I have this code
<html>
<head>
<meta charset="UTF-8">
<title> Game Library</title>
<link href="css/style.css" type="text/css" rel="stylesheet">
</head>
<body>
<div id="wrapper">
<div id="search">
<img id="del" src="img/delete.jpg" alt="Error">
<a class="A" href="index.php"><img class="AR" src="img/src.jpg" title="Search library"></a>
<a class="A" href="insert.php"><img class="AR" src="img/add.png" title="Add game"></a>
<div id="results">
<?php
require("inc/connection.php");
$query = "SELECT * FROM Library";
$result = mysqli_query($conn,$query);
if(mysqli_num_rows($result)>0){
while($row = mysqli_fetch_assoc($result)){
?><div id="results">
<form action="inc/delete.php" method="GET">
<input type="checkbox" name="checkbox[]" value="<?php echo $row['ID'] ?>">
<p id="p1">Name: <?php echo $row['Name']?>;</p>
<p>Genre: <?php echo $row['Genre']?></p>
<p>Release date: <?php echo $row['Release_date']?></p>
<p>Publisher: <?php echo $row['Publisher']?>:</p>
<p>Platforms: <?php echo $row['Platforms']?></p>
</div>
<?php
}?>
<input type="submit" name = "submit" value = "Submit"></form>
<?php
}else{
echo "No results!";
}
</div>
</div>
</div>
</body>
</html>
And I am supposed to create action="inc/delete.php" script which will allow me to delete multiple rows using checkbox and submit button. Can someone help me out with this please ? I honestly have no clue on how to complete this ....
<?php
if(isset($_POST['submit']))
{
if(!empty($_POST['checkbox']))
{
foreach ($_POST['checkbox'] as $key => $check)
{
// delete query
}
}
}
?>
NOTE: You failed to close PHP at line no. 40. Please make sure your PHP tag is closed or not.
Need to first set form method to `POST'
So, modify
<form action="inc/delete.php" method="GET">
To:
<form action="inc/delete.php" method="post">
And in inc/delete.php,
if (isset($_POST['submit'])){ // Check if form is submitted.
if (! empty($_POST['checkbox'])){ // Check if Checkbox is checked.
// Loop over the checkbox and get selected ids.
foreach($_POST['checkbox'] as $checkbox_id){
echo $checkbox_id."<br/>";
// Add your delete query here.
}
}
}
So the resolution is that you get at your delete script array of ids. You can parse that array and execute DELETE query for each id:
if(isset($_POST['checkbox']))
{
foreach($_POST['checkbox'] as $val)
{
$stmt = $conn->prepare("DELETE FROM Library WHERE id = ?");
$stmt->bind_param('i', $val);
$stmt->execute();
}
}
inc/delete.php
<?php
if(isset($_POST['submit'])){//to run PHP script on submit
if(!empty($_POST['checkbox'])){
// Loop to store and display values of individual checked checkbox.
foreach($_POST['checkbox'] as $row_id){
echo $row_id."</br>";
// DELETE QUERY HERE
}
}
}
?>
Related
I need help on my code. When data is entered on my site, it does not show up in the mySQL data table. The insert function that I used might be the problem, but I cannot figure out how to get it to actually insert and show up in my table in my database. Can someone please guide me in the right direction with my code?
