Edit row in php - php

I'm trying to build a forum. I have forum.php with a table that includes a row for the title, a row for the edit link and a row for the delete link. When I click the edit link, I am taken to edit.php where I have another form to insert the new topic title. When I click on "save new topic" button the row should be updated, however that's my problem, the title remains the same. I've been searching around this website (and the net in general) to find a solution to this but none seem to be working for me.
In forum.php I have this code:
<?php
while ($row = mysqli_fetch_assoc($result))
{
$subject = $row['subject'];
$id = $row['id'];
?>
<tr>
<th> <?php echo $row ['subject']?> </th>
<th>Edit</th>
<th>Delete</th>
</tr>
<?php
}
?>
In edit.php I have this code:
<div id="form">
<form id='edit' action='edit.php' method='post' >
<fieldset>
<legend>Edit Topic</legend>
<br />
<label for='name' >New Subject</label><br/>
<input type="text" id="subject" name="newsubject" /><br />
<br/>
<input type='hidden' name='id' value='<?php echo $id ?>'/>
<input type="submit" name="save" value="Save New Topic" />
</fieldset>
</form>
</div>
<?php
if (isset($_POST['save']))
{
$subject_save = $_POST['newsubject'];
require_once("db_connection.php");
$conn = connectToMySQL();
$id =$_POST['id'];
$query = "UPDATE tbl_topic SET subject = '$subject_save' WHERE id = 'id'";
$result = mysqli_query($conn, $query)
or die("Error in query: ".mysqli_error($conn));
header("Location: forum.php");
die();
}
?>

Write the query as
$query = "UPDATE tbl_topic SET subject = '$subject_save' WHERE id = $id";

if this not a mistake in typing
change $query = "UPDATE tbl_topic SET subject = '$subject_save' WHERE id = 'id'";
to $query = "UPDATE tbl_topic SET subject = '$subject_save' WHERE id = $id";

<div id="form">
<form id='edit' action='edit.php' method='post' >
<fieldset>
<legend>Edit Topic</legend>
<br />
<label for='name' >New Subject</label><br/>
<input type="text" id="subject" name="newsubject" value=<?php isse($_POST['newsubject']) ? "{$_POST['newsubject']}" : ""?> /><br />
<br/>
<input type='hidden' name='id' value='<?php echo $id ?>'/>
<input type="submit" name="save" value="Save New Topic" />
</fieldset>
</form>
</div>
<?php
if (isset($_POST['save']))
{
$subject_save = $_POST['newsubject'];
require_once("db_connection.php");
$conn = connectToMySQL();
$id =$_POST['id'];
$query = "UPDATE tbl_topic SET subject = '$subject_save' WHERE id = $id";
$result = mysqli_query($conn, $query)
or die("Error in query: ".mysqli_error($conn));
header("Location: forum.php");
die();
}

