I want to upload an image and insert it into the database from different page. I want to create an admin panel where you can upload an image to image sider but the image slider is in a different page. It's working when my form method is inside the index.php, but when I put it to my admin.php it's not working. I think I need a GET method?
Can someone give me idea what method, requirements to do that? I'm new to php and sql.
Here is my index.php code this is where I want to show the slide.
<?php
//for connecting db
include('connect.php');
if (!isset($_FILES['image']['tmp_name'])) {
echo "";
}
else
{
$file=$_FILES['image']['tmp_name'];
$image= addslashes(file_get_contents($_FILES['image']['tmp_name']));
$image_name= addslashes($_FILES['image']['name']);
move_uploaded_file($_FILES["image"]["tmp_name"],"gallery/" . $_FILES["image"]["name"]);
$photo="gallery/" . $_FILES["image"]["name"];
$query = mysqli_query($mysqli, "INSERT INTO images(photo)VALUES('$photo')");
$result = $query;
echo '<script type="text/javascript">alert("image successfully uploaded ");window.location=\'index.php\';</script>';
}
?>
<!DOCTYPE html>
<html>
<head>
<link href="css/style.css" rel="stylesheet" />
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.1/jquery.min.js"></script>
<script src="js/slider.js"></script>
<script>
$(document).ready(function () {
$('.flexslider').flexslider({
animation: 'fade',
controlsContainer: '.flexslider'
});
});
</script>
</head>
<body>
<div class="container">
<div class="flexslider">
<ul class="slides">
<?php
// Creating query to fetch images from database.
$query = mysqli_query($mysqli, "SELECT * from images order by id desc limit 5");
$result = $query;
while($r = mysqli_fetch_array($result)){
?>
<li>
<img src="<?php echo $r['photo'];?>" width="400px" height="300px"/>
</li>
<?php
}
?>
</ul>
</div>
</div>
</body>
</html>
here is my connect.php code.
<?php
// hostname or ip of server
$servername='localhost';
// username and password to log onto db server
$dbusername='root';
$dbpassword='';
// name of database
$dbname='pegasus';
////////////// Do not edit below/////////
$mysqli = new mysqli($servername,$dbusername,$dbpassword,$dbname);
if($mysqli->connect_errno){
printf("Connect failed: %s\n", $mysql->connect_error);
exit();
}
?>
and here is my admin.php code this is where i want to upload the image.
<form class="form" action="" method="POST" enctype="multipart/form-data">
<div class="image">
<p>Upload images and try your self </p>
<div class="col-sm-4">
<input class="form-control" id="image" name="image" type="file" onchange='AlertFilesize();'/>
<input type="submit" value="image"/>
</div>
</div>
</form>
here is my index.php this is where i want to show the slide.
and this is my admin.php where i want to upload the image of my image slider.
I solved it myself by putting the php code in my admin.php
<?php
//for connecting db
include('connect.php');
if (!isset($_FILES['image']['tmp_name'])) {
echo "";
}
else
{
$file=$_FILES['image']['tmp_name'];
$image= addslashes(file_get_contents($_FILES['image']['tmp_name']));
$image_name= addslashes($_FILES['image']['name']);
move_uploaded_file($_FILES["image"]["tmp_name"],"gallery/" . $_FILES["image"]["name"]);
$photo="gallery/" . $_FILES["image"]["name"];
$query = mysqli_query($mysqli, "INSERT INTO images(photo)VALUES('$photo')");
$result = $query;
echo '<script type="text/javascript">alert("image successfully uploaded ");window.location=\'admin.php\';</script>';
}
?>
<form class="form" action="" method="POST" enctype="multipart/form-data">
<div class="image">
<p>Upload images and try your self </p>
<div class="col-sm-4">
<input class="form-control" id="image" name="image" type="file" onchange='AlertFilesize();'/>
<input type="submit" value="image"/>
</div>
</div>
</form>
Related
I am trying to make a CRUD application. on the Create page I have to have three fields (title, text, category). the problem is that I have to make a method / function in PHP or JS that chooses a random picture from the "images" file and automatically loads it in the database along with the other 3 fields. then it has to appear on the admin.php page together with the other 3 fields.