<?php
session_start();
include("db_connect.php");
if(isset($_POST['submit'])){
$item = $_POST['item'];
if(empty($item)) {
$errors = "you must enter something";
}
else{
mysqli_query("INSERT INTO a4_todolist (item) VALUES ('$item')");
header('location: index.php');
}
}
$a4_todolist = mysqli_query("SELECT * FROM a4_todolist");
?>
<!DOCTYPE html>
<html>
<head>
<title> Assignment 4 - To Do List </title>
<link rel ="stylesheet" type ="text/css" href="style.css">
</head>
<body>
<div class "head">
<h2> To Do </h2>
</div>
<form method= "POST" action = "index.php">
<?php if (isset($errors)) { ?>
<p><?php echo $errors; ?></p>
<?php } ?>
Item <input type = "text" name= "item" class="item_input">
Author <input type = "text" name= "author" class="author_input">
<button type = "submit" class="add-btn" name="submit"> Add Task
</button>
</form>
<table>
<tbody>
<?php while ($row = mysqli_fetch_array($a4_todolist)) { ?>
<tr>
<td class="id"> <?php print $row['id']; ?> </td>
<td class="item"> <?php echo $row['item']; ?> </td>
</tr>
<?php } ?>
</tbody>
</table>
</thread>
</body>
</html>
You have missed the sql connection variable which is coming from db_connect.php file inside your mysqli_query. Your mysqli_query() should be like this
mysqli_query($connection,"INSERT INTO a4_todolist (item) VALUES ('$item')");
Also this
$a4_todolist = mysqli_query($connection,"SELECT * FROM a4_todolist");
It seems that you are a beginner so I recommend you to learn Prepared statements which is more efficient and safe to use.
You should pass the connection link identifier as well as you can check for errors.
$con = mysqli_connect("localhost","my_user","my_password","my_db");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
Also after executing query you can again check for error.
if (!mysqli_query($con,"INSERT INTO a4_todolist (item) VALUES ('$item')")) {
echo("Error description: " . mysqli_error($con));
}
I am creating a quiz builder using SQL, HTML and PHP, whereby the user can insert their question and four answer choices for that question (A, B, C or D).
In the same form they can then choose what choice will be the correct answer from a dropdown. This should then assign the 'is correct' boolean value of that certain answer row to 1 in the database.
[my database structure and what my form should look like is attached]
<?php
include ("nav.php");
include("connect.php");
$quiz_id = htmlentities($_GET["quizid"]);
$quiz = "SELECT * FROM `q_quiz` WHERE q_quiz.id ='$quiz_id'";
$quizresult= $conn->query($quiz);
if(!$quizresult){
echo $conn->error;
}
?>
<html>
<head>
<title></title>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<link rel="stylesheet" type="text/css" href="ui/styles.css">
<!-- UIkit CSS -->
<link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/uikit/3.0.3/css/uikit.min.css" />
<!-- UIkit JS -->
<script src="https://cdnjs.cloudflare.com/ajax/libs/uikit/3.0.3/js/uikit.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/uikit/3.0.3/js/uikit-icons.min.js"></script>
</head>
<body>
<div class ='uk-container'>
<?php
if (isset($_POST['submit'])) {
include("connect.php");
$newquest = $conn->real_escape_string($_POST['question']);
$insertquest = "INSERT INTO q_questions(quiz_id, question)
VALUES ('$quiz_id', '$newquest')";
$resultquest = $conn->query($insertquest);
$quest_id = $conn -> insert_id;
if (!$resultquest) {
echo $conn->error;
}
foreach ($_POST['ans'] as $ans) {
$answers = $conn->real_escape_string($ans);
$insertans = "INSERT INTO q_answers (question_id, answer) VALUES ('$quest_id', '$answers')";
$resultans = $conn ->query($insertans);
if (!$resultans) {
echo $conn->error;
} else {
header("Location: managequiz.php?quizid=$quiz_id");
}
} //END OF FOREACH LOOP
}
?>
<?php
while ($row = $quizresult->fetch_assoc()) {
$titledata = $row["title"];
}
?>
<div uk-grid>
<form class="uk-form-horizontal" method='POST' action='#'>
<input type='hidden' name='questionid' value=''>
<fieldset class="uk-fieldset">
<legend class="uk-legend">New question for quiz: <?php echo $titledata?></legend>
<div class="uk-margin">
<label class="uk-form-label">Question</label>