Related

Updating a field from an sql table from php

I have a table with more columns. On one column I have 3 buttons for each row with different queries associated. When I click one of the buttons, field in the table from my database should update.The problem is that when I click on a button on any row in the table, it updates only the field from the first row.For example, I have a table with 10 rows.If I click one of the buttons on the 10th row IT will update my first row, not the 10th one as I want. Is there any possibility to solve this?
This is the code..I'm sorry it is too long:
<?php
$query = "SELECT * FROM masculin ORDER BY id_concurent";
$result = mysqli_query($link ,$query);
$row = mysqli_fetch_assoc($result);
if(isset($_POST['locul1']))
{
$sql = "UPDATE masculin SET Premiu=1 WHERE CNP='".$row['CNP']."'";
mysqli_query($link,$sql);
header("Location:administrare.php");
}
else if(isset($_POST['locul2']))
{
$sql = "UPDATE masculin SET Premiu=2 WHERE CNP='".$row['CNP']."'";
mysqli_query($link,$sql);
header("Location:administrare.php");
}
else if(isset($_POST['locul3']))
{
$sql = "UPDATE masculin SET Premiu=3 WHERE CNP='".$row['CNP']."'";
mysqli_query($link,$sql);
header("Location:administrare.php");
} ?>
<u><i><h1 align="center">Administrare concurenti</h1></i></u>
<u><i><h2>MASCULIN</h2></i></u>
<?php
$query = "SELECT * FROM masculin ORDER BY id_concurent";
$result = mysqli_query($link ,$query);
if (mysqli_num_rows($result) == 0) {
echo 'Inca nu s-a inscris niciun concurent.';
} else {
?>
<table width="100%">
<tr>
<th>Nr.<br />concurs</th>
<th>Nume</th>
<th>Prenume</th>
<th>CNP</th>
<th>Categoria</th>
<th>Varsta</th>
<th>Premiu</th>
<th>Modifica<br />rezultat</th>
<th>Descalifica</th>
</tr>
<?php while($row = mysqli_fetch_assoc($result)){ ?>
<tr>
<?php
$query1="SELECT id_concurent FROM concurenti WHERE CNP='".$row['CNP']."'";
$result1=mysqli_query($link,$query1);
$nr_conc=mysqli_fetch_assoc($result1);
?>
<td> <?php echo $nr_conc['id_concurent'] ?> </td>
<td> <?php echo $row['Nume'] ?> </td>
<td> <?php echo $row['Prenume'] ?> </td>
<td> <?php echo $row['CNP'] ?> </td>
<td> <?php echo $row['Categorie'] ?> </td>
<td> <?php echo $row['Varsta'] ?> </td>
<td> <?php echo $row['Premiu'] ?> </td>
<td>
<form action="administrare.php" method="post">
<input type="submit" name="locul1" value="Premiul 1">
<input type="submit" name="locul2" value="Premiul 2">
<input type="submit" name="locul3" value="Premiul 3">
</form>
</td>
</tr>
<?php } ?>
</table> <?php } ?>
for updated the da tbale row when you submit the values related to a html table row you should add and hidden input with the value that let you reach the correct row eg:
<form action="administrare.php" method="post">
<input type="submit" name="locul1" value="Premiul 1">
<input type="submit" name="locul2" value="Premiul 2">
<input type="submit" name="locul3" value="Premiul 3">
<input type="hidden" name="CNP" value="<?php echo $row['CNP'] ?>">
</form>
and in your sql updated query add the POST value related to the row
$sql = "UPDATE masculin SET Premiu=1 WHERE CNP='".$_POST['CNP']. "'";

Can't edit data in the database and cant fetch name with id?

