PHP Delete event ID - php

I have made an Event calender but now I'm facing a problem when the user deletes a event.
Screenshot: http://i.stack.imgur.com/LBLJ9.png
<h2>Date <?php echo "$day"."/"."$month"."/"."$year"."<br>"; ?></h2>
<?php
// event weer gegeven
while ($events = mysql_fetch_array($resultEvents)){
echo "<strong>Event ID:</strong> ".$events['id']."<br>";
echo "Added: ".$events['added']."<br>";
echo "Title: ".$events['titel']."<br>";
echo "Detail: ".$events['content']."<br>";
?>
<form method="POST" action="<?php $_SERVER['PHP_SELF']; ?>">
<input type="submit" name="delete" value="Delete"><br >
<input type="submit" name="edit" value="Edit"><br >
</form>
<?php
if(isset($_POST['delete'])){
$user = $_POST['delete'];
$delet_query = mysql_query("DELETE FROM kalender_contents WHERE `id` = '$events'") or die(mysql_error());
if($delet_query) {
echo "event deleted";
echo $events;
}
}
}
}
}
?>
But the WHEREid= '$events' does not work. How can I specify the ID on the button click.

Try to put its relevant event id into the form like
<form method="POST" action="<?php $_SERVER['PHP_SELF']; ?>">
<input type="hidden" name="event_id" id="event_id" value="<?php echo $events['id'];?>">
<input type="submit" name="delete" value="Delete"><br >
<input type="submit" name="edit" value="Edit"><br >
</form>
And in your submit query use like
$delet_query = mysql_query("DELETE FROM kalender_contents
WHERE `id` = $_POST['event_id']");
Whenever you are submitting the form then its event id will be passed and you need to pass that post variable to your delete query.
Further you need to escape the id to prevent the sql injection with mysql_real_escape_string at your query like
$delet_query = mysql_query("DELETE FROM kalender_contents
WHERE `id` = ".mysql_real_escape_string($_POST['event_id']));

Create a hidden field with value of event id.
<h2>Date <?php echo "$day"."/"."$month"."/"."$year"."<br>"; ?></h2>
<?php
// event weer gegeven
while ($events = mysql_fetch_array($resultEvents)){
echo "<strong>Event ID:</strong> ".$events['id']."<br>";
echo "Added: ".$events['added']."<br>";
echo "Title: ".$events['titel']."<br>";
echo "Detail: ".$events['content']."<br>";
?>
<form method="POST" action="<?php $_SERVER['PHP_SELF']; ?>">
<input type="hidden" name="event_id" value="<?php echo $events['id'];?>">
<input type="submit" name="delete" value="Delete"><br >
<input type="submit" name="edit" value="Edit"><br >
</form>
<?php
if(isset($_POST['delete'])){
$user = $_POST['delete'];
$delet_query = mysql_query("DELETE FROM kalender_contents WHERE `id` = '$events'") or die(mysql_error());
if($delet_query) {
echo "event deleted";
echo $events;
}
}
}
}
}
?>

Related

How i can get value from radio button form in php?

