php- upload image and upload to database

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php- upload image and upload to database

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#2 by (0 votes)

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I'm trying to insert an image into the database but when I click upload image, it gives the following error: Warning: getimagesize (): Filename can not be empty in /home/unn_w17015779/public_html/upload.php on line 9 File is not an image.;

HTML code:

<body>
<?php include 'addDataAdmin.php';?>
<form name="ContactForm" action="addDataAdmin.php" method="POST" enctype="multipart/form-data" autocomplete="off" onsubmit="return ValidateContactForm();">
    ISBN:<input type="text" name="ISBN">
    Author's Name:<input type="text" name="Authorsname">
    Title:<input type="text" name="Title">
    Edition:<input type="number" name="edition" >
    Year:<input type="text" name="year" onkeypress="return justNumber(event)" >

    Category:
    <select name="category" size="1">
        <option value="computing">Computing</option>
        <option value="Romance">Romance</option>
        <option value="Fiction">Fiction</option>
        <option value="Non-Fiction">Non-Fiction</option>
    </select>
    <br />

    Publisher:<input type="text" name="publisher">
    Quantity-in-stock:<input type="number" name="quantityinstock" >
    Price:<input type="text" name="price" onkeypress="return justNumber(event)">
    <input type="file" name="fileToUpload" id="fileToUpload">
    <input type="submit" value="Upload Image" name="submit" formaction="/upload.php">
    <input type="submit" value="Send" name="send">
    <input type="reset" value="Clear">
</form>

PHP code:

<?php
include('config.php');
$target_dir = "uploads/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$uploadOk = 1;
$imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);
// Check if image file is a actual image or fake image
if(isset($_POST["submit"])) {
    $check = getimagesize($_FILES["fileToUpload"]["tmp_name"]);
    if($check !== false) {
        echo "File is an image - " . $check["mime"] . ".";
        $uploadOk = 1;
    } else {
        echo "File is not an image.";
        $uploadOk = 0;
    }
}

if (!empty($_FILES["fileToUpload"]["name"])) {

 $Image =$conn->real_escape_string($_POST['Image']);

    $sql="INSERT INTO books (Image) VALUES('$Image')";

    if(mysqli_query($conn,$sql))
     {
        echo '<h3><font color="red">You have successfully updated </font></h3>';

        }
            else
            {
                echo 'Error';
                echo $sql;
            }
        }




?>;

What am I doing wrong? Thank you in advance.

php sql html

asked by anonymous 21.04.2017 / 18:01

2 answers

0

Here are my examples below:

Upload_Picture.php

 <h1> Carregar a foto</h1>
<form method="POST" action="proc_upload.php" enctype="multipart/form-data">
	Imagem: <input name="arquivo" type="file"><br><br>
	
	<input type="submit" value="Cadastrar">
</form>

proc_upload.php

<!DOCTYPE html>
<html lang="pt-br">
    <head>
        <meta charset="utf-8">
    </head>
    </body>
        <?php
            include_once("conexao.php");
            $arquivo    = $_FILES['arquivo']['name'];

            //Pasta onde o arquivo vai ser salvo
            $_UP['pasta'] = 'foto/';

            //Tamanho máximo do arquivo em Bytes
            $_UP['tamanho'] = 1024*1024*100; //5mb

            //Array com a extensões permitidas
            $_UP['extensoes'] = array('png', 'jpg', 'jpeg', 'gif');

            //Renomeiar
            $_UP['renomeia'] = false;

            //Array com os tipos de erros de upload do PHP
            $_UP['erros'][0] = 'Não houve erro';
            $_UP['erros'][1] = 'O arquivo no upload é maior que o limite do PHP';
            $_UP['erros'][2] = 'O arquivo ultrapassa o limite de tamanho especificado no HTML';
            $_UP['erros'][3] = 'O upload do arquivo foi feito parcialmente';
            $_UP['erros'][4] = 'Não foi feito o upload do arquivo';

            //Verifica se houve algum erro com o upload. Sem sim, exibe a mensagem do erro
            if($_FILES['arquivo']['error'] != 0){
                die("Não foi possivel fazer o upload, erro: <br />". $_UP['erros'][$_FILES['arquivo']['error']]);
                exit; //Para a execução do script
            }