I am trying to edit data into the database I don't know why I cant do. I have tried something till now. maybe someone can have a look please. I am trying to built a update where i can change name, surname blah blah blah, but i cant config even just for a name first..
Home file
Managament System
<body>
<h1>TU Chemnitz Student managament system</h1>
<br>
ADD Person
Edit Person
Manage Boards
Manage Departments
Search N&S
Triple Search
Membership
<br>
<br>
<?php
include_once('coneksioni.php');
// create query
$querys = "SELECT * FROM tblperson";
// execute query
$result = mysql_query($querys) or die ("Error in query: $query. ".mysql_error());
$res = mysql_query("SELECT * FROM tblperson");
echo "<table border=1 align=center>
<tr>
<th>Personal ID</th>
<th>First Name</th>
<th>Last Name</th>
<th>Deparment</th>
<th>Board</th>
<th>Marticulation Number</th>
<th>Reg Date</th>
<th>Action</th>
</tr>";
while($row = mysql_fetch_array($res)) {
?>
<tr>
<td><?=$row['personid']?></td>
<td><?=$row['personname']?></td>
<td><?=$row['personsurname']?></td>
<td><?=$row['persondepartment']?></td>
<td><?=$row['personboard']?></td>
<td><?=$row['martinumber']?></td>
<td><?=$row['personregdate']?></td>
<td>
edit |
del
</td>
</tr>
<?php
}
?>
</body>
</html>
and edit20.php
<?php
include_once('coneksioni.php');
if( isset($_GET['edit']) )
{
$personid = $_GET['edit'];
$res= mysql_query("SELECT * FROM tblperson WHERE personid='$personid'");
$row= mysql_fetch_array($res);
}
if( isset($_POST['personname']) )
{
$personname = $_POST['personname'];
$personid = $_POST['personid'];
$sql = "UPDATE tblperson SET personname='$personname' WHERE personid='$personid'";
$res = mysql_query($sql)
or die("Could not update".mysql_error());
echo "<meta http-equiv='refresh' content='0;url=home.php'>";
}
?>
<form action="edit20.php" method="POST">
Name: <input type="text" name="personname" value="<?php echo $row[1]; ?>"><br />
<input type="hidden" name="personid" value="<?php echo $row[0]; ?>">
<input type="submit" value=" Update "/>
</form>
and in the database primary key in my table is personid name field personname (not Primary key).
Please use Prepared Statement for reduce the risk of SQL Injection
check the coneksioni.php
$conn = new mysqli(HOST, USER, PASS, DBNAME);
the edit.php
require_once ('coneksioni.php');
$edit_person = $conn->prepare("
UPDATE tblperson SET
personname = ? WHERE personid = ?
");
$edit_person->bind_param(
"si",
$personname, $personid
);
if(isset($_POST['personname']) && isset($_POST['personid']) ) {
$personname = $_POST['personname'];
$personid = $_POST['personid'];
if (!$edit_person->execute()) {
// action if failed
} else {
// action if success
}
$edit_person->close();
}
the form.html
<form action="edit.php" method="POST">
Name: <input type="text" name="personname" value="<?php echo $row[1]; ?>"><br />
<input type="hidden" name="personid" value="<?php echo $row[0]; ?>">
<input type="submit" value=" Update "/>
</form>
Cheers

when opening accordion, GET url by ID from that article

I want to display the article ID in my URL when i press the a href in code beneath
Update, i included the php section and the while loop with the article ID
require_once("inc/connection.php");
