I have a php scripts whit a table thats sorts after time.
in that table i have a check box to mark if somthings is deleverd and i have a hidden input thats get the id. it worked yesterday but know it gets the id lowest on the table.
$sql = "SELECT * FROM `bestalning` WHERE lev=0
ORDER BY tid";
$result = mysqli_query($conn, $sql) or die(mysqli_error($conn));
echo "<form method='POST' action='ID_change.php'>";
echo "<div class='continer bg-vit table-rsponsive-sm '><table class='table'><thead class='thead-dark'>
<tr><th >".'Lev'."</th><th>".' '."</th><th >".'Artikelnr'."</th><th >".'Antal'."</th>
<th >".'Singnatur'."</th><th >".'Önskad ankomst'."</th><th >".'Skickad'."</th></tr></thead></div>";
while($row = mysqli_fetch_array($result)){
echo "<tbody><tr><td><input style='margin-right:5px;' name='levJA' type='checkbox' value='1'></td><td>
<input style='margin-right:5px;' name='RowID' type='hidden' value='".$row['ID']."'></td><td>".$row['artikelnr']."</td>
<td>".$row['antal']."</td><td>".$row['ovrigt']."</td><td>".$row['Tid']."</td><td>".$row['date']."</td></tr></tbody>";
}
echo " <input class='btn btn-dark m-1' type='submit' value='submit'></form>";
mysqli_close($conn);
ID_change.php
$Lev = $_POST['levJA'];
$ID = $_POST['RowID'];
$sql = "UPDATE bestalning SET lev='".$Lev."' WHERE ID='".$ID."'";
echo $sql;
if (mysqli_query($conn, $sql,)) {
echo "Record updated successfully";
} else {
echo "Error updating record: " . mysqli_error($conn);
}
mysqli_close($conn);
?>
Related
This is my code which shows current events and lets the user change the date, name or venue of the event.
I keep getting a 500 error for some reason. I think it is due to the information being passed to and from the database.
database set up is :userid ,eventname, venue, date, name ,eventid... respectivley
<div class="current events">
<h1>Your Current Events:</h1>
<?php
$sql = "SELECT * FROM events WHERE userid='{$_SESSION['u_id']}';";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0){
while ($row = mysqli_fetch_assoc($result)){
echo "<b>Event name: </b>";
echo " ";
echo $row['eventname'];
echo " ";
echo "<b>Event Venue: </b>";
echo " ";
echo $row['venue'];
echo " ";
echo "<b>Event Date: </b>";
echo " ";
echo $row['date'];
echo "
<form method='POST' action='editevent.php'>
<input type='hidden' name='eventname' value='" .$row['eventname']. "'>
<input type='hidden' name='venue' value='" .$row['venue']. "'>
<input type='hidden' name='date' value='" .$row['date']. "'>
<input type='hidden' name='name' value='" .$row['name']. "'>
<button>Edit</button>
</form>
";
}
}else{
echo "No Upcoming Events";
}
?>
</div>
I then have another file in my includes directory which allows changes to the information.
<?php
session_start();
if (isset($_POST['eventsubmit'])) {
$eventname = $_POST['eventname'];
$venue = $_POST['venue'];
$date = $_POST['date'];
$name = $_POST['name'];
$eventname = mysqli_real_escape_string($conn, $_POST['eventname']);
$venue = mysqli_real_escape_string($conn, $_POST['venue']);
$date = mysqli_real_escape_string($conn, $_POST['date']);
$name = mysqli_real_escape_string($conn, $_POST['name']);
$sql = "UPDATE events SET eventname='$eventname' WHERE userid='2' ";
mysqli_query($conn, $sql);
header("Location: ../members.php?event=success");
exit();
} else {
header("Location: ../signup.php");
exit();
}
}
I check your code in second php file you put one extra this } please remove it.
please use mysqli_error instruction to get exactly which error you get
mysqli_query($conn, $sql)or die( mysqli_error($conn));
or you can use to show php error if there any error in php syntax in start of page
error_reporting(E_ALL);
ini_set('display_errors', 1);
table.php
<?php
include('../connections/conn.php');
include('../php/login.php');
$sql = "SELECT * FROM person";
$records = mysqli_query($conn, $sql)
?>
<html>
<head>
<title>Table</title>
</head>
<body>
<table>
<tr>
<th>Name</th>
<th>Age</th>
<th>Salary</th>
</tr>
<?php
while($row = mysqli_fetch_array($records)){
$name = $row['Name'];
$age = $row['Age'];
$salary = $row['Salary'];
$id = $row['id'];
echo "<tr><form action=update.php method=post>";
echo "<td><input type=text name=pname value='$name'></td>";
echo "<td><input type=text name=age value='$age'></td>";
echo "<td><input type=text name=salary value='$salary'></td>";
echo "<input type=hidden name=id value='$id'></td>";
echo "<td><input type=submit>";
echo "</form></tr>";
}
?>
</table>
</body>
</html>
(this part of the code displays the table and its values)
Update.php
<?php
include('../connections/conn.php');
include('../php/login.php');
$sql = "UPDATE person SET
Name='$_POST[pname]',Age='$_POST[age]',Salary='$_POST[salary]' WHERE
id='$_POST[id]'";
if(mysqli_query($conn, $sql)){
header("refresh:1 url=table.php");
}
else{
echo"Not Update";
}
$records = mysqli_query($conn, $sql)
?>
(this part is for updating the table)
I have got the code to update the contents of a table using buttons however I would like to just have one button that will update the whole table. At the moment I use a button per row to update that particular row.