<div class="uk-form-controls">
<textarea class="uk-textarea" rows="5" placeholder="Enter your question here..." name='question' required></textarea>
</div>
</div>
<div class="uk-margin">
<div class="uk-form-label">A</div>
<div class="uk-form-controls uk-form-controls-text">
<input class="uk-input" id="form-horizontal-text" type="text" name='ans[]'>
</div>
<div class="uk-form-label">B</div>
<div class="uk-form-controls uk-form-controls-text">
<input class="uk-input" id="form-horizontal-text" type="text" name='ans[]'>
</div>
<div class="uk-form-label">C</div>
<div class="uk-form-controls uk-form-controls-text">
<input class="uk-input" id="form-horizontal-text" type="text" name='ans[]'>
</div>
<div class="uk-form-label">D</div>
<div class="uk-form-controls uk-form-controls-text">
<input class="uk-input" id="form-horizontal-text" type="text" name='ans[]'>
</div>
<div class="uk-form-label">Correct Answer:</div>
<div class="uk-form-controls">
<select class="uk-select" id="form-stacked-select">
<option value=''>A</option>
<option value=''>B</option>
<option value=''>C</option>
<option value=''>D</option>
</select>
</div>
<div uk-form-custom>
<input class="uk-button uk-button-default" type="submit" name='submit' tabindex="-1" value='Save & Continue'>
</div>
</div>
</fieldset>
</form>
</div>
</body>
</html>
as kiks73 mentioned, you can assign values to the option of the Select and POST the selected one to your database as correct answer ID.
<html>
<select class="uk-select" id="form-stacked-select" name="correctAnswer">
<option value='1'>A</option>
<option value='2'>B</option>
<option value='3'>C</option>
<option value='4'>D</option>
</select>
</html>
If you select A, the value "1" would be sent to your database as correct answer (as long as you correct the SQL part in your php.
EDIT:
I got your question a bit late, sorry.
you need to POST the select aswell in order to get the nessecary information about the correct answer. You could try something like that:
<?php
//just a counter
$i = 1;
//correct boolean 1 or 0
$bool = 0;
if (isset($_POST['submit'])) {
include("connect.php");
$newquest = $conn->real_escape_string($_POST['question']);
$insertquest = "INSERT INTO q_questions(quiz_id, question)
VALUES ('$quiz_id', '$newquest')";
$resultquest = $conn->query($insertquest);
$quest_id = $conn -> insert_id;
if (!$resultquest) {
echo $conn->error;
}
foreach ($_POST['ans'] as $ans) {
//get the value of the selected answer
$correctAnswer = $_POST['correctAnswer'];
//check if the selected one matches with our counter
if($i == $correctAnswer)
{
$bool = 1;
}
else
{
$bool = 0;
}
$answers = $conn->real_escape_string($ans);
$insertans = "INSERT INTO q_answers (question_id, answer, correct) VALUES ('$quest_id', '$answers', $bool)";
$resultans = $conn ->query($insertans);
if (!$resultans) {
echo $conn->error;
} else {
header("Location: managequiz.php?quizid=$quiz_id");
}
$i ++;
} //END OF FOREACH LOOP
}
?>
If you POST the whole form, you can use anything in it. i made an ugly loop, which checks each time you instert an answer to your SQL DB, if this one is selected.
So I have a script using HTML, PHP, and mysql, and I want to display a button under certain circumstances.
Here is my script:
<?php
include_once('dbconnect.php');
$q = $_POST['q'];
$q = $_GET['query'];
$query = mysqli_query($conn,"SELECT * FROM `Persons` WHERE `id` LIKE '%$q%'");
$count = mysqli_num_rows($query);
if($count != "1"){
$output = '<h2>No result found!</h2>';
}else{
while($row = mysqli_fetch_array($query)){
$s = $row['name'];
$output .= '<h2>Found: '.$s.'</h2><br>';
}
}
?>
<!DOCTYPE html>
<html>
<head>
<title>Search</title>
</head>
<body>
<form method="POST" action="index.html">
<input type="submit" name="return" value="Return">
</form>
<?php echo $output; ?>
</body>
</html>
Specifically, I want to display the return button only when the output is "No results found", when the amount of rows in the SQL database matching the given query is not 1. How could I go about accomplishing this? I'm relatively new to PHP and mySQLi, but from my research I couldn't figure out how to do such a task, any ideas?
