I had inserted the data from html and it is stored in mysql database, and the data is retrieved from database to html,now I had done how to delete the data that are displayed in html,my task is how to update the displayed.And my deleted code is:
Below is my HTML code:
<html>
<head>
<title>STUDENT_DATA</title>
</head>
<body>
<form action="tab.php" method="post" >
<center>
sname: <input type="text" name="sname" required><br></br>
sno:<input type="text" name="sno"><br></br>
marks:<input type="text" name="marks"><br></br>
class:<input type="text" name="class"><br></br>
phno:<input type="text" name="phno" onkeypress='return event.charCode >
= 48 && event.charCode <= 57'><br></br>
DOB:<input type="date" placeholder="DD-MM-YYYY"
required pattern="(0[1-9]|1[0-9]|2[0-9]|3[01]).(0[1-9]|1[012]).[0-9]{4}"
name="DOB"/><br></br>
<button>submit</button></br>
</center>
</form>
</body>
</html>
Below is my PHP code for displaying the data:
<!DOCTYPE html>
<html>
<head>
</head>
<body>
<?php
$connection = mysql_connect('localhost', 'root','');
if (!$connection)
{
die("Database Connection Failed" . mysql_error());
}
$select_db = mysql_select_db( "student",$connection);
if (!$select_db)
{
die("Database Selection Failed" . mysql_error());
}
$sql = "SELECT * FROM hello1 ";
$result = mysql_query($sql) or die(mysql_error());
?>
<table border="2" style= " margin: 0 auto;" id="myTable">
<thead>
<tr>
<th>sname</th>
<th>sno</th>
<th>marks</th>
<th>class</th>
<th>phno</th>
<th>DOB</th>
</tr>
</thead>
<tbody>
<?php
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['sname'] . "</td>";
echo "<td>" . $row['sno'] . "</td>";
echo "<td>" . $row['marks'] . "</td>";
echo "<td>" . $row['class'] . "</td>";
echo "<td>" . $row['phno'] . "</td>";
echo "<td>" . $row['DOB'] . "</td>";
echo "<td><a href='delete.php?did=".$row['sname']."'>Delete</a></td>";
echo "</tr>";
}
?>
</tbody>
</table>
</body>
</html>
Below is my PHP code for deleting:
<?php
$connection = mysql_connect('localhost', 'root','');
if (!$connection)
{
die("Database Connection Failed" . mysql_error());
}
$select_db = mysql_select_db( "student",$connection);
if (!$select_db)
{
die("Database Selection Failed" . mysql_error());
}
?>
<?php
if(isset($_GET['did'])) {
$delete_id = $_GET['did'];
$sql = mysql_query("DELETE FROM hello1 WHERE sname = '".$delete_id."'");
if($sql) {
echo "<br/><br/><span>deleted successfully...!!</span>";
}
else
{
echo "ERROR";
}
}
?>
Below is what I have tried in update.php
<?php
$connection = mysql_connect('localhost', 'root','');
if (!$connection) { die("Database Connection Failed" . mysql_error());
}
$select_db = mysql_select_db( "emp",$connection);
if (!$select_db) { die("Database Selection Failed" . mysql_error()); } ?>
<?php if(isset($_GET['did']))
{
$update_id = $_GET['did']; $sql = mysql_query("UPDATE FROM venu WHERE id = '".$update_id."'");
if($sql) {
echo "<br/><br/><span>updated successfully...!!</span>";
} else {
echo "ERROR";
}
} ?>
Related
So, I am trying to figure out how do this this and it boggling me. THIS WILL NOT BE USED ONLINE LIVE SO SQL INJECTION I DONT' CARE ABOUT. What am I doing wrong/right?
<?php
$db = mysql_connect("localhost", "root", "root");
if (!$db) {
die("Database connect failed: " . mysql_error());
}
$db_select = mysql_select_db("UNii", $db);
if (!$db_select) {
die("Database selection failed: " . mysql_error());
}
$comment = $_GET['comment'];
$id = $_GET['id'];
$sql = "UPDATE Dbsaved SET comment = '{$comment}' WHERE id = $id";
$comment1 = mysql_query($sql);
if (!$comment1) {
die("did not save comment: " . mysql_error());
}
echo $sql;
The main problem is with the statement itself, the connection is fine. I am trying to read $comment, and then update that into a MYSQL table and then have it read back in a different file.
EDIT: Mark up for the form I'm taking $comment from.
<!DOCTYPE html>
<html lang="en">
<LINK href="stylesheet.css" rel="stylesheet" type="text/css">
<script src ="js/validateform.js"></script>
<head>
<meta charset="UTF-8">
<title>UniHelp Home</title>
</head>
<body>
<div id="headeruni">
<h1>Welcome <?php echo $_GET["name"]; ?> to UniHelp!</h1>
</div>
<div id ="infouni">
<h3>Welcome to UniHelp. The social Network getting you connected to other people all over the University for any help you require!</h3>
</div>
<div id ="nameandemail">
<form action="formsend.php" method="post">
First name: <br> <input type="text" name="name"><br>
Email: <br> <input type="text" name="email"><br>
Comment: <br> <input type="text" name="message"><br>
<input type="submit" name="submit">
</form>`enter code here`
</div>
<div id="grabphpdiv">
<?php
$db = mysql_connect("localhost", "root", "root");
if (!$db) {
die("Database connect failed: " . mysql_error());
}
$db_select = mysql_select_db("UNii", $db);
if (!$db_select) {
die("Database selection failed: " . mysql_error());
}
$result = mysql_query("SELECT * FROM Dbsaved", $db);
if (!$result) {
die ("Database query failed: " . mysql_error());
}
$comment = $_POST['$comment'];
while ($row = mysql_fetch_array($result)) {
echo "<div id='posts'>";;
echo "<h2>";
echo $row[1] . "";
echo "</h2>";
echo "<p>";
//echo $timestamp = date('d-m-y G:i:s ');
echo "<br>";
echo "<br>";
echo $row[2] . "";
echo "</p>";
echo "<p>";
echo $row[3] . "";
echo "</p>";
echo 'Delete';
echo "<br>";
echo "<br>";
echo 'Comment: <br>
<input type=text name=comment><br>
<a href=addcomment.php?id=' . $row[0]. '&comment='. $row['$comment'].'>Comment</a>';
echo "<p>";
echo $row['comment'] . "";
echo "</p>";
echo "</div>";
echo "<br>";
}
?>
</div>
</body>
<div id="footer">Copyright © James Taylor 2016</div>
</html>
I just ran this code:
$comment = "Hello World!";
$id = 1;
$sql = "UPDATE Dbsaved SET comment = '{$comment}' WHERE id = {$id}";
echo $sql;
and saw:
UPDATE Dbsaved SET comment = 'Hello World!' WHERE id = 1
which is a correct SQL statement, so if it is not working, you might want to play with SQL directly to get something working. Hope that helps!