mysql_select_db("nieuws");
if ($_SERVER['REQUEST_METHOD'] === 'POST') {
$id = $_GET['id'];
$datum = $_POST["datum"];
$titel = $_POST["titel"];
$artikel = $_POST["artikel"];
$checkbox = $_POST["checkbox"];
$titel = mysql_real_escape_string(nl2br(htmlentities($_POST["titel"])));
$artikel = mysql_real_escape_string($_POST["artikel"]);
$id = mysql_real_escape_string($_POST['id']);
date_default_timezone_set('GMT');
$datum = date('Y-m-d', strtotime(str_replace('-', '/', $datum)));
if(isset($_POST['add'])){
if(!empty($_POST['titel']) && !empty($_POST['artikel']) && !empty($_POST['datum'])){
$query="INSERT INTO nieuws (id,datum,titel,artikel) VALUES ('$id','$datum','$titel','$artikel')";
$datum = date('Y-m-d', strtotime(str_replace('-', '/', $datum)));
str_replace('<br />', "\n", $textarea);
$result=mysql_query($query);
$juist1 = true;
}else{
$fout1 = true;
}
}if(isset($_POST['delete'])){
foreach($_POST['checkbox'] as $del_id){
$sql="DELETE FROM nieuws WHERE id='$del_id'";
$result = mysql_query($sql);
$juist2 = true;
}
}
}
This is the accordion script that opens on click
$('.acc_container').hide(); //Hide/close all containers
$('.acc_trigger').click(function(){
if( $(this).next().is(':hidden') ) { //If immediate next container is closed...
$('.acc_trigger').removeClass('active').next().slideUp();
$(this).toggleClass('active').next().slideDown();
}
return false;
});
});
This is the FORM part
<form method="POST" action="<?php echo htmlspecialchars($_SERVER['PHP_SELF']); ?>">
<table style="width:950px; margin-bottom:5px;">
<tr>
<td>Datum:<br />
<input type="text" name="datum" size="20" style="width:100px;" value="<?php echo $datum; ?>" id="datepicker" placeholder="Kies datum"/><br /></td>
<td>Titel:<br />
<input type="text" name="titel" size="200" style="width:500px;" maxlength="45" value="<? php echo $titel; ?>" placeholder="Max. 50 characters toegelaten"/><br /></td> </tr>
</table>
Artikel: <br />
<textarea id="textarea" name="artikel" style="width:500px; height:150px;" value="<?php echo $artikel; ?>" ></textarea><br />
<input type="submit" name="add" value="Artikel toevoegen" />
</form>
This is the while loop where i ADD the row ID
$query="SELECT id,datum,titel,artikel FROM nieuws ORDER BY id DESC";
$result=mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result)){
echo ("<div id=\"artikeltitel\" align=\"center\">
<div id=\"containerdatum\">".$row['datum']."</div>
<div id=\"containertitel\">".$row['titel']."</div>
<div id=\"container3\" style=\"font-size:12px;\">".$row['id']."
<input type=\"checkbox\" name=\"checkbox[]\" id=\"checkbox\" value=\"".$row['id']."\" />
</div>
</div>
<div class=\"container\" align=\"center\">
<h2 class=\"acc_trigger\"> ยป </h2>
<div class=\"acc_container\">
<div class=\"block\">".$row['artikel']."</div>
<div class=\"fb-comments\" data-href=\"http://www.zpb-polonez.be/user.php\" data-num- posts=\"10\" data-width=\"678\" style=\"margin-top:2px;\"></div>
</div>
</div>
");
}
I don't know what to add more to explain what i'm trying to achive
We lack some information here to be able to help but let's try anyway.
Let's assume that your article id is $row['id'] you could build your link this way:
echo 'Link';

