I am trying to upload an image of a plane and then the user needs to fill the registration, airport, and company and then upload it, the problem is that the information is not going to the database, any suggestion?
<?php
include("connection.php");
?>
<html>
<body>
<form action="" method="post" enctype="multipart/form-data">
Matricula: <input type="text" name="registration" value=""/><br><br>
Aeroporto: <input type="text" name="airport" value=""/><br><br>
Companhia: <input type="text" name="company" value=""/><br><br>
Upload da foto: <input type="file" name="uploadfile" value=""/><br><br>
<input type="submit" name="submit" value="Submit"/>
</form>
<?php
if ($_POST['submit'])
{
$mat = $_POST["registration"];
$air = $_POST["airport"];
$comp = $_POST["company"];
$filename = $_FILES["uploadfile"]["name"];
$tempname = $_FILES["uploadfile"]["tmp_name"];
$folder = "fotos/".$filename;
move_uploaded_file($tempname, $folder);
if($mat!="" && $air!="" && $comp!="" && $filename!="")
{
$query = "INSERT INTO aviacao VALUES
('$mat','$air','$comp','$folder')";
$data = mysql_query($conn, $query);
if($data)
{
echo "Foto inserida na Base de Dados";
}
}
else
{
echo "Preencher todos os campos";
}
}
?>
</body>
</html>
Database
1 matricula varchar(35)
2 aeroporto varchar(100)
3 companhia varchar(100)
4 foto varchar(100) (here should go the path to the folder where
the photos are beign stored)
Related
I am having a trouble to saving data into the database. My connection details and sql insert query everything is correct and image is also uploading to folder but I do not know why data along with image is not saving into an database when i hit upload button.Can anyone help me please?
My php code
<?php
include('server.php');
$userID = 1;
if(isset($_SESSION['username']))
{
$userName = $_SESSION['username'];
$queryID = "SELECT id from users WHERE username = '$userName'";
$resultID = $db->query($queryID);
$row=$resultID->fetch_assoc();
$userID = $row['id'];
}
if(isset($_POST['submit']))
{
$image = $_FILES['image']['name'];
$target = "images/".basename($image);
$eventName = $_POST['eventName'];
$eventDetail = $_POST['eventDetail'];
$eventDate = $_POST['eventDate'];
$eventTime = $_POST['eventTime'];
$queryImage = "INSERT INTO event_detail(eventName,eventDetails,eventDate,eventTime,imagePath,userID) VALUES('$eventName','$eventDetail','$eventDate','$eventTime','$image','$userID')";
mysqli_query($db,$queryImage);
if(move_uploaded_file($_FILES['image']['tmp_name'],$target))
{
$msg = "Image uploaded successfully";
}
else
{
$msg = "There is problem";
}
}
?>
html
<form method="post" enctype="multipart/form-data">
<label for="eventName">Event Name:<label>
<input type="text" id="eventName" name="eventName" ><br><br>
<label for="eventDetail">Event Detail:<label>
<textarea id="eventDetail" name="eventDetail" ></textarea><br><br>
<label for="eventDate">Event Date:<label>
<input type="text" id="eventDate" name="eventDate" ><br><br>
<label for="eventTime">Event Time:<label>
<input type="text" id="eventTime" name="eventTime" ><br><br>
<input type="file" id="image" name="image"><br><br>
<button type="submit" id="submit" name="submit" >Submit</button>
</form>
Change this
$queryImage = "INSERT INTO event_detail(eventName,eventDetails,eventDate,eventTime,imagePath,userID) VALUES ('$eventName','$eventDetail','$eventDate','$eventTime','$image','$userID')";
to
$queryImage = "INSERT INTO event_detail(eventName,eventDetails,eventDate,eventTime,imagePath,userID) VALUES ($eventName,$eventDetail,$eventDate,$eventTime,$image,$userID)";
my application wont insert into database, and its not giving me any errors.
If my question is a duplicate, please redirect me to the original question, thanks.