SOLUTION:
$comment = $_GET['$comment'];
$id = $_GET['$id'];
while ($row = mysql_fetch_array($result)) {
echo "<div id='posts'>";;
echo "<h2>";
echo $row[1] . "";
echo "</h2>";
echo "<p>";
//echo $timestamp = date('d-m-y G:i:s ');
echo "<br>";
echo "<br>";
echo $row[2] . "";
echo "</p>";
echo "<p>";
echo $row[3] . "";
echo "</p>";
echo 'Delete';
echo "<br>";
echo "<br>";
echo $row[4] . "";
echo "<br>";
echo 'Comment: <br>
<form action="addcomment.php?id=' . $row[0]. '" method="post">
<input type=text name=comment><br>
<input type=submit name="submit">
</form>';
echo "<p>";
echo $row['comment'] . "";
echo "</p>";
echo "</div>";
echo "<br>";
}
?>
and:
<?php
$db = mysql_connect("localhost", "root", "root");
if (!$db) {
die("Database connect failed: " . mysql_error());
}
$db_select = mysql_select_db("UNii", $db);
if (!$db_select) {
die("Database selection failed: " . mysql_error());
}
$comment = $_POST['comment'];
$id = $_GET['id'];
$sql = "UPDATE Dbsaved SET comment = '$comment' WHERE id = $id ";
$comment1 = mysql_query($sql);
echo $sql;
if (!$comment1) {
die("did not save comment: " . mysql_error());
}
else {
header("location: UniHelpindex.php");
}
It was to do with mainly needing to get the id which was used in $row[0]' in the form created in the while loop. And actually using the correct syntax for the update Dbsaved... bit.
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);
}
?>
this is starva i need your help for storing values into database here is my code i want to store the values fetched from list into database '$post[]'
<html>
<head>
<title>order</title>
</head>
<body>
<table>
<form action="order2.php" method="post">
<tr><td><input type="submit" name="submit" value="show"></td>
<td><input type="submit" name="order" value="Order"></td>
</tr>
</table>
<!________________________________!>
<?php
$servername="localhost";
$username="root";
$password="";
$database="mess_db";
$con = mysql_connect($servername,$username,$password);
if(!$con)
{
Die('Could not connect:'.mysql_error());
}
if($con)
{
echo "DB Connected<br>";
}
mysql_select_db($database,$con);
if(isset($_POST["submit"]))
{
$sql="SELECT * FROM menu";
$result= mysql_query($sql,$con);
echo '<select name="roti">';
while($row = mysql_fetch_array($result))
{
echo '<option value="'. $row['roti'].'">' . $row['roti'].'';
}
echo '</select><br><br>';
}
if(isset($_POST["submit"]))
{
$sql="SELECT * FROM menu";
$result= mysql_query($sql,$con);
echo '<select name="sabji">';
while($row = mysql_fetch_array($result))
{
echo '<option value=" ' . $row['sabji'].'">' . $row['sabji'].'';
}
echo '</select><br><br>';
}
if(isset($_POST["submit"]))
{
$sql="SELECT * FROM menu";
$result= mysql_query($sql,$con);
echo '<select name="daal">';
while($row = mysql_fetch_array($result))
{
echo '<option value=" ' . $row['daal'].'">' . $row['daal'].'';
}
echo '</select><br><br>';
}
if(isset($_POST["submit"]))
{
$sql="SELECT * FROM menu";
$result= mysql_query($sql,$con);
echo '<select name="sweet">';
while($row = mysql_fetch_array($result))
{
echo '<option value=" ' . $row['sweet'].'">' . $row['sweet'].'';
}
echo '</select><br><br>';
}
if(isset($_POST["submit"]))
{
$sql="SELECT * FROM menu";
$result= mysql_query($sql,$con);
echo '<select name="starter">';
while($row = mysql_fetch_array($result))
{
echo '<option value=" ' . $row['starter'].'">' . $row['starter'].'';
}
echo '</select><br><br>';
}
if(isset($_POST["order"]))
{
echo "Hi...";
$servername="localhost";
$username="root";
$password="";
$database="mess_db";
$con = mysql_connect($servername,$username,$password);
if(!$con){
Die('Could not connect:'.mysql_error());
}
if($con){
echo "DB Connected<br>";
}
mysql_select_db($database,$con);
if(!$_POST=="")
{
$sql="INSERT INTO order VALUES('$_POST[roti]')";
$result=mysql_query($sql,$con);
echo "Hello";
}
}
mysql_close($con);
?>
please help me on this topic
i ma not at all getting any error and database is still empty
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 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