Question.php
<?php
include 'Pre-function.php'
?>
<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" href="CSS/Start.css">
</head>
<body>
<div class="nav">
Home
News
Contact
</div>
<div class="question">
<div class="A4">
<form action="Answer.php" method="post">
<?php getQuestion($conn); ?>
<input type="submit" name="Submit" value="Submit">
</form>
</div>
</div>
</body>
</html>
Its html page to ask question
Pre-function.php
<?php
include 'conn.php';
function getQuestion($conn) {
$query = "SELECT * FROM question ";
$result = mysqli_query($conn, $query);
if($result){
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
$question_id = $row['question_id'];
$question_body = $row['question_body'];
$option_a = $row['option_a'];
$option_b = $row['option_b'];
echo '
<h2 class="qtitle">'.$question_body.'</h2>
<label for='.$question_body.'>Yes</label>
<input type="radio" name="'.$question_id.'" value="Yes">
<input type="hidden" name="option_a" value="'.$option_a.'">
<label for="'.$question_body.'">No</label>
<input type="radio" name="'.$question_id.'" value="No">
<input type="hidden" name="option_b" value="'.$option_b.'">
<input type="hidden" name="submitted" value="submitted"><hr>';
}
}
}
?>
Basically this form asked question whether yes or no using radio button. $option_a == 'Yes' and $option_b == 'No'. The question is like this "Are you have fever ?". So when i submit the value did not pass to the 'Answer.php' page.
'Answer.php' page.
<?php
include 'conn.php';
if(isset($_POST['Submit']) && !empty($_POST['Submit'])){
echo $_POST['option_a'];
echo 'succeed';
}
else{
echo 'no data';
}
?>
In this page have error undefined_index value but still echo succeed.
Your HTML code should look like the following:
<input type="radio" name="whatevername" value="Yes">
<input type="hidden" name="whatevername" value="No">
You can use PHP to insert whatever values you want but you need the radio button name to be the same.
Then if you want to echo that in PHP you'd use:
echo $_POST['whatevername']; //same name you used in the form
You are passing value as name for radio buttons
name="'.$option_a.'"
This will result you in name ="Yes"
And you are trying to fetch echo $_POST['option_a']; where option_a is not defined.
Try this
<input type="radio" name="'.$question_id.'" value="Yes">
<input type="hidden" name="option_a" value="'.$option_a.'">
Same for other radio button
Try this one :
<?php
include 'conn.php';
function getQuestion($conn) {
$query = "SELECT * FROM question ";
$result = mysqli_query($conn, $query);
if($result){
echo '<div class="A4">';
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
$question_id = $row['question_id'];
$question_body = $row['question_body'];
$option_a = $row['option_a'];
$option_b = $row['option_b'];
echo '
<div class="A4">
<h2 class="qtitle">'.$question_body.'</h2>
<form action="Answer.php" method="post">
<label for='.$question_body.'>Yes</label>
<input type="radio" name="radioQuestions[]" value="'.$question_id.'-'.$option_a.'">
<label for="'.$question_body.'">No</label>
<input type="radio" name="radioQuestions[]" value="'.$question_id.'-'.$option_b.'">
<input type="hidden" name="submitted" value="submitted"><hr>';
}
echo'
<input type="submit" name="Submit" value="Submit">
</form>
</div>';
}
}
?>
<?php
include 'conn.php';
if(isset($_POST['submitted']) && !empty($_POST['submitted'])){
$questionAndOptions = $_POST['radioQuestions'];
foreach ($questionAndOptions as $questionAndOption) {
$arrQuestionAndOption = explode("-", $questionAndOption);
echo $arrQuestionAndOption[0]; //question
echo $arrQuestionAndOption[1]; //option
}
echo 'succeed';
}
else{
echo 'no data';
}
?>

How to add caption field for every image preview and post all caption through one submit button