Images have almost the same name except the last digit which differs (1-2-3)
I have no idea how to make this method/function.
my create.php page
// Include config file
require_once "config.php";
// Define variables and initialize with empty values
$title = $text = $category = "";
$title_err = $text_err = $category_err = "";
// Processing form data when form is submitted
if($_SERVER["REQUEST_METHOD"] == "POST"){
// Validate title
$input_title = trim($_POST["title"]);
if(empty($input_title)){
$title_err = "Please enter a title.";
} else{
$title = $input_title;
}
// Validate text
$input_text = trim($_POST["text"]);
if(empty($input_text)){
$text_err = "Please enter an text.";
} else{
$text = $input_text;
}
// Validate category
$input_category = trim($_POST["category"]);
if(empty($input_category)){
$category_err = "Please enter the category.";
} else{
$category = $input_category;
}
// Check input errors before inserting in database
if(empty($title_err) && empty($text_err) && empty($category_err)){
// Prepare an insert statement
$sql = "INSERT INTO informatii (title, text, category) VALUES (?, ?, ?)";
if($stmt = $mysqli->prepare($sql)){
// Bind variables to the prepared statement as parameters
$stmt->bind_param("sss", $param_title, $param_text, $param_category, );
// Set parameters
$param_title = $title;
$param_text = $text;
$param_category = $category;
// Attempt to execute the prepared statement
if($stmt->execute()){
// Records created successfully. Redirect to landing page
header("location: admin.php");
exit();
} else{
echo "Oops! Something went wrong. Please try again later.";
}
}
// Close statement
$stmt->close();
}
}
?>
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Create Record</title>
<link rel="stylesheet" href="https://stackpath.bootstrapcdn.com/bootstrap/4.5.2/css/bootstrap.min.css">
<style>
.wrapper {
width: 600px;
margin: 0 auto;
}
</style>
</head>
<body>
<div class="wrapper">
<div class="container-fluid">
<div class="row">
<div class="col-md-12">
<h2 class="mt-5">Create Record</h2>
<p>Please fill this form and submit to add employee record to the database.</p>
<form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>" method="post">
<div class="form-group">
<label>title</label>
<input type="text" name="title"
class="form-control <?php echo (!empty($title_err)) ? 'is-invalid' : ''; ?>"
value="<?php echo $title; ?>">
<span class="invalid-feedback"><?php echo $title_err;?></span>
</div>
<div class="form-group">
<label>Text</label>
<textarea name="text"
class="form-control <?php echo (!empty($text_err)) ? 'is-invalid' : ''; ?>"><?php echo $text; ?></textarea>
<span class="invalid-feedback"><?php echo $text_err;?></span>
</div>
<div class="form-group">
<label>Category</label>
<textarea name="category"
class="form-control <?php echo (!empty($category_err)) ? 'is-invalid' : ''; ?>"><?php echo $category; ?></textarea>
<span class="invalid-feedback"><?php echo $category_err;?></span>
</div>
<input type="submit" class="btn btn-primary" value="Submit">
Cancel
</form>
</div>
</div>
</div>
</div>
</body>
</html>
this should get you in the right direction (saving the image src is enough), you of course will have to adapt the path to your image folder, and image name
$nr_images = 3;
$random_nr_index = random_int(1,$nr_images);
$random_image_src = '/images/image-'.$random_nr_index.'.jpg';
To do it you need more than one step creating:
A simple html page to post 3 fields value and the image
A php file that receive the post fields and the image and save into mysql
A simple admin.PHP page that shows 3 fields and image
if you already have the images on the server please specify it in a comment
STEP 1:
<html>
<body>
<form method="POST" action="post.php">
f1:<input type="text" name="field1"><br>
f2:<input type="text" name="field2"><br>
f3:<input type="text" name="field3"><br>
im:<input type="file" name="image"><br>
<input type="submit" value="Save">
</form>
</body>
</html>
STEP 2: post.php
<?php
$f1=$_POST["field1"];
$f2=$_POST["field2"];
$f3=$_POST["field3"];
$im=$_POST["image"];
if ($f1 == "" || $f2 == "" || $f3 == "" ){
die("Errors: fields can't be empty! Go back check the fields and try Again");
}
//Saving image on Server's file system if any image
if(isset($_POST["image"])) {
//Saving image with no checking nothing: filetype, mime , extention (it may be very dangerous in a real server exposed to the public)
$where_save = "images/";
$im_name = basename($_FILES["image"]["name"]);
$tmp_name = $_FILES["image"]["tmp_name"];
move_uploaded_file ( $tmp_name , $where_save.$im_name );
}
$h = "localhost";
$u = "username";
$p = "password";
$db = "yourDB";
// Creating connection to mysql server
$conn = mysqli_connect($h, $u, $p, $db);
// Checking connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
// WARNINGS ------------------------------------------------
// I do not care about security , please pay attention to it .
