how to save runtime created table values into database - php

I create one form when i enter number of Rows & Columns then that number of rows & Column table would be generated i want save values entered into that table into database.Please anyone can help me..
My PHP CODE:
<?php
global $Hostname;
global $Username;
global $Password;
global $Database_name;
function getConnection()
{
$Hostname = "localhost";
$Username ="root";
$Password ="";
$Database_name="labdata";
$oMysqli = new mysqli($Hostname,$Username,$Password,$Database_name);
return($oMysqli);
}
if(isset($_POST['submit']))
{
echo "<table border='1' align='center'>";
for($iii = 0;$iii <$_POST['column'];$iii++)
{
echo "<tr>".$jjj."</tr>";
for($jjj = 0; $jjj <$_POST['rows'];$jjj++) //loop for display no. of rows.
{
echo "<td>" ."<input type=\"text\" name='$iii'>"."</td>";
}
}
echo "</table>";
echo"<form name=\"aa\" method=\"post\">";
echo "<input type=\"submit\" name=\"save\" value=\"save\">";
echo "</form>";
}
$TestName = $_POST['testname'];
$Result = $_POST['result'];
$Unit = $_POST['unit'];
$NormalRange = $_POST['NormalRange'];
if(isset($_POST['save']))
{
$InsertQuery = "INSERT INTO rct(testname,result,unit,NormalRange) VALUES('$TestName','$Result','$Unit','$NormalRange')";
$oMysqli= getConnection();
$oMysqli->query($InsertQuery);
print_r($InsertQuery);exit();
while($Row = $InsertQuery->fetch_array())
{
$TestName = $Row['testname'];
$Result = $Row['result'];
$Unit = $Row['unit'];
$NormalRange = $Row['NormalRange'];
}
}
?>
<html>
<head>
<title>Rct</title>
</head>
<body>
<form name='abc' method="post">
<label for='Table'>Define Table</label>
<label for='rows'>Row</label>
<input type="text" name="column"></input>
<label for='column'>Column</label>
<input type="text" name="rows"></input>
<input type="submit" name="submit" value="submit" onclick="aaa()">
</form>
</body>
</html>

There are many problems with your code:
The table containing <input/>s should be inside <form name='aa'>.
The inputs that are in the table should be named something like $iii[].
The [] tells PHP to create an array of all the rows in $_POST, so you can access $_POST[0][0] for row 0, col 0, $_POST[1][2] for row 2, col 1, etc.
You seem to have rows and columns backwards.
All your <input/> tags are missing the /.
There's no form with inputs named testname, result, unit, NormalRange, so why are you accessing these $_POST values?
fetch_array() can only be used after a SELECT query. INSERT doesn't return any values.
Probably other things I missed.

First of all you must correct the input tag
Replace
<input type="text" name="column"></input>
To
<input type="text" name="column" />
Second the upper loop must be row loop instead of colum as column sit inside the row
if(isset($_POST['submit']))
{
echo"<form name=\"aa\" method=\"post\">";
echo "<table border='1' align='center'>";
for($iii = 0;$iii <$_POST['rows'];$iii++)
{
echo "<tr>";//start Row here
for($jjj = 0; $jjj <$_POST['column'];$jjj++) //loop for display no. of rows.
{
echo "<td>" ."<input type=\"text\" name='".$iii.$jjj."'>"."</td>";//all TDs must be inside the row
}
echo "</tr>";//end row here
}
echo "</table>";
echo "<input type=\"submit\" name=\"save\" value=\"save\">";
echo "</form>";
}
And of course table must be inside the form.
Let me know if this helps

Related

PHP: Generating a html table through a form [duplicate]