MY PHP CODE:
<?php
include_once("dbconfig.php");
if(isset($_POST['submit'])){
/*$targetDir = "slike/";
$allowTypes = array('jpg','png','jpeg','gif');
$statusMsg = $errorMsg = $insertValuesSQL = $errorUpload = $errorUploadType = '';*/
$image = $_FILES['files']['tmp_name'];
$file = addslashes(file_get_contents($image));
$name = mysqli_real_escape_string($mysqli, $_POST['ime']);
$group=mysqli_real_escape_string($mysqli,$_POST['grupa']);
$semi=mysqli_real_escape_string($mysqli,$_POST['podgrupa']);
$price=mysqli_real_escape_string($mysqli,$_POST['cena']);
if(empty($name) || empty($group) || empty($semi)||empty($price)) {
echo "<font color='red'>Namesteno je da sve moraju da budu popunjene.</font><br/>";
if(empty($name)) {
echo "<font color='red'>Ime je prazno.</font><br/>";
}
if(empty($group)) {
echo "<font color='red'>Grupa je prazna.</font><br/>";
}
if(empty($semi)) {
echo "<font color='red'>Podgrupa je prazna.</font><br/>";
}
if(empty($price)){
echo "<font color='red'>Cena je prazna.</font><br/>";
}
}
else{
$result = mysqli_query($mysqli, "INSERT INTO slike(ime,grupa,podgrupa,cena,slika) VALUES('$name','$group','$semi','$price','$file')");
echo "<font color='green'>Data added successfully.";
}
}
?>
My html:
</div>
<div class="content">
<form action="" method="post" enctype="multipart/form-data">
Slike biras ovde:
<input type="file" name="files">
<p>IME</p>
<input type="text" name="ime">
<p>GRUPA</p>
<input type="text" name="grupa">
<p>PODGRUPA</p>
<input type="text" name="podgrupa">
<p>CENA</p>
<input type="text" name="cena">
<input type="submit" name="submit" value="UPISI">
</form>
If you need some more codes i will post it in replies.
Problem is that i get the message "Data added successfully"
But when i go to the database, its empty.
THANKS to anyone who helps me.
<form action="/your_php_file.php" method="post" enctype="multipart/form-data">
Please set action to your form which should go to your php file.
I'm having trouble with posting values typed into textarea. everything else works well, any idea how to make it work?
HTML:
<form id="formData2" action="artistuploader.php" method="post"
enctype="multipart/form-data">
<input type="hidden" name="size" value="1000000"></input>
<br/>
<input id="inputField" type="text" name="actname" placeholder="Act Name" >
<br>
<input id="inputField" type="text" name="fullname" placeholder="Full Name" >
<br>
<input id="inputField" type="text" name="genre" placeholder="Genre" >
<br>
<textarea id="inputField" name="biography" form="formData2" placeholder="Biography"<?php echo $biography; ?>></textarea>
<br>
<input id="inputField" type="file" name="artistImage" placeholder="Artwork" >
<br>
<input id="inputField" type="text" name="imagepath" placeholder="Image path URL" >
<br>
<input id="submitButton" type="submit" name="uploadArtist" value="Register Artist">
</form>
PHP
<?php
$msg = "";
//if Upload button is pressed
if (isset($_POST['uploadArtist'])){
$target = "uploads/artistPics".basename($_FILES['artistImage']['name']);
//connecting to our database
$db = mysqli_connect("127.0.0.1", "user", "pass", "tablename");
$tmp_name = $_FILES['artistImage']['tmp_name'];
$name = $_FILES['artistImage']['name'];
//getting the submitted form data
$ActName = $_POST['actname'];
$FullName = $_POST['fullname'];
$Genre = $_POST['genre'];
$ArtistPhoto = $_FILES['artistImage']['name'];
$imageURLpath = $_POST['imagepath'];
$Biography = $_POST['biography'];//having problem with this line here
//saving submitted data into database table songsDB
$sql = "INSERT INTO artistsdb (ActName,FullName,Genre,ArtistPhoto,Biography,imageURLpath) VALUES ('$ActName','$FullName','$Genre','$ArtistPhoto','$Biography','$imageURLpath')";
mysqli_query($db, $sql); //stores the submitted data into table
//now moving the uploaded image to uploads folder
if(move_uploaded_file($_FILES['artistImage']['tmp_name'], $target)){
$msg = "Uploaded Successful";
}else{
$msg = "There was a problem uploading Data";
}
}
//header("refresh:1; url=index.php"); ?>
Replace your textarea with
<textarea id="inputField" name="biography" form="formData2" placeholder="Biography"><?php echo $biography; ?></textarea>
I´ve been working on this project for a quite while and, the other day I was doing some code when I walk by this piece of code that keeps giving me headaches. So i got my html and php code all right but whenever i try to upload an image to my database, the image goes null, what am i doing wrong?