Just use this
$sql = "UPDATE person SET
Name='$_POST[pname]',Age='$_POST[age]',Salary='$_POST[salary]' WHERE
id in('$_POST[id]')";
I am using the following code to display certain rows from my database table:
<?php
$searchtype=$_POST['searchtype'];
$searchterm=$_POST['searchterm'];
$searchterm= trim($searchterm);
if (!$searchtype || !$searchterm)
{
echo 'Error';
exit;
}
if (!get_magic_quotes_gpc())
{
$searchtype = addslashes($searchtype);
$searchterm = addslashes($searchterm);
}
$db = include "connect2db.php";
$query = "select * from notes where ".$searchtype." like '%".$searchterm."%'";
$result = $db->query($query);
$num_results = $result->num_rows;
echo '<p>Number of rows found: '.$num_results.'</p>';
for ($i=0; $i <$num_results; $i++)
{
$row = $result->fetch_assoc();
echo '<i>';
echo stripslashes($row['date']);
echo '</i><br /> ';
echo '<b>';
echo stripslashes($row['notetitle']);
echo '</b><br /> ';
echo stripslashes($row['note']);
echo '<br /><br /> ';
echo '</p>';
}
$result->free();
$db->close();
?>
Now I would like to display an edit-link for each row displayed, that can open a new page in which it is possible to edit a specific row. I already have the code that lets you edit the row:
<?php
if ($_REQUEST['save']=="Save") { // is data submitted?
// create variables
$noteid = $_REQUEST['noteid'];
$coursename = $_REQUEST['coursename'];
$notetitle = $_REQUEST['notetitle'];
$note = $_REQUEST['note'];
$query = "UPDATE notes SET ";
$query .= "coursename='$coursename', ";
$query .= "notetitle='$notetitle', ";
$query .= "note='$note' ";
$query .= "WHERE noteid='$noteid'";
$result = $db->query($query);
} elseif ($_REQUEST['delete']=="Delete") { // is data to be removed?
$noteid = $_REQUEST['noteid'];
$query="DELETE FROM notes WHERE noteid='$noteid'";
$result = $db->query($query);
}
?>
<div class="formular">
<div class="row1">
<p>Id</p>
<p>Notetitle</p>
<p>Note</p>
</div>
<?php
$query = "SELECT * FROM notes ORDER BY noteid DESC";
$result = $db->query($query);
while ($row = mysqli_fetch_array($result)) {
echo "<form ".$_SERVER['PHP_SELF']." name='edit-form' method='post' class='row1'>\n";
echo "<p class='align_top padding_top'>".$row['noteid']."<input type='hidden' name='noteid' value='".$row['noteid']."' /></p>\n";
echo "<p class='align_top'><input type='text' name='notetitle' value='".$row['notetitle']."' /></p>\n";
echo "<p><textarea name='note' rows='10' cols='50'>".$row['note']."</textarea></p>\n";
echo "<p><input type='submit' name='save' value='Save' /></p>";
echo "<p><input type='submit' name='delete' value='Delete' /></p>";
echo "</form>\n";
}
echo '</div>';
$result->free();
$db->close();
?>
What I am struggling with is how to display an edit-link for each row that lets you open a page where you can edit/delete the content of only that row.
I hope someone can help, I am very new at this.
Thank you!
Add a button next to each row that opens an edit page (or modal) with the id inside, example: <button onclick="edit('randomId')">Edit RandomId </button>
You could implement something different that accepts the unique id of that specific row and open a new page or modal with it.
This code looks horrible and I know. I don't know how to fix it though. When I try and update the table using the edit web page, only the first row in the first column updates but the subtitle is not updating in the second column, first row. Is there a way to change this? Sorry for the terrible explanation.