// use some mysql_escape_string , or real_mysql_escape_string
// could mitigate the violence of some sqlinjection attack
$sql = "INSERT INTO yourtable (field1, field2, field3,im_name)
VALUES ('$f1', '$f2', '$f3',$im_name)";
//executing mysql query to save data into it
if (!mysqli_query($conn, $sql)) {
die("Error: " . $sql . "<br>" . mysqli_error($conn));
}
//closing connection
mysqli_close($conn);
//Now we can redirect the user to admin.php where we show data
header("Location: admin.php");
?>
STEP 3:
<?php
$where_are_images="images/";
$h = "localhost";
$u = "username";
$p = "password";
$db = "yourDB";
// Again creating connection to mysql server
$conn = mysqli_connect($h, $u, $p, $db);
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
//now we want to read the data from mysql
$sql = "SELECT * FROM yourtable LIMIT 1"; //just limit to the first record
$result = mysqli_query($conn, $sql);
?>
<html>
<body>
<h2>Admin page</h2>
<em> hey every one can see top secret data here , Needs soma care about security!</em>
<?php while($d = mysqli_fetch_assoc($result)){ // LOOPING ?>
<br>
f1:<?= $d["field1"] ?><br>
f2:<?= $d["field2"] ?><br>
f3:<?= $d["field3"] ?><br>
<img src="<?=$where_are_images.$d['im_name']?>">
<br>
<br>
<?php } ?>
</body>
</html>
<php? // CLOSING AND FREE RESOURCES
mysqli_free_result($result);
mysqli_close($conn); ?>
Now you have all you need . Have fun editing it with random images part ...
I hope there are no error (i have not tested it)
Hello im still a beginner but i have a uni project that i have to complete.I have to make a php/mysql image gallery and i have to add a DELETE button on each image, when clicked it needs to delete the image from the page and from the database.I have looked up everywhere but im not sure how exactly to write the code in order to delete the images, i have created the button and i have linked it to delete-image.php.
Any recommendations ?
view-album.php
<?php
include 'connection.php';
if (isset($_GET['album_id'])) {
$album_id = $_GET['album_id'];
$get_album = $mysqli->query("SELECT * FROM gallery_albums WHERE album_id = $album_id");
$album_data = $get_album->fetch_assoc();
} else {
header("Location: index.php");
}
?>
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<title><?php echo $album_data['album_name'] ?></title>
</head>
<body>
<?php
$photo_count = $mysqli->query("SELECT * FROM gallery_photos WHERE album_id = $album_id");
?>
Home | <?php echo $album_data['album_name'] ?> (<?php echo $photo_count->num_rows; ?>)<br><br>
<form method="post" action="upload-photo.php?album_id=<?php echo $album_id ?>" enctype="multipart/form-data">
<label>Add photo to this album:</label><br>
<input type="file" name="photo" /> <input type="submit" name="upload-photo" value="Upload" />
</form>
<?php
if (isset($_GET['upload_action'])) {
if ($_GET['upload_action'] == "success") { ?>
<br><br>Photo successfully added to this album!<br><br>
<?php }
}
?>
<?php
$photos = $mysqli->query("SELECT * FROM gallery_photos WHERE album_id = $album_id");
while($photo_data = $photos->fetch_assoc()) { ?>
<img src="<?php echo $photo_data['photo_link'] ?>" width="200px" height="200px" />
<form method="POST" action="delete-image.php">
<button name='delete'>Delete file </button>
</form>
<?php }
?>
</body>
</html>
delete-image.php
<?php
include 'connection.php';
?>
(1.)First you have to create input hidden field with row id with every button Tag.
(2.)Get id from hidden field in delete_image.php by $_POST['']; POST method.
(3.)then delete the image from that id.
<?php
$photos = $mysqli->query("SELECT * FROM gallery_photos WHERE album_id = $album_id");
while($photo_data = $photos->fetch_assoc()) {
$album_id = $photo_data['album_id'];
?>
<img src="<?php echo $photo_data['photo_link']; ?>" width="200px" height="200px" />
<form method="POST" action="delete-image.php">
<input type="hidden" name="delete" value="$album_id"> // hidden input field with id.
<button name='delete'>Delete file </button>
</form>
<?php }
?>
delete-image.php
<?php
include 'connection.php';
$idget = $_POST['delete']; // get here hidden id by POST method.