This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 5 months ago.
This is practice coursework for my Informatics course. We've just started programming in PHP and our task is to program a website that generates tables for the user.
password prompt
ask the user how many rows and columns the table should have
based on the last question; create a form with the same amount of input boxes
generate the table with the input of step 3
I've accomplished everything until step 4. The user can input data in the form, but I the problem is that when I try to generate the table, PHP will show me this error message: "Undefined index: rows on line 70".
As I described earlier I'm just about to learn PHP, so there may be many "not so very nice programming approaches"; therefore I'm open to all kinds of recommendations.
<!DOCTYPE html>
<html>
<body>
<form method="post" target="">
<label for="login">User: </label>
<input name="login">
<br />
<label for="password">Password: </label>
<input name="password" type="password">
<br />
<input type="submit" name="generate" value="Login" />
</form>
<?php
if (isset($_POST['generate'])) {
$username = $_POST['login'];
$password = $_POST['password'];
$hashed_username = sha1($username);
$hashed_password = sha1($password);
$correct_username = '9d6035e25958ec12fca7ec76d68c8daaf4815b9b'; //wims
$correct_password = 'a94a8fe5ccb19ba61c4c0873d391e987982fbbd3'; //test
if ($hashed_username != $correct_username)
die("Wrong user name!");
if ($hashed_password != $correct_password)
die("Wrong password!");
echo "How many rows and columns should the table have? <br />";
echo('
<form method="POST" target="">
Rows: <input type="number" name="rows" min="1" max="100"/><br/>
columns: <input type="number" name="columns" min="2" max="100"/><br/>
<input type="submit" name="generate1" value="Generate Table" />
</form>');
}
if (isset($_POST['generate1'])) {
$rows = $_POST['rows'];
$columns = $_POST['columns'];
global $rows, $columns;
if ($rows > 100 || $rows < 1)
die("Nope!");
if ($columns > 100 || $columns < 2)
die("Nope!");
echo '<form method="POST" target="">';
echo "<table>";
for ($a=1;$a<=$rows;$a++) {
echo "<tr>";
for ($b=0;$b<=$columns;$b++) {
if ($b==0)
echo "<td>$a. Row</td>";
else {
$c = $a . $b;
echo "<td><input type='text' name='$c' /></td>";
}
}
echo "</tr>";
}
echo "</table>";
echo "<input type='submit' name='generate2' value='Generate' />";
echo "</form>";
}
if (isset($_POST['generate2'])) {
echo "<table>";
for ($a=1;$a<=$GLOBALS['rows'];$a++) {
echo "<tr>";
for ($b=0;$b<=$GLOBALS['columns'];$b++) {
if ($b==0)
echo "<td>$a. row</td>";
else {
$c = $a . $b;
echo "<td>$_POST[$c]</td>";
}
echo "</tr>";
}
echo "</table>";
}
}
?>
</body>
</html>
You need to store your $rows and $columns in $_SESSION variables. With $Globals, I assume you cannot reach to that point, and you get the warning at this point: for ($a=1;$a<=$GLOBALS['rows'];$a++), because $GLOBALS are not declared the second time you reload the page by submitting the second form.
In fact, as W3Schools states, "$GLOBALS is a PHP super global variable which is used to access global variables from anywhere in the PHP script", while a "session is a way to store information (in variables) to be used across multiple pages." When you submit the pages for the second time, you are in fact refreshing the page, and here Globals are not a pick for you access your rows and columns. Instead you should use sessions to store your $_POST['rows'] and $_POST['columns'].
So, try the following instead. Start Session and then declare new $_Session variables for your $_POST['rows'] and $_POST['columns']. Then voila, the problem is solved.
IMPORTANT: add session_start(); at the top of your page. The very first line.
if (isset($_POST['generate'])) {
$username = $_POST['login'];
$password = $_POST['password'];
$hashed_username = sha1($username);
$hashed_password = sha1($password);
$correct_username = '9d6035e25958ec12fca7ec76d68c8daaf4815b9b'; //wims
$correct_password = 'a94a8fe5ccb19ba61c4c0873d391e987982fbbd3'; //test
if ($hashed_username != $correct_username)
die("Wrong user name!");
if ($hashed_password != $correct_password)
die("Wrong password!");
echo "How many rows and columns should the table have? <br />";
echo('
<form method="POST" target="">
Rows: <input type="number" name="rows" min="1" max="100"/><br/>
columns: <input type="number" name="columns" min="2" max="100"/><br/>
<input type="submit" name="generate1" value="Generate Table" />
</form>');
}
if (isset($_POST['generate1'])) {
$rows = $_POST['rows'] ?? '';
$columns = $_POST['columns'] ?? '';
$_SESSION['rows'] = $rows;
$_SESSION['columns'] = $columns;
global $rows, $columns;
if ($rows > 100 || $rows < 1)
die("Nope!");
if ($columns > 100 || $columns < 2)
die("Nope!");
echo '<form method="POST" target="">';
echo "<table>";
for ($a = 1; $a <= $rows; $a++) {
echo "<tr>";
for ($b = 0; $b <= $columns; $b++) {
if ($b == 0)
echo "<td>$a. Row</td>";
else {
$c = $a . $b;
echo "<td><input type='text' name='$c' /></td>";
}
}
echo "</tr>";
}
echo "</table>";
echo "<input type='submit' name='generate2' value='Generate' />";