Update Page
//Home Title
$homeTitleUpdate = $_POST["homeTitleChange"];
$editRow = $_POST["rowID"];
$query = " UPDATE Home SET title = '$homeTitleUpdate' WHERE homeID = '$editRow' ";
$result = mysqli_query($conn, $query) or die(mysqli_error($conn));
if ($result) {
echo "<p> - Title updated succesfully to $homeTitleUpdate.</p>";
} else {
echo "<p> - Title did not update. Something went wrong</p>";
}
//Home Subtitle
$homeSubtitleUpdate = $_POST["homeSubtitleChange"];
$query1 = " UPDATE Home SET subtitle = '$homeSubtitleUpdate' ";
$result1 = mysqli_query($conn, $query) or die(mysqli_error($conn));
if ($result) {
echo "<p> - Subtitle updated successfully to $homeSubtitleUpdate.</p>";
} else {
echo "<p> - Subtitle did not update. Something went wrong</p>";
}
Edit Page
<?php
echo "<h2 style='color:black'>";
echo "<input type'text' name='homeTitleChange' value=$homeTitle>";
echo "<input type='hidden' name='rowID' value=$getID>";
echo "</h2>";
echo "<h4 style='color:black'>";
echo "<input type'text' name='homeSubtitleChange' value=$homeSubtitle>";
echo "<input type='hidden' name='rowID' value=$getID>";
echo "</h4>";
?>
<input type="submit" value="save" />
<?php
echo "<h2 style='color:black'>";
echo "<form action="change to your file" method="post">
echo "<input type'text' name='homeTitleChange' value=$homeTitle>";
echo "<input type='hidden' name='rowID' value=$getID>";
echo "</h2>";
echo "<h4 style='color:black'>";
echo "<input type'text' name='homeSubtitleChange' value=$homeSubtitle>";
echo "<input type='hidden' name='rowID' value=$getID>";
echo "</h4>";
?>
<input type="submit" name="submit" value="save" />
</form>
You did not have a form
//Home Title
if(isset($_POST['submit'])){
if
(
!empty($_POST["homeTitleChange"])
&&
!empty($_POST["homeSubtitleChange"]) &&
!empty($_POST["rowID"])
)
{
$homeTitleUpdate = $_POST["homeTitleChange"];
$homeSubtitleUpdate = $_POST["homeSubtitleChange"];
$editRow = $_POST["rowID"];
$query = "UPDATE Home SET title = '$homeTitleUpdate', subtitle ='$homeSubtitleUpdate' WHERE homeID = '$editRow' ";
$result = mysqli_query($conn, $query) or die(mysqli_error($conn));
if ($result) {
echo "<p> - Title/Subtitle updated succesfully to $homeTitleUpdate.</p>";
} else {
echo "<p> - Title/Subtitle did not update. Something went wrong</p>";
}
}
}
You can change your php and do it all within one query
You need to add 'where' condition while updating subtitle as well
$query1 = " UPDATE Home SET subtitle = '$homeSubtitleUpdate' WHERE homeID = '$editRow' ";
on other hand, you can update both of them in single query, like this
$query = " UPDATE Home
SET title = '$homeTitleUpdate', subtitle = '$homeSubtitleUpdate'
WHERE homeID = '$editRow' ";
wouldn't this be better? unless you have some specific reason
I am coding some video upload script and I am with the admin panel right now.
There I have a List with all Videos. And each video has one delete button on the right side.
When I click the button then this video should be deleted from database but its not working after I click the button nothing happens.
<?php
$query = mysql_query("SELECT * FROM `videos`");
while($row = mysql_fetch_assoc($query))
{
$id = $row['id'];
$name = $row['name'];
echo "<a href='watch.php?id=$id'>$name</a><br />
<input type='submit' name='remove' value='Delete'<br />";
}
if (isset($_POST['remove']))
{
foreach ($_POST['id'] as $the_id)
{
if (!mysql_query("DELETE FROM videos WHERE id = '$the_id'"))
{
echo mysql_error();
}
}
}
?>
Of course on the header I have the mysql connect query. This is just the php code for listing all videos and try to delete.
Here is an example of doing this with MySQLi, including binding parameters to avoid SQL injection -
if (isset($_POST['remove'])) {
$remove = $mysqli->prepare("DELETE FROM `videos` WHERE `id` = ?");
$vid_id = $_POST['vid_id'];
$remove->bind_param('s', $vid_id);
if(!$remove->execute() === true) {
echo $mysqli->error;
}
}
$query = "SELECT * FROM `videos`";
if ($result = $mysqli->query($query)) {
while($row = $result->fetch_object()){
$id = $row->id;
$name = $row->name;
echo "<a href='watch.php?id=$id'>$name</a><br />";
echo "<form name='delete_vid' method='post'>";
echo "<input type='hidden' name='vid_id' value='$id'>";
echo "<input type='submit' name='remove' value='Delete'<br />";
echo "</form>";
}
} else {
echo mysqli_error($connection);
}
$result->close();
Of course you will have to provide a $connection` to the database, but thsi should get you started not only with MySQLi but with adding a form for each video.
More on SQL Injection
Maybe the problem is in the html, each delete button must be in and independent form, with a hidden input with the id too.
echo "<a href='watch.php?id=$id'>$name</a><br />
<form method='post'><input type='hidden' value='$id'><input type='submit' name='remove' value='Delete'<br /></form>";
<form method="post" >
<?php
$query = mysql_query("SELECT * FROM `videos`");
while($row = mysql_fetch_assoc($query))
{
$id = $row['id'];
$name = $row['name'];
echo "<a href='watch.php?id=$id'>$name</a><br />
<button name='id' value='".$id."' type='submit' >Delete</button>
<br />";
}
if (!mysql_query("DELETE FROM videos WHERE id = '".$_POST['id']."'"))
{
echo mysql_error();
} else {
echo 'successfully deleted';
}
?>
</form>