<?php
if ($count==0) {
echo '<input type="submit" name="return" value="Return">';
}
?>
If you want a much cleaner html code, do this:
<form method="POST" action="index.html">
<?php if ($count!= "1") : ?>
<input type="submit" name="return" value="Return">
<?php else : ?>
<!-- put your other button here -->
<?php endif; ?>
</form>
You can read more about escaping from HTML here.
<?php
include_once('dbconnect.php');
$q = $_POST['q'];
$q = $_GET['query'];
$query = mysqli_query($conn,"SELECT * FROM `Persons` WHERE `id` LIKE '%$q%'");
$results = mysqli_fetch_array($query);
?>
<!DOCTYPE html>
<html>
<head>
<title>Search</title>
</head>
<body>
<form method="POST" action="index.html">
<input type="submit" name="return" value="Return">
</form>
<?php if(0 < count($results)) ?>
<?php foreach($results AS $row) : ?>
<h2><?= $row['name'] ?></h2>
<?php endforeach; ?>
<?php else : ?>
<H2> No results found!</h2>
<?php endif; ?>
</body>
</html>
I displayed a set of questions.
On selecting one of those questions for which I required the answers to be posted on the same page, but the problem is I am not getting id parameter from url which I am passing on selectiong the question.
Now when I am trying to answer the question which is selected, I will need the id of question to post the answer of that perticular question which I already have in url as a parameter for Example: id=1.
Here is the body section of html page:
<?php
include("menu/menu.php");
$sqli = "SELECT * FROM forum_question where id='$id'";
$result=mysqli_query($conn,$sqli);
?>
<form action="submit_answer.php" method="post" name="answers">
<br> <br> <br>
<?php
while($row = mysqli_fetch_array($result))
echo "Q".$row['detail'];
?>
<br>
answers:<br>
<textarea class="tinymce" name="answers"></textarea>
<input type="hidden" name="id" value="<?php echo $id;?>">
<br> <br>
<input type="submit" value="submit" name="submit">
After submit the page "submit_answer.php", Code is:
<?php
include'config.php';
if($conn){
if ($_SERVER["REQUEST_METHOD"] == "POST") {
$answers = $_REQUEST['answers'];
$id= $_GET ['id'];
}
$sqli= "INSERT INTO answers (answers)
VALUES ('$answers')";
if (mysqli_query( $conn,$sqli))
{
echo "New record created successfully";
header("location:answer.php?id='$id'");
} else {
echo "Error: " . $sqli . "<br>" . $conn->error;
}
}else{
}
mysqli_close($conn);
?>
Basically I am very much fresher in Php I just want to know how should I get the id of question and submit it to the "submit_answer.php" with the answer content.
just take a hidden field below the answer field and get the url parameter to that hidden field on page load, as said by user buivankim2020 and submit submit_answer.php,
after submit get the value of that field in variable like what you do for getting the answer..