echo "</form>";
}
if (isset($_POST['generate2'])) {
echo "<table>";
$row = $_SESSION['rows'] ?? '';
$columns = $_SESSION['columns'] ?? '';
for ($a = 1; $a <= $row; $a++) {
echo "<tr class='border: 1px solid #BDBDBD'>";
for ($b = 0; $b <= $columns; $b++) {
if ($b == 0)
echo "<td style='border: 1px solid #BDBDBD'>$a. row</td>";
else {
$c = $a . $b;
echo "<td style='border: 1px solid #BDBDBD'>$_POST[$c]</td>";
}
echo "</tr>";
}
echo "</table>";
}
session_destroy();
}
Your code here is the problem
if (!isset($_POST['generate1']))
die('');
Here you are checking if $_POST['generate1] is set, if it is not then die (halt/terminate execution of the script)
php die();
So when you submit your second form (submit has name of generate2) then the above check will fail (it is not set so it will die(); and end execution of your script.
if (isset($_POST['generate1'])) {
// Show the form....
}
Do this for both the generate1 and generate2 and it will only execute that code if the if statements evaluates to true.
instead using "if (!isset($_POST['generate1']))" change it into "if(isset($_POST['generate1']))".When you click generate2 it will be die because that condition is not fulfilled.
<!DOCTYPE html>
<html>
<body>
<form method="post" target="">
<label for="login">User: </label>
<input name="login">
<br />
<label for="password">Password: </label>
<input name="password" type="password">
<br />
<input type="submit" name="generate" value="Login" />
</form>
<?php
if (isset($_POST['generate'])) {
$username = $_POST['login'];
$password = $_POST['password'];
$hashed_username = sha1($username);
$hashed_password = sha1($password);
$correct_username = '9d6035e25958ec12fca7ec76d68c8daaf4815b9b'; //wims
$correct_password = 'a94a8fe5ccb19ba61c4c0873d391e987982fbbd3'; //test
if ($hashed_username != $correct_username)
die("Wrong user name!");
if ($hashed_password != $correct_password)
die("Wrong password!");
echo "How many rows and columns should the table have? <br />";
echo('
<form method="POST" target="">
Rows: <input type="number" name="rows" min="1" max="100"/><br/>
columns: <input type="number" name="columns" min="2" max="100"/><br/>
<input type="submit" name="generate1" value="Generate Table" />
</form>');
}
if (isset($_POST['generate1'])){
$rows = $_POST['rows'];
$columns = $_POST['columns'];
if ($rows > 100 || $rows < 1)
die("Nope!");
if ($columns > 100 || $columns < 2)
die("Nope!");
echo "<form method='POST' target=''>";
echo "<input type='hidden' name='row' value='$rows'/>";
echo "<input type='hidden' name='column' value='$columns'/>";
echo "<table>";
for ($a=1;$a<=$rows;$a++) {
echo "<tr>";
for ($b=0;$b<=$columns;$b++) {
if ($b==0)
echo "<td>$a. Row</td>";
else {
$c = $a . $b;
echo "<td><input type='text' name='$c' /></td>";
}
}
echo "</tr>";
}
echo "</table>";
echo "<input type='submit' name='a' value='Generate' />";
echo "</form>";
}
if (isset($_POST['a'])) {
$rows = $_POST['row'];
$columns = $_POST['column'];
echo "<table border='1'>";
for ($a=1;$a<=$rows;$a++) {
echo "<tr>";
for ($b=0;$b<=$columns;$b++) {
if ($b==0){
echo "<td>$a. row</td>";
}else {
$c = $a . $b;
echo "<td>$_POST[$c]</td>";
}
}
echo "</tr>";
}
echo "</table>";
}
?>
</body>
</html>
First
Your primary need is to READ THE MANUAL for all the things you're doing, thus you will see that the sha1() Manaul page states:
Warning
It is not recommended to use this function to secure passwords, due to the fast nature of this hashing algorithm.
You REALLY should be fixing this issue.
Anyway, your issue is:
Undefined index: rows on line 70
Which (I guess, because you didn't indicate in your question) is this line:
for ($a=1;$a<=$GLOBALS['rows'];$a++) {
This means that $GLOBALS key rows doesn't exist. Why? All PHP data is generated when a script is executed; before ANY script starts, the PHP knows NOTHING, there is never any incoming data at the start if the PHP script.
Some people here might shout and scream "SESSIONS!!" but even the $_SESSION array is empty at the start of the script, until the PHP code has read the stored session data in the cookie key.
So how do you populate $GLOBALS? What you did was not far off, but you ran the form and submitted the data to $_POST['generate1'] which worked, and this populated the data, but this presented a form to the end user so that user then had to resubmit the form, and by default that reloads the page, therefore restarting the PHP script from zero again, so all data in $GLOBALS is forgotten.
How do you make PHP "remember" data when loading a page? In general there are several ways; all of them have positive and negative sides:
Database. Read and write data to a third party
Sessions. Read and write data to a file/database associated with that specific client only.
Form data, reading data from a submitted form or via URL parameters (GET/POST).
Using (3) is probably easiest for you; so when you run the $_POST['generate1'] you need to add hidden inputs to your form so your "part 2" form can then pass on this data to "part3" form ($_POST['generate2']) .
And that's the data you need to read, not the GLOBALS.