This is my code I want to add one input field for every image preview and save it to db.. the field is coming but I'm not getting any data.. can anyone suggest how can I post them???
$fetch_imgid=$con->prepare("SELECT * FROM attempt010 where link='$rand'");
$fetch_imgid->setFetchMode(PDO:: FETCH_ASSOC);
$fetch_imgid->execute();
?>
<ul class="reorder_ul reorder-photos-list" id="previewImg">
<?php
while($row = $fetch_imgid->fetch()):
$delid = $row['id'];
//echo $row['id'].' '.$row['name'].'<br/>';?>
<li id="image_li_<?php echo $row['id']; ?>" class="ui-sortable-handle" data-image-id="<?php echo $delid; ?>">
<img src="uploads/<?php echo $row['name']; ?>" alt="">
<input type="submit" class="del_btn" value="Delete Image" />
<input type="text" id="cap" name="cap[]" placeholder="Enter Caption Here" />
<input type="hidden" id="cap_id" value="<?php echo $row['id']; ?>" />
<?php
endWhile;
?>
</ul>
<input type="submit" value="Add Caption" name="addcap" /> <?php include('addcap.php'); ?>
and this is addcap.php
<?php
error_reporting(E_ALL);
ini_set('display_errors',1);
if(isset($_POST['addcap'])){
foreach($_POST['cap'])
{
$imgcap = $_POST['cap'];
if($imgcap!=empty())
{
try
{
$con=new PDO("mysql:host=localhost;dbname=newimg","root","");
$sql=$con->prepare("UPDATE attempt010 SET caption='$imgcap' WHERE id='$cap_id'");
$sql->execute();
}
catch(PDOException $e)
{
echo $sql . "<br>" . $e->getMessage();
}
}
}
}
?>
<input type="hidden" id="cap_id" value="<?php echo $row['id']; ?>" />
Must have unique id. You can't send multiple fields with same id. You will get only last one.
For example:
$fetch_imgid=$con->prepare("SELECT * FROM attempt010 where link='$rand'");
$fetch_imgid->setFetchMode(PDO:: FETCH_ASSOC);
$fetch_imgid->execute();
?>
<form action="addcap.php" method="post">
<ul class="reorder_ul reorder-photos-list" id="previewImg">
<?php
$id_array="";
while($row = $fetch_imgid->fetch()):
$id_array = $id_array.$row['id'].",";
$delid = $row['id'];
//echo $row['id'].' '.$row['name'].'<br/>';?>
<li id="image_li_<?php echo $row['id']; ?>" class="ui-sortable-handle" data-image-id="<?php echo $delid; ?>">
<img src="uploads/<?php echo $row['name']; ?>" alt="">
<input type="text" id="cap_<?php echo $row['id']; ?>" placeholder="Enter Caption Here" />
<?php
endWhile;
$id_array = substr($id_array, 0, -1);
?>
<input type="hidden" id="cap_ids" value="<?php echo $id_array ; ?>" />
</ul>
<input type="submit" value="Add Caption" name="addcap" />
</form>
<!--addcap.php-->
<?php
error_reporting(E_ALL);
ini_set('display_errors',1);
if(isset($_POST['addcap'])){
if(isset($_POST['cap_ids'])){
$ids_array = explode(",", $_POST['cap_ids']);
foreach($ids_array as $ids)
{
$idcap = 'cap_'.$ids;
$imgcap = $_POST[$idcap];
if($imgcap!=empty())
{
try
{
$con=new
PDO("mysql:host=localhost;dbname=newimg","root","");
$query = "UPDATE attempt010 SET
caption='$imgcap' WHERE id='$ids'";
echo $query;
$sql=$con->prepare($query);
$sql->execute();
}
catch(PDOException $e)
{
echo $sql . "<br>" . $e->getMessage();
}
}
}
}
}
?>
This code looks like it can't work. Because you have submit and form handling code in same page. The idea behind form is to post data to different page(set in form action) and this page will do something with this data and display results to the user. For your example to work make form in your first file like:
<form action="addcap.php">
<inputs here>
</form>
Nowadays it is common that database operations are done asynchronic on server side, when user can continue using the page/app.
So learn how to use jQuery and AJAX. Maybe nodeJS or other new stuff.

A simple operation but it giving an error I want to update each id

It is simple but I am not getting values of radio buttons for updation when I press submit.
It is giving an error in foreach statement like :Warning: Invalid argument supplied for foreach() Please say how to get the values of radio button after pressing submit button,
here I am updating the status of state .
<?php
include("db.php");
?>
<html>
<body>
<?php
$state=$_POST['state'];
if(isset($_POST['submit']))
{
$result1 = mysql_query("select state,id,status from states ");
while($rr1=mysql_fetch_array($result1))
{
//getting values of radio buttons
$myval=$rr1['id'];
echo $myval;
$val=$_POST['yes.$rr1["id"]'];
echo $val;
$val1=$val.$myval;
$vall=yes.$rr1['id'];
foreach($vall as $values)
{
echo $values;
$update=mysql_query("UPDATE states SET status='". $values."'
WHERE id='$myval' ");
}
}
}
echo "ok";
?>
<form action="" name="form" id="form" method="post" >
<?php
//session_start();
include("db.php");
$result = mysql_query("select state,id,status from states ");
while($info1=mysql_fetch_assoc( $result))
{
echo $info1['city'];
echo "<br>";
/*echo "<br>";
echo "company Name:".$info1['company_name'];
echo "<br>";
echo "salary:".$info1['maxsalary'];
echo "<br>";
//echo $info1['company_name'];*/
?>
<label><?php echo $info1['state']; ?></label>
<input type="radio" <?php if($info1['status']=="yes"){ echo
"checked='checked'"; } ?> name="yes.<?php echo $info1[ 'id']; ?>[]"
value="yes">Yes
<input type="radio" <?php if($info1['status']=="no"){ echo
"checked='checked'"; } ?> name="yes.<?php echo $info1[ 'id']; ?>[]"
value="no">no
<br/>
<?php } ?>
<br />
<input type="submit" name="submit" value="submit" />
</form>
</body>
</html>
I think you need to write
$vall=yes.$rr1['id'];
to
$vall="yes".$rr1['id'];
Thanks

