php
if (isset($_POST['Search'])) {
$con = mysql_connect("localhost", "root", "");
mysql_select_db("myDB", $con);
if (!$con) {
die("Could not connect: " . mysql_error());
}
$where = '';
if ($_POST['name']) {
$where .="name like '%" . $_POST['name'] . "%'";
}
if ($_POST['EmailAddress']) {
if (!empty($where))
$where.=" or ";
$where .="email like '%" . $_POST['EmailAddress'] . "%'";
}
if ($_POST['Qualification']) {
if (!empty($where))
$where.=" or ";
$where .="qualification like '%" . $_POST['Qualification'] . "%'";
}
$sql = "select id, name, email, qualification from student where " . $where;
echo $sql;
$res = mysql_query($sql) or die("Error in query " . mysql_error());
while ($row = mysql_fetch_array($res, MYSQL_ASSOC)) {
echo "<br><br>";
echo 'No. : ' . $row['id'];
echo '<br /> Name: ' . $row['name'];
echo '<br /> Email: ' . $row['email'];
echo '<br /> Qualification: ' . $row['qualification'];
}
mysql_close($con);
}
?>
I hope this will work for you as you want.
I have gone over my code many many MANY times and added any missing brackets or semi-colons but still whenever I upload this code to my website and load the page I still get a completely blank screen. The code was from an O'Reilly book so I went and checked the website if there are any reported errors in the book but found nothing related to this particular example.
I don't feel like it's an issue with permissions because I think the page would at least report one of the errors I coded into it. Could it have to do with the versions of PHP or MySQL I am using? I was able to connect to the database in the past and query it but writing just isn't happening. I am at a complete loss at this point. All I want to do is write to my MySQL database and party :(
Here is the code:
<?php
require_once 'login.php';
// Create connection
$db_server = mysql_connect($db_hostname, $db_username, $db_password);
// Check connection
if (!$db_server) die("Unable to connect to MySQL: " . mysql_error());
mysql_select_db($db_database, $db_server)
or die("Unable to select database: " . mysql_error());
if (isset($_POST['delete']) && isset($_POST['avail']))
{
$avail = get_post('avail');
$query = DELETE FROM test WHERE avail='$avail';
if (!mysql_query($query, $db_server))
echo "DELETE failed: $query<br />" .
mysql_error() . "<br /><br />";
}
if (isset($_POST['id']) &&
isset($_POST['item_name']) &&
isset($_POST['avail']))
{
$id = get_post('id');
$item_name = get_post('item_name');
$avail = get_post('avail');
$query = "INSERT INTO test VALUES" .
"('$id','$item_name','$avail')";
if (!mysql_query($query, $db_server))
{echo "INSERT failed: $query<br />" .
mysql_error() . "<br /><br />";
}
}
?>
<form action="index.php" method="post"><pre>
Line Number <input type="text" name="id" />
Product Name <input type="text" name="item_name" />
Quantity Available <input type="text" name="avail" />
<input type="submit" value="ADD RECORD" />
</pre></form>
<?php
$query = "SELECT * FROM test";
$result = mysql_query($query);
if (!$result) die ("Database access failed: " . mysql_error());
$rows = mysql_num_rows($result);
for ($j = 0 ; $j < $rows ; ++$j)
{
$row = mysql_fetch_row($result);
?>
<pre>
Line Number $row[0]
Product Name $row[1]
Quantity Available $row[2]
</pre>
<form action="index.php" method="post">
<input type="hidden" name="delete" value="yes" />
<input type="hidden" name="avail" value="$row[2]" />
<input type="submit" name="DELETE RECORD" /></form>
<?php
}
mysql_close($db_server);
function get_post($var)
{
return mysql_real_escape_string($_POST[$var]);
}
?>
you have error in the delete statment , try out this code :
<?php
require_once 'login.php';
// Create connection
$db_server = mysql_connect($db_hostname, $db_username, $db_password);
// Check connection
if (!$db_server) die("Unable to connect to MySQL: " . mysql_error());
mysql_select_db($db_database, $db_server)
or die("Unable to select database: " . mysql_error());
if (isset($_POST['delete']) && isset($_POST['avail']))
{
$avail = get_post('avail');
$query = "DELETE FROM test WHERE avail='$avail'";
if (!mysql_query($query, $db_server))
echo "DELETE failed: $query<br />" .
mysql_error() . "<br /><br />";
}
if (isset($_POST['id']) &&
isset($_POST['item_name']) &&
isset($_POST['avail']))
{
$id = get_post('id');
$item_name = get_post('item_name');
$avail = get_post('avail');
$query = "INSERT INTO test VALUES" .