Can't figure out how to process delete button when it is clicked

I am trying to add a 'delete' button on my item table and have a delete button to delete the item and the information about the both item and the seller. I have the delete button on the table but I cannot figure out how to process that button when it is clicked. Please help! Thank you in advance!!
<?php
require 'authentication.inc';
// connect to the server
$connection = sqlsrv_connect( $hostName, $connectionInfo )
or die("ERROR: selecting database server failed");
// prepare SQL query
$UserID = $_SESSION['userID'];
$query = "SELECT * FROM ITEM WHERE userID= '$UserID'";
// Execute SQL query
$query_result = sqlsrv_query($connection, $query)
or die( "ERROR: Query is wrong");
// Output query results: HTML table
echo "<table border=1>";
echo "<tr>";
// fetch attribute names
foreach( sqlsrv_field_metadata($query_result) as $fieldMetadata)
echo "<th>".$fieldMetadata['Name']."</th>";
echo "</tr>";
// fetch table records
while ($line = sqlsrv_fetch_array($query_result, SQLSRV_FETCH_ASSOC)) {
echo "<tr>\n";
foreach ($line as $cell) {
echo "<td> $cell </td>";
}
echo "<td></td>";
echo "</tr>\n";
}
echo "</table>";
// close the connection with database
sqlsrv_close($connection);
?>
To add a delete feature you need two things, a button and secondly the processing of said button. I am not sure what your unique value is in the table, so I am using this fictitious key: itemID. Update with whatever your unique column name is.
Replace your table loop with:
<table border=1>
<tr>
<?php
// fetch attribute names
foreach( sqlsrv_field_metadata($query_result) as $fieldMetadata)
echo "<th>".$fieldMetadata['Name']."</th>"; ?>
</tr>
<?php
// fetch table records
while ($line = sqlsrv_fetch_array($query_result, SQLSRV_FETCH_ASSOC)) { ?>
<tr>
<?php foreach ($line as $cell) {
echo "<td> $cell </td>";
} ?>
<td>
<form method="post">
<input type="hidden" name="itemID" value="<?php echo $line['itemID']; ?>" />
<input type="submit" name="action" value="DELETE" />
</form>
</td>
</tr>
<?php } ?>
</table>
In the processing portion at the top, add processing:
if(!empty($_POST['action']) && ($_POST['action'] == 'DELETE')) {
// Do some sort of validation here
if(is_numeric($_POST['itemID']))
sqlsrv_query($connection, "delete from ITEM where itemID = '".$_POST['itemID']."'");
}