<?php include "connection.php"; ?>
<?php
$n=$_POST["num"];
$t=$_POST["texto"];
$i=$_POST["imagem"];
$img = mysql_query("SELECT imagem2 from segurancaofensiva where nmr=$n");
$file = $_FILES['imagem']['tmp_name'];
$image = addslashes(file_get_contents($file));
$count = $connect->query("SELECT COUNT(DISTINCT nmr) FROM segurancaofensiva")->fetch_row()[0];
if ($connect->connect_error){
die("Connection failed: " . $connect->connect_error);
}
$sql = "UPDATE segurancaofensiva SET texto='$t', imagem='{$image}', imagem2='{$image}' where nmr=$n" ;
$sql1 = "UPDATE segurancaofensiva SET texto='$t', imagem='$img' where nmr=$n";
if ($_FILES['imagem']['name']!=='' && $connect->query($sql) !== false )
{
if ($n <= $count) {
echo "actualizou\n\n";
var_dump($n);
var_dump($i);
var_dump($image);
}
else
{
echo "nao atualizou numero fora dos limites";
}
} else {
if ($connect->query($sql1) !== false){
echo "atualizou\n\n";
} else {
echo "errp";
}
}
$connect->close();
?>
<?php include 'connection.php'; ?>
<?php
$campo = $_POST['selected'];
$query = "SELECT campo FROM segurancaofensiva";
$result1 = mysqli_query($connect, $query);
$stored = $campo;
$obterquery = "SELECT * FROM segurancaofensiva where campo ='$campo'";
$x = $connect->query ($obterquery) or die ("Erro na variavel resultado");
$final = $x->fetch_array (MYSQL_ASSOC);
?>
<html>
<body>
<head>
<link rel="stylesheet" type="text/css" href="css/styleBO.css">
</head>
<div class="formulario" style="width: 100%; height: 100%;">
<form name="form2" method="POST" action="">
<h6>Campo:</h6> <select name ="selected" id="selected" >
<?php while($row1 = mysqli_fetch_array($result1)):;?>
<option value="<?php echo $row1[0];?>"><?php echo $row1[0];?></option>
<?php endwhile;?>
</select>
<input type="submit" id="load" class="load" name="load" value="Carregar">
<input type="hidden" name="selectedValue" value="0"/><br>
</form>
</div>
<div class="formulario" id="form2" style="width: 100%; height: 100%;">
<form name="form1" target="apresenta" method="POST" action="menu3.php">
<label> Atualizar dados </label><br>
<h6>Texto:</h6><textarea name="texto" id="texto"><?php echo htmlspecialchars($final['texto']);?></textarea><br>
<h6>Imagem:</h6><input type="file" name="imagem"><br>
<input type="hidden" value="<?php echo htmlspecialchars($final['nmr']);?>" name="num">
<input type="submit" name="submit" value="Enviar" class="topo">
<input type="reset" value="Limpar" class="topo">
</form>
</div>
</body>
</html>
Change form html like below. PHP will not detect file object without this.
<form enctype="multipart/form-data">
Add form attribute as per your requirement.
You are missing enctype attribute in form tag
enctype='multipart/form-data'
If you post data and trying to upload file you must use enctype
I have following form in html
<form method="post" enctype="multipart/form-data" />
<fieldset>
<legend>Activate Scheme</legend>
<p>Date Of Draw</p>
<p><input type="text" name="dateOfDraw" class="textBox" style="width:150px" /></p>
<p>Time Of Draw</p>
<p><input type="text" class="textBox" name="timeOfDraw" style="width:150px" /></p>
<p>Enter scheme name</p>
<p><input type="text" class="textBox" name="schemeName" style="width:150px" />
</p>
<p>Upload Image</p>
<p><input type="file" name="image" id="image" /></p>
<p><input name="scheme_button" type="submit" class="button1" value="Submit"></p>
</fieldset>
</form>
Problem is that their is no error when I execute the following query
<?php
if(isset($_POST['submit_button']) && count($_POST)>0) {
print_r($_POST);
$dateOfDraw = $_POST['dateOfDraw'];
$timeOfDraw = $_POST['timeOfDraw'];
$schemeName = $_POST['schemeName'];
$imageName = $_FILES['image']['name'];
$destination = '../images/'.$imageName;
$source = $_FILES['image']['tmp_name'];
if(move_uploaded_file($source, $destination)) {
echo 'file uploaded';
} else {
echo ' file not uploaded';
}
$sSQL = "INSERT INTO landing_page(dateOfDraw, timeOfDraw, schemeName, image) VALUES('$dateOfDraw','$timeOfDraw','$schemeName','$imageName')";
echo $sSQL;
if(!$sSQL){
die(mysqli_query($con));
}
mysqli_query($con,$sSQL);
//header("location: list_products.php");
}
?>
but when I check my data in a mysql database the rows are still empty. Please check it am I missing something or the php code is wrong.
Note: I am using php version 5.5.16
The name of your submit button is scheme_button and that's exactly the thing that you need to $_POST.
if(isset($_POST['scheme_button']) && count($_POST) > 0)
You form doesn't have an action and it doesn't know where to go. So add one.
<form method="post" action="index.php" enctype="multipart/form-data" />
firstly you have to name of submit button
if(isset($_POST['scheme_button']) && count($_POST) > 0)
secondly if insert code in same page then
if insert code in another page then
action = "filename"