Displaying mysql data through hidden field values

I am trying to display mysql records through hidden field values, but nothing displays.
A little help!
Here's the code;
Html:
<form name="form11" method="post" action="hpdata.php" enctype="multipart/form-data">
<input name="pro" id="pro" type="hidden" value= "CMS" />
<input name="piror" id="piror" type="hidden" value= "P1" />
<input name="stat" id="stat" type="hidden" value= "In Progress" />
<input type="submit" name="submit" id="submit" class="groovybutton" value="...">
</form>
PHP:
<?php
$project = $_POST["pro"];
$pirority = $_POST["piror"];
$status = $_POST["stat"];
mysql_connect ("one", "two", "three");
mysql_select_db ("wsms");
$rest = mysql_query("SELECT * FROM sheet WHERE project='$project' AND
pirority='$pirority' AND status='$status'");
while($row = mysql_fetch_array($rest))
{
echo $row['id'] . " " . $row['date']; echo "<br>";
}
?>
Put isset into your php code
Example
<?php
if(isset($_POST['submit'])){
echo $project = $_POST["pro"]."<br>";
echo $pirority = $_POST["piror"]."<br>";
echo $status = $_POST["stat"];
/* mysql_connect ("one", "two", "three");
mysql_select_db ("wsms");
$rest = mysql_query("SELECT * FROM sheet WHERE project='$project' AND
pirority='$pirority' AND status='$status'");
while($row = mysql_fetch_array($rest))
{
echo $row['id'] . " " . $row['date']; echo "<br>";
}*/
}
?>
<form name="form11" method="post" action="" enctype="multipart/form-data">
<input name="pro" id="pro" type="hidden" value= "CMS" />
<input name="piror" id="piror" type="hidden" value= "P1" />
<input name="stat" id="stat" type="hidden" value= "In Progress" />
<input type="submit" name="submit" id="submit" class="groovybutton" value="...">
</form>
Output
CMS
P1
In Progress
First of all check that if the data is coming in the post or not:
<?php
echo "<pre>";
print_r($_POST);
exit;
?>
If yes than remove the print code i provided , and use extract($_POST); at the top of your PHP code. You query will become like this:
$rest = mysql_query("SELECT * FROM sheet WHERE project='$pro' AND
pirority='$piror' AND status='$stat'");

Updating Post not working

I recently made a code that updates my posts on my blog. It worked perfectly on localhost. But when i uploaded it online it did not work any more. The weird thing is it doesn't even display a error so i have no idea where to look. Can someone please help me ?
require('config.php');
$query = "SELECT * FROM project ORDER BY idproject DESC";
$result = mysqli_query($verbinding, $query ) or die (mysqli_error('kan geen verbinding maken met de database'));
if(isset($_POST['editBut'])){
$editTitle = $_POST['editName'];
$editThis = mysqli_query($verbinding, "SELECT * FROM project WHERE title = '".$editTitle."'");
$values = mysqli_fetch_assoc($editThis);
}
if(isset($_POST['update'])){
$editedTitle = $_POST['newTitle'];
$editedText = $_POST['newTekst'];
$oldTitle = $_POST['oldTitle'];
$date = $_POST['datum'];
$updater = mysqli_query($verbinding, "UPDATE Project SET title='".$editedTitle."', content='".$editedText."' WHERE title='".$oldTitle."' AND datum='".$date."'");
echo $updater;
header('location:editPost.php?id=1');
}
if(isset($_GET['id'])){
echo 'post has been succesfully updated';
}
<?php if(isset($_POST['editBut'])){ ?>
<form action="" method="post">
Title: <input type="text" name="newTitle" value="<?php echo $values['title'] ?>"><br>
Text: <textarea type="text" name="newTekst" id="newTekst"><?php echo $values['content'] ?></textarea><br>
<input type="hidden" value="<?php echo $values['title'] ?>" name="oldTitle">
<input type="hidden" value="<?php echo $values['datum'] ?>" name="datum">
<input type="submit" name="update" value="Edit post">
</form>
<?php } else { ?>
<p>Find the post you want to edit:</p>
<form action="" method="post">
<select name="editName">
<?php
while ($row = mysqli_fetch_assoc($result)) {
?> <option value="<?php echo $row['title'] ?>"><?php echo $row['title'] ?></option>
<?php } ?>
</select>
<input type="submit" name="editBut" value="Choose">
</form>
<?php } ?>
In update query replace your table name with small letter.
replace Project with project

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