Deleting something in a mysqli database with a button?

So I'm just making a simple program that puts names into a database. I got that part down, I can enter a name into a form, then display it on the page, but now I'd like to know how to delete them from the database, and no longer show them on the page.
I added a button next to each name that triggers the third if statement (with the commented out query), and from what I can tell it's best to run a query based on the element's id (my primary key that auto increments), but I have no idea how to get the id from the element who's button I'm clicking on.
How do I get the id from one of the elements in my while loop? Or if there's a better way to delete them, what's that?
if (mysqli_connect_errno()) {
die('could not connect');
}
if (isset($_POST['first_name'], $_POST['last_name'])){
$first_name = trim($_POST['first_name']);
$last_name = trim($_POST['last_name']);
$putitin = mysqli_query($db, "INSERT INTO names (first_name, last_name) VALUES ('$first_name', '$last_name')");
}
if (isset($_POST['del'])){
//$takeitout = mysqli_query($db, "DELETE FROM names WHERE id = ");
}
?>
<html>
<head>
</head>
<body>
<form action='' method='post'>
<div>
<label for "first_name">First name</label>
<input type="text" name="first_name">
</div>
<div>
<label for "last_name">Last name</label>
<input type="text" name="last_name">
</div>
<div>
<input type="submit" value="Insert">
</div>
</form>
<hr>
<?php
$resultset = $db->query('SELECT * FROM names');
if($resultset->num_rows != 0){
while($rows = $resultset->fetch_assoc()) {
$fname = $rows['first_name'];
$lname = $rows['last_name'];
$id = $rows['id'];
echo "<form action='' method='post'><p>Name: $fname $lname $id<input type='submit' name='del'></form></p>";
}
} else {
echo 'No results';
}
?>
</body>
</html>
This is one way.
change your html part to
<form action='' method='post'>
<input type='hidden' name='id' value='$id' />
<p>Name: $fname $lname $id
<input type='submit' name='del' value=''>
</form></p>
and your php
if (isset($_POST['del'])){
$id = $_POST['id'];
$takeitout = mysqli_query($db, "DELETE FROM names WHERE id = '$id'");
}
Note:
What you can do is to put all your input fields inside your while loop. Then assign values to each of them, but we have to use array to store them accordingly.
We can use checkbox to store the IDs.
What will happen, is user can select from the list of names they wanted to delete by ticking the corresponding checkbox, then pressing the Delete button below.
Your code
<form action="" method="POST">
<?php
$resultset = $db->query('SELECT * FROM names');
if($resultset->num_rows != 0){
while($rows = $resultset->fetch_assoc()) {
$fname = $rows['first_name'];
$lname = $rows['last_name'];
$id = $rows['id'];
echo '<input type="checkbox" name="id[]" value="'.$id.'">'.$fname.' '.$lname.'<br>';
} /* END OF WHILE LOOP */
?>
<input type="submit" value="Delete" name="delete">
</form>
And your PHP that will process the form:
<?php
if(isset($_POST["delete"])){
$counter = count($_POST["id"]);
for($x = 0; $x<$counter; $x++){
if(!empty($_POST["id"][$x])){ /* CHECK IF AN ITEM IS SELECTED */
/* DELETE QUERY */
if($stmt = $db->prepare("DELETE FROM names WHERE id = ?")){
$stmt->bind_param("i",$_POST["id"][$x]);
$stmt->execute();
$stmt->close();
} /* END OF PREPARED STATEMENT */
} /* END OF IF; CHECKING IF IT IS SELECTED */
} /* END OF FOR LOOP */
} /* END OF ISSET DELETE */
?>