$photosDelete = $mysqli->query("DELETE FROM gallery_photos WHERE album_id = $idget");
?>
Note:- For More information regarding hidden field see this:-
https://www.w3schools.com/tags/att_input_type_hidden.asp
Currently, I developed a project that requires user to upload image from Computer. I do this from an example on the internet. For now, the column image at the database uses "BLOB" and the upload successful.
But, I don't want to uses BLOB, I want to change to LONGTEXT. When I change this format, the image saved to the database is successful but the image display is broken. The encode text and column image also looks weird. Can I know what is the problem? Below is the code
<?php
$connect = mysqli_connect("localhost", "root", "", "imagephp");
if(isset($_POST["insert"]))
{
$file = addslashes(file_get_contents($_FILES["image"]["tmp_name"]));
$query = "INSERT INTO tbl_images(name) VALUES ('$file')";
if(mysqli_query($connect, $query))
{
echo '<script>alert("Image Inserted into Database")</script>';
}
}
?>
<!DOCTYPE html>
<html>
<head>
<title>Image Upload</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.2.0/jquery.min.js"></script>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css" />
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/js/bootstrap.min.js"></script>
</head>
<body>
<br /><br />
<div class="container" style="width:500px;">
<h3 align="center">Insert and Display Images From Mysql Database in PHP</h3>
<br />
<form method="post" enctype="multipart/form-data">
<input type="file" name="image" id="image" />
<br />
<input type="submit" name="insert" id="insert" value="Insert" class="btn btn-info" />
</form>
<br />
<br />
<table class="table table-bordered">
<tr>
<th>Image</th>
</tr>
<?php
$query = "SELECT * FROM tbl_images ORDER BY id DESC";
$result = mysqli_query($connect, $query);
while($row = mysqli_fetch_array($result))
{
echo '
<tr>
<td>
<img src="data:image/jpeg;base64,'.base64_encode($row['name'] ).'" height="200" width="200" class="img-thumnail" />
</td>
</tr>
';
}
?>
</table>
</div>
</body>
</html>
<script>
$(document).ready(function(){
$('#insert').click(function(){
var image_name = $('#image').val();
if(image_name == '')
{
alert("Please Select Image");
return false;
}
else
{
var extension = $('#image').val().split('.').pop().toLowerCase();
if(jQuery.inArray(extension, ['gif','png','jpg','jpeg']) == -1)
{
alert('Invalid Image File');
$('#image').val('');
return false;
}
}
});
});
</script>
Hope someone can help me. Thanks!
I'm following a tutorial by mmtuts on youtube to show how to post comments to a myphpadmin database. All of my code is exactly the same as his, but I'm working from a different starting point becuase I already had a website I was working on and I just wanted to add the new code.
Basically, the video showed the code working flawlessly and my posts do not show up in the database like his did.
https://www.youtube.com/watch?v=4pPGOF5MI4U
".setComments($conn)." on the second document of code is blue instead of white like in the video.
<?php
require 'includes/dbh.inc.php';
session_start();
?>
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<meta name="description" content="This is an example of a meta description. This will often show up in search results.">
<meta name=viewport content="width=device-width, initial-scale=1">
<title>TAG</title>
<link rel="stylesheet" type="text/css" href="style.css">
</head>
<div id="headerContainer">
<?php
if (isset($_SESSION['userID'])) {
$id = $_SESSION['userID'];
$sqlImg = "SELECT * FROM profileimg WHERE userid='$id'";
$resultImg = mysqli_query($conn, $sqlImg);
while ($rowImg = mysqli_fetch_assoc($resultImg)) {
if ($rowImg['status'] == 0) {
$filename = "profilepics/profile".$id."*";
$fileinfo = glob($filename);
$fileext = explode(".", $fileinfo[0]);
$fileactualext = $fileext[1];
echo "<div class=userPicture><img src='profilepics/profile".$id.".".$fileactualext."?".mt_rand()."'></div>";
}
else {
echo "<div class='userPicture'><img src='profilepics/noUser.png'></div>";
}
}
echo '
<div class="userName">'. $_SESSION['userUserName'] .'</div>
<div id="logoutForm">
<form action="includes/logout.inc.php" method="post">
<button type="Submit" name="logout-submit">Logout</button>
</form>
</div>
<div class="upload">
<form action="upload.php" method="POST" enctype="multipart/form-data">
<input type="file" name="file">
<button type="submit" name="submit">Profile Image</button>
</form>
</div>
';
}
else {
echo '
<div class="userPicture"><img src="profilepics/noUser.png"></div>
<div class="userName">You are not logged in!</div>
<div id="loginForm">
<form action="includes/login.inc.php" method="post">
<input type="text" name="mailuid" placeholder="Username/E-mail">
<input type="password" name="password" placeholder="Password">
<button type="Submit" name="login-submit">Login</button>
</form>
</div>
<div id="signupForm">
or Signup
</div>
';
}
?>
</div>
<?php
require "header.php";
date_default_timezone_set('America/Chicago');
include 'includes/comments.inc.php';
?>
<div class="homeBody">
<p>Starting Filler</p>
<p>-</p>
<p>-</p>
<p>-</p>
<video width="320" height="240" controls>
<source src="videos/sample.mp4" type="video/mp4">
Your browser does not support the video tag.