You should change it (add hidden input for id in markup html)
<?php
include'config.php';
//session_start();
$id= $_GET ['id'];
?>
<html>
<head>
<link rel="stylesheet" type="text/css" href="css/style.css">
<script type="text/javascript" src="tinymce/js/jquery.min.js"></script>
<script type="text/javascript" src="tinymce/plugin/tinymce/tinymce.min.js"></script>
<script type="text/javascript" src="tinymce/plugin/tinymce/init-tinymce.js"></script>
</head>
<body>
<div id="container">
<div id="main">
<?php
include("menu/menu.php");
$sqli = "SELECT * FROM forum_question where id='$id'";
$result=mysqli_query($conn,$sqli);
?>
<form action="submit_answer.php" method="post" name="answers">
<br> <br> <br>
<?php
while($row = mysqli_fetch_array($result))
echo "Q".$row['detail'];
?>
<br>answers:<br>
<textarea class="tinymce" name="answers"></textarea>
<input type="hidden" name="id" value="<?php echo $id;?>">
<br> <br>
<input type="submit" value="submit" name="submit">
</form>
</body>
</html>
submit_answer.php
<?php
include'config.php';
if($conn){
if (isset($_POST['answers']) && isset($_POST['id'])) {
$answers = $_POST['answers'];
$id= $_POST['id'];
$sqli= "INSERT INTO answers (answers) VALUES ('$answers')";
if (mysqli_query( $conn,$sqli))
{
echo "New record created successfully";
header("location:answer.php?id='$id'");
} else {
echo "Error: " . $sqli . "<br>" . $conn->error;
}
}
mysqli_close($conn);
}
?>
I am trying to make another page to appear when I click certain button, I basically know how to do this, but this time I am in trouble.I have following code:
<html>
<head>
<meta charset="UTF-8"/>
</head>
<body>
<!--See siin all tekstiväli-->
<H3>Minu küsitlused </H3>
<hr>
<br>
<br>
<br>
<ol>
<?php
include_once 'init/init.funcs.php';
$result = mysql_query('SELECT * from katse_kysimustik_pealkiri');
while($row = mysql_fetch_assoc($result)) {
$titles[] = $row['pealkiri'];
}
foreach($titles as $title) {
?>
<li>
<?php echo $title ?>
<form action='Minu_kysitlused_1.php'>
<input type="button" name = "saada" value="saada">
<input type="button" value="tulemused">
<input type="button" value="lõpeta ennetähtaegselt">
<input type="button" value="X">
</li>
</form>
<?php
}
?>
</ol>
</body>
</html>
<?php
if(isset($_POST['saada'])){
echo "<meta http-equiv='refresh' content='0;url=http://localhost/Praks/saada.html'>";
}
?>
Everything works just fine but when I click button 'saada' nothing happens. What should I do to make saada.html appear on this click?
replace this:
<form action='Minu_kysitlused_1.php' method="post">
Missing:
method="post"
why you try with meta tag
You just put header("location:yourpagename");
just like below
if(isset($_POST['saada'])){
header("Location:http://localhost/Praks/saada.html");
}
try like this
header function must be First thing to sent as html otherwise it will not work
<?php
if(isset($_POST['saada'])){
header( "Location : localhost/Praks/saada.html");
die();
}
?>
<html>
<head>
<meta charset="UTF-8"/>
</head>
<body>
<!--See siin all tekstiväli-->
<H3>Minu küsitlused </H3>
<hr>
<br>
<br>
<br>
<ol>
<?php
include_once 'init/init.funcs.php';
$result = mysql_query('SELECT * from katse_kysimustik_pealkiri');
while($row = mysql_fetch_assoc($result)) {
$titles[] = $row['pealkiri'];
}
foreach($titles as $title) {
?>
<li>
<?php echo $title ?>
<form action='Minu_kysitlused_1.php'>
<input type="button" name = "saada" value="saada">
<input type="button" value="tulemused">
<input type="button" value="lõpeta ennetähtaegselt">
<input type="button" value="X">
</li>
</form>
<?php
}
?>
</ol>
</body>
</html>
second approach
<form action="your-url">
<input type="submit" name="Submit" value="saada"/>
</form>
in first time, I don't understand why do you use the form tag to call a page, if there aren't any data passed via POST?
Anyway, in your foreach cicle you can use a very simple collection of link:
foreach($titles as $title) {
echo $title;
echo 'Link 1';
echo 'Link 2';
echo 'Link 3';
echo 'Link 4';
}
This solution is too clean.
Bye
Marco