How to write an update SQL statement to update multiple records

I have this code so far, which reads a simple table with 3 varchar fields:
<?php
//db connection code...
// select database
mysql_select_db($db) or die ("Unable to select database!");
// create query
$query = "SELECT * FROM Sheet1";
// execute query
$result = mysql_query($query) or die ("Error in query: $query. ".mysql_error());
// see if any rows were returned
if (mysql_num_rows($result) > 0) {
// yes
// see if any rows were returned
if (mysql_num_rows($result) > 0) {
// yes
// print them one after another
echo "<html><body><table cellpadding=10 border=1>";
while($row = mysql_fetch_assoc($result)) {
echo "<tr>";
echo "<td>".$row['stickerID']."</td>";
echo "<td>" .$row['stickerName']."</td>";
echo "<td>".$row['stickerSection']."</td>";
echo "<td>"?>
<form name="some form" action="editform.php" method="post">
<input type="checkbox" name="<?php echo $row['stickerID'] ?>" value=" <?php echo $row['stickerStatus'] ?> ">
<?php "</td>";
echo "</tr>";
}
echo "</table></body></html>";
echo " " ?>
<input type="submit" name="editWish" value="Edit">
</form>
<?php " ";
} else {
// no
// print status message
echo "No rows found!";
}
// free result set memory
mysql_free_result($result);
// close connection
mysql_close($connection);
?>
The database has 4 fields, 3 varchar and 1 int with current value of 0. I checked the page source code and confirmed each checkbox name is the stickerID. Now I will post this to the editform.php which I must create. What Im wondering is how should I write the update sql so that it takes into account each new value selected by the user in the form?
This is my idea, but how to I do it for every checkbox?
editform.php
<?php
//update multiple records
//UPDATE user_items SET stickerStatus = $_POST["stickerStatus"] WHERE stickerID = $_POST["stickerID"];
?>
First question: use mysql_fetch_assoc() instead of mysql_fetch_row(). That will return an associative array instead of an enumerated one.
Second question: read up on HTML forms and form handling.
The answer to the question in the comments:
// The <form> tag should only be echoed once.
echo '<form name="some form" action="editform.php" method="post">';
while($row = mysql_fetch_assoc($result)) {
echo "<tr>";
echo "<td>".$row['stickerID']."</td>";
echo "<td>" .$row['stickerName']."</td>";
echo "<td>".$row['stickerSection']."</td>";
echo "<td>"?>
<input type="hidden" name="status_<?php echo $row['stickerID"; ?>" value="0">
<input type="checkbox" name="status_<?php echo $row['stickerID'] ?>" value="<?php echo $row['stickerStatus'] ?> ">
<?php "</td>";
echo "</tr>";
}
// You need a submit button to send the form
echo '<input type="submit">';
// Close the <form> tag
echo '</form>';
Using a hidden input with the same name as the checkbox makes sure a value for the given input name is sent to the server. The value of a checkbox that's not checked will not be sent. In that case the hidden input will be used.
You can get the submitted values in editform.php as follows:
<?php
foreach ($_POST as $field => $value) {
if (strpos($field, 'status_')) {
// Using (int) makes sure it's cast to an integer, preventing SQL injections
$stickerID = (int) str_replace('status_', '', $field);
// Again, preventing SQL injections. If the status could be a string, then use mysql_real_escape_string()
$stickerStatus = (int) $value;
// Do something with the results
}
}
Do
print_r($row)
to find out exactly how your row arrays are constructed and work from there.
For your comparison operator, use
$row[3] === 0
instead of
$row[3] == 0
This will return true if both the value and data type match rather than just the value.
0 can mean the Boolean false aswell as the numeric value 0