</video>
<?php
echo "<form method='POST' action='".setComments($conn)."'>
<input type='hidden' name='uid' value='Anonymous'>
<input type='hidden' name='date' value='".date('Y-m-d H:i:s')."'>
<textarea name='message'></textarea><br>
<button type='submit' name='commentSubmit'>Comment</button>
</form>";
?>
<?php
function setComments($conn) {
if (isset($POST['commentSubmit'])) {
$uid = $_POST['uid'];
$date = $_POST['date'];
$message = $_POST['message'];
$sql = "INSERT INTO comments (uid, date, message) VALUES ('$uid', '$date', '$message')";
$result = $conn->query($sql);
}
}
<?php
$servername = "localhost";
$dBUsername = "root";
$dBPassword = "thisisnotmyactualpassword";
$dBName = "tagloginsystem";
$conn = mysqli_connect($servername, $dBUsername, $dBPassword, $dBName);
if (!$conn) {
die("Connection failed: ".mysqli_connect_error());
}
All I want is for the posts to make it into the database.
Try making a separate PHP-file and have your action attribute inside the form tag point to it. Right now it looks like you are running the function in the action attribute. In the PHP-file you can run your PHP function and write the PHP you need.
EX:
<form method='POST' action="includes/comments.php">
I was missing the "_" in $_POST on the 3rd page of code
I am trying to execute below code but it gives me an error that 'trying to get property of non-object in c....' .
This code should pull up info about 'images' and 'texts' to be displayed on the index.php page. I have tried in all means but couldn't figure out what is the problem; I am a beginner in PHP by the way :) .I will appreciate if you please help me.
<!DOCTYPE html>
<?php
$alert = "";
//if upload button is pressed
if(isset($_POST['upload'])){
//the path to store the uploaded image
$target = "images/".basename($_FILES['image']['name']);
//connect to the database
$conn = new mysqli('localhost', 'imgcms', '', '');
//Get all the submitted data from thye form
$image = $_FILES['image']['name'];
$text = $_POST['text'];
$sql = "INSERT INTO images (image, text) VALUES ('$image', '$text')";
//Move the uploaded image into the folder: images
if(move_uploaded_file($_FILES['image']['tmp_name'], $target)){
$alert = "Image uploaded successfully";
}else{
$alert = "There was a problem uploading the image";
}
}
?>
<html>
<head>
<title>ImageBlogger</title>
<link rel="stylesheet" href="style.css">
</head>
<body>
<div id="content">
<?php
//connect to the database to display image from the database
$conn = new mysqli('localhost', 'imgcms', '', '');
$sql = "SELECT * FROM images";
$result = $conn->query($sql);
if($result->num_rows > 0){
//output data of each row: image and text
while($row = $result->fetch_assoc()){
echo "<div id='img_div'>";
echo "<img src='images/".$row['image']."'>";
echo "<p>".$row['text']."</p>";
echo "</div>";
}
}else{
echo "0 results";
}
$conn->close();
?>
<form action="index.php" method="post" autocomplete="off" enctype="multipart/form-data">
<input type="hidden" name="size" value="1000000">
<div>
<input type="file" name="image">
</div>
<div>
<textarea name="text" cols"40" rows="4" placeholder="Content..."></textarea>
</div>
<div>
<input type="submit" name="upload" value="Post the content">
</div>
</form>
</div>
</body>
</html>
Copying your code into an editor it looks like line 47 is;
if($result->num_rows > 0)
Before this line add the following and see if you get an error.
if (!$result) {
echo 'Query Error is: ' . $conn->error;}