Adding to database with multiple text boxes

What I am trying to do with this script is allow users to update a url for their websites, and since each user isn't going to have the same amount of websites is is hard for me to just add $_POST['website'] for each of these.
Here is the script
<?php
include("config.php");
include("header.php");
include("functions.php");
if(!isset($_SESSION['username']) && !isset($_SESSION['password'])){
header("Location: pubs.php");
}
$getmember = mysql_query("SELECT * FROM `publishers` WHERE username = '".$_SESSION['username']."'");
$info = mysql_fetch_array($getmember);
$getsites = mysql_query("SELECT * FROM `websites` WHERE publisher = '".$info['username']."'");
$postback = $_POST['website'];
$webname = $_POST['webid'];
if($_POST['submit']){
foreach ( $_POST['website'] as $key => $value )
{
$update = mysql_query("UPDATE `websites` SET `postback` = '".mysql_real_escape_string($postback[$value])."' WHERE id = '$webname'");
}
}
print"
<div id='center'>
<span id='tools_lander'><a href='export.php'>Export Campaigns</a></span>
<div id='calendar_holder'>
<h3>Please define a postback for each of your websites below. The following variables should be used when creating your postback.<br />
cid = Campaign ID<br />
sid = Sub ID<br />
rate = Campaign Rate<br />
status = Status of Lead. 1 means payable 2 mean reversed<br />
A sample postback URL would be <br />
http://www.example.com/postback.php?cid=#cid&sid=#sid&rate=#rate&status=#status</h3>
<table class='balances' align='center'>
<form method='POST' action=''>";
while($website = mysql_fetch_array($getsites)){
print"
<tr>
<input type ='hidden' name='webid' value='".$website['id']."' />
<td style='font-weight:bold;'>".$website['name']."'s Postback:</td>
<td><input type='text' style='width:400px;' name='website[]' value='".$website['postback']."' /></td>
</tr>";
}
print"
<td style='float:right;position:relative;left:150px;'><input type='submit' name='submit' style='font-size:15px;height:30px;width:100px;' value='Submit' /></td>
</form>
</table>
</div>";
include("footer.php");
?>
What I am attempting to do insert the what is inputted in the text boxes to their corresponding websites, and I cannot think of any other way to do it, and this obviously does not works and returns a notice stating Array to string conversion
If there is a more logical way to do this please let me know.
UPDATE
I added a foreach statement, but this still doesn't seem to solve the problem. It doesn't update anything in the database.
I was able to fix the problem with some trial and error, Lawrence helped with the informing me to use a foreach statement. This is what I have ended up with.
<?php
include("config.php");
include("header.php");
include("functions.php");
if(!isset($_SESSION['username']) && !isset($_SESSION['password'])){
header("Location: pubs.php");
}
$getmember = mysql_query("SELECT * FROM `publishers` WHERE username = '".$_SESSION['username']."'");
$info = mysql_fetch_array($getmember);
$getsites = mysql_query("SELECT * FROM `websites` WHERE publisher = '".$info['username']."'");
$postback = $_POST['website'];
$webname = $_POST['webid'];
if($_POST['submit']){
$i = -1;
foreach ($postback as $key => $value)
{
$i ++;
print_r($webname[$i]);
$update = mysql_query("UPDATE `websites` SET `postback` = '".cleanQuery($postback[$key])."' WHERE `id` = '".$webname[$i]."'") or die("MySQL ERROR: ".mysql_error());
}
}
print"
<div id='center'>
<span id='tools_lander'><a href='export.php'>Export Campaigns</a></span>
<div id='calendar_holder'>
<h3>Please define a postback for each of your websites below. The following variables should be used when creating your postback.<br />
cid = Campaign ID<br />
sid = Sub ID<br />
rate = Campaign Rate<br />
status = Status of Lead. 1 means payable 2 mean reversed<br />
A sample postback URL would be <br />
http://www.example.com/postback.php?cid=#cid&sid=#sid&rate=#rate&status=#status</h3>
<table class='balances' align='center'>
<form method='POST' action=''>";
while($website = mysql_fetch_array($getsites)){
print"
<tr>
<input type ='hidden' name='webid[]' value='".$website['id']."' />
<td style='font-weight:bold;'>".$website['name']."'s Postback:</td>
<td><input type='text' style='width:400px;' name='website[]' value='".$website['postback']."' /></td>
</tr>";
}
print"
<td style='float:right;position:relative;left:150px;'><input type='submit' name='submit' style='font-size:15px;height:30px;width:100px;' value='Submit' /></td>
</form>
</table>
</div>";
include("footer.php");
?>