how to print the checklist array value

I want to print the checklist array value, here checklist is dynamically created for each row of Food table. How I can do that ? If I click submit no value is submitted I guess so, How to use submit button? In fact where to use ..plz help me
<?php // File: anyco.php
require('anyco_ui.inc.php');
// Create a database connection
$conn = oci_connect('system','123','localhost/orcl');
ui_print_header('FoodItemList');
//session_start();
//$cid=$_SESSION['cid'];
//do_query($conn, 'SELECT Fooditem_ID,Food_item_name,price,day_available,time_available,discount_percentage,start_date,deadline FROM Food_Item');
ui_print_footer(date('Y-m-d H:i:s'));
function do_query($conn, $query)
{
$stid = oci_parse($conn, $query);
$r = oci_execute($stid,OCI_DEFAULT);
print '<table border="1">';
print '<tr>';
print '<td>Food_ID<td>Food_Name<td>Price(tk)<td>Dvailable_day<td>Avaliable_time<td>Discount<td>Dis_start date<td>Dis_finish date<td>selected item<td>quanity';
while ($row = oci_fetch_array($stid, OCI_ASSOC+OCI_RETURN_NULLS))
{
print '<tr>';
$num=1;
$val="";
foreach ($row as $item)
{
if($num==1)
{
$val = $item;
$num=2;
}
print '<td>'.($item!== null ? htmlentities($item) : ' ').'</td>';
}
//echo$val;
echo '<td><input type="checkbox" name = "invite[]" value="$val>" </td>';
echo '<td><input type="number" name = "name[]" ></td>';
print '</tr>';
}
print '</table>';
}
if(isset($_POST['submit']))
{
if (is_array($_POST['invite']))
{
foreach($_POST['invite'] as $key=>$name)
{
echo $key, '=>', $name,'<br/>';
//Here $key is the array index and $name is the value of the checkbox
}
}
}
?>
<html>
<style>
body
{
background:orange;
}
</style>
<body text="green">
<br><br>
<form method="post">
<?php do_query($conn, 'SELECT Fooditem_ID,Food_item_name,price,day_available,time_available,discount_percentage,start_date,deadline FROM Food_Item'); ?>
<input type ="submit" value="submit" name="submit"><br><br>
</form>
</body>
</html>
Your form doesn't have any inputs inside it because you are calling your function way before you print out the form tags.
This line needs to move to within the form:
do_query($conn, 'SELECT Fooditem_ID,Food_item_name,price,day_available,time_available,discount_percentage,start_date,deadline FROM Food_Item');
So that the code looks like this, and the whole table with its inputs is printed inside the form tags. I also added a name for the submit button:
<form method="post">
<?php do_query($conn, 'SELECT Fooditem_ID,Food_item_name,price,day_available,time_available,discount_percentage,start_date,deadline FROM Food_Item'); ?>
<input type ="submit" value="submit" name="submit"><br><br>
</form>
Finally, when you echo values into your inputs, there are two syntax errors in this line:
echo '<td><input type="checkbox" name = "invite[]" value="$val>" </td>';
You can echo a variable into a string in several ways - here are two. 1) if the variable is between single quotes, PHP will automatically convert the variable to its value. In this case, it's double quotes, so no go. 2) You start your echo command with single quotes, so you can break out of the string with a single quote, concatenate the variable with a period, and then break back in to the string with another single quote, like this:
echo '<td><input type="checkbox" name = "invite[]" value="' . $val . '"> </td>';

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