"('$id','$item_name','$avail')";
if (!mysql_query($query, $db_server))
{echo "INSERT failed: $query<br />" .
mysql_error() . "<br /><br />";
}
}
?>
<form action="index.php" method="post"><pre>
Line Number <input type="text" name="id" />
Product Name <input type="text" name="item_name" />
Quantity Available <input type="text" name="avail" />
<input type="submit" value="ADD RECORD" />
</pre></form>
<?php
$query = "SELECT * FROM test";
$result = mysql_query($query);
if (!$result) die ("Database access failed: " . mysql_error());
$rows = mysql_num_rows($result);
for ($j = 0 ; $j < $rows ; ++$j)
{
$row = mysql_fetch_row($result);
?>
<pre>
Line Number $row[0]
Product Name $row[1]
Quantity Available $row[2]
</pre>
<form action="index.php" method="post">
<input type="hidden" name="delete" value="yes" />
<input type="hidden" name="avail" value="$row[2]" />
<input type="submit" name="DELETE RECORD" /></form>
<?php
}
mysql_close($db_server);
function get_post($var)
{
return mysql_real_escape_string($_POST[$var]);
}
?>
If no errors are being displayed add the following to the top of your file, it will allow for errors to be shown:
error_reporting(E_ALL);
ini_set('display_errors', '1');
Please ensure that you are displaying errors.
ini_set("display_errors", "1");
You can also create a new page with just
phpinfo();
to check that PHP is running (and what PHP configuration you have).
I'm trying to let the user check off which item to be deleted. When the user check off one or many items and click the Delete button, those data will be erased from the database. I've also added a search box to search for the dvd. The search box works, but the deleting doesn't. This is what it looks like in the browser.
My PHP looks like this (I took out the searching code):
<form action="" method="post">
<p><input type="text" name="search"> <input type="submit" value="Search"></p>
<p><input type="submit" name="deleting" value="Delete"></p>
</form>
<?php
$link = mysqli_connect( $host, $user, $password, $dbname);
if (!$link) {
die('Could not connect: ' . mysqli_connect_error());
}
echo 'Connected successfully<br/>';
//searching code goes here
if (isset ($_POST['deleting']) && isset ($_POST['deleteThese']) )
{
$deleteThese = implode(",", $_POST['deleteThese']);
$queryTwo = "DELETE FROM `$dbname`.`dvds` WHERE `dvds`.`DvdID` IN ($deleteThese)";
$resultTwo = mysqli_query($link, $queryTwo);
}
echo "<table border=\"1\"><tr><th>DvdTitle</th><th>RunningTime</th><th>Delete</th></tr>";
if (mysqli_num_rows($result) == 0)
echo "<tr><td colspan='2'>No records found.</td></tr>";
else {
while ($row = mysqli_fetch_assoc($result)) {
echo "<tr><td>" . $row['DvdTitle'] . "</td>";
echo "<td>" . $row['RunningTime'] . "</td>";
echo "<td>" . "<form>" . "<input type='checkbox' name='deleteThese[]' value='" . $row['DvdID'] . "' >" . "</form>" . "</td></tr>\n";
}
}
echo "</table>";
mysqli_free_result($result);
mysqli_close($link);
?>
Each DvdTitle has an unique Dvd ID, hence the value of each row is the dvd's ID $row['DvdID'].
Adding the parentheses will allow for those ID's to be selected for deletion.
IN($deleteThese)
EDIT
Do not close the form after the submit button. Put that at the end of the code. This will allow the form to include the checkbox values.
<form action="" method="post">
<p><input type="text" name="search"> <input type="submit" value="Search"></p>
<!-- YOUR PHP CODE -->
<p><input type="submit" name="deleting" value="Delete"></p>
</form>
2nd Edit [requested to improve code]
Move the isset on top of the form.
<?php
if (isset ($_POST['deleting']) && isset ($_POST['deleteThese']) )
{
$deleteThese = implode(",", $_POST['deleteThese']);
$queryTwo = "DELETE FROM `$dbname`.`dvds` WHERE `dvds`.`DvdID` IN ($deleteThese)";
$resultTwo = mysqli_query($link, $queryTwo);
}
?>
<form>....
$deletethese might need to have quotes around it.
The following code is not working as the page displays nothing, and I am not exactly sure why. It gets a few things from the URL and then the final Album name from the database. Here is the code:
<?php
$cart1 = rawurldecode($_GET["path"]);
list( , , , , , $cart2) = explode ("\\", $cart1);
$cart3 = $cart2;
list($cart4) = explode (" ", $cart3);
$con = mysql_connect("SERVER","USER","PASS");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("cartmatch", $con);
$result = mysql_query("SELECT * FROM cartmatch WHERE CARTNO='$cart4'");
while($row = mysql_fetch_array($result))
{
echo '<form enctype="multipart/form-data" action="albumgo.php" method="POST"><input name="ID" type="hidden" value=';
echo $_GET["ID"];
echo ' ><input name="enabled" type="hidden" value=';
echo $_GET["enabled"];
echo ' ><input name="artist" type="hidden" value=';
echo $_GET["artist"];
echo ' ><input name="title" type="hidden" value="';
echo $_GET["title"];
echo '" >Name:<br/><input name="album" type="text" autofocus="autofocus" value="';
echo $row['ALBUM'];
echo '" ><input type="submit" name="edit" value="Save"></form>';
}
mysql_close($con);
?>
Try to put the following code to see if there is an error in your script.
ini_set('display_errors', 1);
ini_set('log_errors', 1);
ini_set('error_log', dirname(__FILE__) . '/error_log.txt');
error_reporting(E_ALL);
Probably there is a database connection error or something preventing PHP from displaying the rest of the content.
list($cart4) = explode (" ", $cart3);
Should have been
list($cart4) = explode ("+", $cart3);