Editing data from MySQL via PHP

I am running into a frustrating problem with a PHP script that's supposed to allow me to edit individual rows within my MySQL database.
This is the file where all of the rows from the database are displayed; it works just like it's supposed to.
<table cellpadding="10">
<tr>
<td>ID</td>
<td>First Name</td>
<td>Last Name</td>
<td>E-mail</td>
<td>Phone</td>
</tr>
<?php
$username="username here";
$password="password here";
$database="database name here";
mysql_connect(localhost,$username,$password);
#mysql_select_db($database) or die( "Unable to select database");
$query="SELECT * FROM students";
$result=mysql_query($query);
mysql_close();
while ($row=mysql_fetch_array($result)){
echo ("<tr><td>$row[id]</td>");
echo ("<td>$row[first]</td>");
echo ("<td>$row[last]</td>");
echo ("<td>$row[email]</td>");
echo ("<td>$row[phone]</td>");
echo ("<td>Edit</td></tr>");
}
echo "</table>";
?>
As you can see, each row has an "Edit" link that is supposed to allow the user to edit that individual student's data. Here, then, is StudentEdit.php:
<?php
$username="username";
$password="password";
$database="database";
mysql_connect(localhost,$username,$password);
$student_id = $_GET[id];
$query = "SELECT * FROM students WHERE id = '$student_id'";
$result = mysql_query($query);
$row = mysql_fetch_array($result);
mysql_close();
?>
<form method="post" action="EditStudentData.php" />
<table>
<tr>
<td><input type="hidden" name="id" value="<? echo "$row[id]" ?>"></td>
</tr>
<tr>
<td>First Name:</td>
<td><input type="text" name="first" value="<? echo "$row[first]" ?>"></td>
</tr>
<tr>
<td>Last Name:</td>
<td><input type="text" name="last" value="<? echo "$row[last]" ?>"></td>
</tr>
<tr>
<td>Phone Number:</td>
<td><input type="text" name="phone" value="<? echo "$row[phone]" ?>"></td>
</tr>
<tr>
<td>E-mail:</td>
<td><input type="text" name="email" value="<?echo "$row[email]" ?>"></td>
</tr>
</table>
</form>
When I execute this, however, I get the following error message:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home4/lukaspl1/public_html/StudentEdit.php on line 12
Any ideas what's wrong, and how to fix it?
Thank you in advance!
Remove the mysql_close from here
mysql_connect(localhost,$username,$password);
#mysql_select_db($database) or die( "Unable to select database");
$query="SELECT * FROM students";
$result=mysql_query($query);
mysql_close();
The code should mysql_connect(localhost,$username,$password);
#mysql_select_db($database) or die( "Unable to select database");
$query="SELECT * FROM students";
$result=mysql_query($query);
And moreover,you are going to use only key based resultset.. simply have mysql_fetch_assoc.
And another suggestion would be instead of $row[id]..replace it with $row['id'].
StudentEdit.php: you forgot to call #mysql_select_db($database) or die( "Unable to select database"); before you executed the query
This part of the code is wrong:
$student_id = $_GET[id];
the correct code is
$student_id = $_GET['id'];
code from expertsnote.com
Try...
echo ("<td>Edit</td></tr>");
instead of
echo ("<td>Edit</td></tr>");
this code was missing
$select_db = mysql_select_db("$db_name");
if (!$select_db) {echo "Error Selecting Database";}
this is the cod for edit the details dynamically
<?php
include('db.php');
$id=$_REQUEST['id'];
$query="SELECT * FROM `camera details` WHERE id='".$id."'";
$result=mysqli_query($db,$query) or die(mysqli_error());
$row1=mysqli_fetch_assoc($result);
if(isset($_POST['submit'])&&(isset($_POST['new'])&&($_POST['new'])==1))
{
$id=$_REQUEST['id'];
foreach($_POST as $key=>$values)
{
if($key!="submit"){
$names[]=$key;
$val[]= "'".$values."'";
if($key!="new"){
$k[] = "`".$key."` = '".$values."'";
}
}
}
$output=implode(",",(array)($k));
//$v=implode(",",(array)($val));
// `name` = 'san'
$query="UPDATE `camera details` SET $output WHERE id='".$id."'";
$output=mysqli_query($db,$query) or die(mysqli_error($db));
if($output)
{
header('location:cameralist.php');
}
}
else{
?>
I recommend doing this in studentEdit.php
$student_id = mysql_real_escape_string($_GET[id]);
$query = "SELECT * FROM students WHERE id = '$student_id'";
$result = mysql_query($query) or die(mysql_error() . ' ' . $query);
$row = mysql_fetch_array($result);
mysql_close();
Two things I've changed here is firstly to escape the data being passed in the url and secondly I've added or die(mysql_error() . ' ' . $query); If something is going wrong in the sql statement you should now see the error and hopefully you'll be able to fix it from there.
What looks incorrect to me is the way you are displaying the value retrieved from the database:
<input type="hidden" name="id" value="<? echo "$row[id]" ?>">
It should be
<input type="hidden" name="id" value="<?php echo $row['id']; ?>">
This code gives the option to add, search, edit and delete options. Thought it might to see all the options in one code.
$searchedUsername = "";
$searchedEmail = "";
//registration (Add) function
if ( isset($_POST['stdregister'])){
$username = $_POST['stdusername'];
$password = $_POST['stdpassword'];
$email = $_POST['stdemail'];
$hashedPassword = md5($password);
$connection = mysqli_connect("localhost","root","","std");
$query = "INSERT INTO student VALUES ('$username','$hashedPassword','$email')";
if ( mysqli_query($connection,$query) == 1 ){
echo "Successfully saved";
}
else{
echo "<p style='color: #f00;'>There is an error</p>";
}
mysqli_close($connection);
}
//delete function
if ( isset($_POST['stddelete'])){
$username = $_POST['stddelusername'];
$connection = mysqli_connect("localhost","root","","std");
$query = "DELETE FROM student WHERE username LIKE '$username'";
mysqli_query($connection,$query);
echo mysqli_error($connection);
mysqli_close($connection);
}
//update function
if ( isset($_POST['stdupdate'])){
$username = $_POST['stdusername'];
$stdpass = md5($_POST['stdpassword']);
$stdemail = $_POST['stdemail'];
$connection = mysqli_connect("localhost","root","","std");
$query = "UPDATE student SET password='$stdpass', email='$stdemail' WHERE username LIKE '$username'";
mysqli_query($connection,$query);
echo mysqli_error($connection);
mysqli_close($connection);
}
if ( isset($_POST['stdsearch']) ){
$searchUsername = $_POST['stdeditusername'];
$connection = mysqli_connect("localhost","root","","std");
$query = "SELECT * FROM student WHERE username LIKE '$searchUsername' ";
$result = mysqli_query($connection, $query);
while( $row = mysqli_fetch_array($result) ){
$searchedUsername = $row['username'];
$searchedEmail = $row['email'];
}
}
?>
<html>
<head>
</head>
<body>
<h1>Student Registration</h1>
<form name="stdregistration" action="forms.php" method="post">
<label>Username :</label>
<input name="stdusername" required="required" type="text" /><br /><br />
<label>Password :</label>
<input name="stdpassword" type="password" /><br /><br />
<label>E-mail :</label>
<input name="stdemail" type="email" /><br /><br />
<input name="stdregister" type="submit" value="Save" />
</form>
<h2>Delete Students</h2>
<form name="stddeletion" action="forms.php" method="post">
<label>Select the Username :</label>
<select name="stddelusername" required>
<option value="">Select One</option>
<?php
$connection2 = mysqli_connect("localhost","root","","std");
$query2 = "SELECT username FROM student";
$result = mysqli_query($connection2,$query2);
while( $row = mysqli_fetch_array($result) ){
echo "<option value='".$row['username']."'>".$row['username']."</option>";
}
mysqli_close($connection2);
?>
</select>
<input name="stddelete" type="submit" value="Delete" />
</form>
<h2>Edit Students</h2>
<form name="stdedition" action="forms.php" method="post">
<label>Select the Username :</label>
<select name="stdeditusername" required>
<option value="">Select One</option>
<?php
$connection2 = mysqli_connect("localhost","root","","std");
$query2 = "SELECT username FROM student";
$result = mysqli_query($connection2,$query2);
while( $row = mysqli_fetch_array($result) ){
echo "<option value='".$row['username']."'>".$row['username']."</option>";
}
mysqli_close($connection2);
?>
</select>
<input name="stdsearch" type="submit" value="Search" />
</form>
<form name="stdedit" action="forms.php" method="post">
<label>Username :</label>
<input name="stdusername" required="required" type="text" readonly value="<?php echo $searchedUsername; ?>" /><br /><br />
<label>Password :</label>
<input name="stdpassword" type="password" /><br /><br />
<label>E-mail :</label>
<input name="stdemail" type="email" value="<?php echo $searchedEmail; ?>" /><br /><br />
<input name="stdupdate" type="submit" value="Update" />
</form>
</body>
</html>

Categories