            //Faz a verificação da extensao do arquivo
            $extensao = strtolower(end(explode('.', $_FILES['arquivo']['name'])));
            if(array_search($extensao, $_UP['extensoes'])=== false){        
                echo "
                    <META HTTP-EQUIV=REFRESH CONTENT = '0;URL=http://localhost/teste/upload_imagem.php'>
                    <script type=\"text/javascript\">
                        alert(\"A imagem não foi cadastrada extesão inválida.\");
                    </script>
                ";
            }

            //Faz a verificação do tamanho do arquivo
            else if ($_UP['tamanho'] < $_FILES['arquivo']['size']){
                echo "
                    <META HTTP-EQUIV=REFRESH CONTENT = '0;URL=http://localhost/teste/upload_imagem.php'>
                    <script type=\"text/javascript\">
                        alert(\"Arquivo muito grande.\");
                    </script>
                ";
            }

            //O arquivo passou em todas as verificações, hora de tentar move-lo para a pasta foto
            else{
                //Primeiro verifica se deve trocar o nome do arquivo
                if($UP['renomeia'] == true){
                    //Cria um nome baseado no UNIX TIMESTAMP atual e com extensão .jpg
                    $nome_final = time().'.jpg';
                }else{
                    //mantem o nome original do arquivo
                    $nome_final = $_FILES['arquivo']['name'];
                }
                //Verificar se é possivel mover o arquivo para a pasta escolhida
                if(move_uploaded_file($_FILES['arquivo']['tmp_name'], $_UP['pasta']. $nome_final)){
                    //Upload efetuado com sucesso, exibe a mensagem
                    $query = mysqli_query($conn, "INSERT INTO usuarios (
                    nome_imagem) VALUES('$nome_final')");
                    echo "
                        <META HTTP-EQUIV=REFRESH CONTENT = '0;URL=http://localhost/teste/upload_imagem.php'>
                        <script type=\"text/javascript\">
                            alert(\"Imagem cadastrada com Sucesso.\");
                        </script>
                    ";  
                }else{
                    //Upload não efetuado com sucesso, exibe a mensagem
                    echo "
                        <META HTTP-EQUIV=REFRESH CONTENT = '0;URL=http://localhost/teste/upload_imagem.php'>
                        <script type=\"text/javascript\">
                            alert(\"Imagem não foi cadastrada com Sucesso.\");
                        </script>
                    ";
                }
            }


        ?>

    </body>
</html>

Change in proc_upload.php to lines containing <META HTTP-EQUIV=REFRESH CONTENT = '0;URL=http://localhost/teste/upload_imagem.php'> to your URL, this has to be done, if not, it will not work

connection.php

<?php
    $servidor = "localhost";
    $usuario = "root";
    $senha = "";
    $dbname = "info";

    //Criar a conexão
    $conn = mysqli_connect($servidor, $usuario, $senha, $dbname);
    if(!$conn){
        die("Falha na conexao: " . mysqli_connect_error());
    }else{
        //echo "Conexao realizada com sucesso";
    }
?>

Test these 3 files in a Test / folder, with chmod 777

22.04.2017 / 22:08

How to make a timer from a set date? Error to close form

score 0 · Accepted Answer

Your code works (with the observation that in my environment there was no execution of the functions in javascript, because you did not put them). With the form you posted:

ClickingUploadimage:

Fileisanimage-image/png.

Isuggestyoucallyourupload.phpfilewiththislittlebitofcode(var_dumpwillactuallyshowanimageinthefilesarray):

<?phpvar_dump($_FILES);$target_dir="uploads/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$uploadOk = 1;
$imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);
// Check if image file is a actual image or fake image
if(isset($_POST["submit"])) {
    $check = getimagesize($_FILES["fileToUpload"]["tmp_name"]);
    if($check !== false) {
        echo "File is an image - " . $check["mime"] . ".";
        $uploadOk = 1;
    } else {
        echo "File is not an image.";
        $uploadOk = 0;
    }
}

Save to database The written code should be saving something, except the image you want. Use the file_get_contents function to get upload image content.

if (!empty($_FILES["fileToUpload"]["name"])) {
    $Image =$conn->real_escape_string(
    file_get_contents($_FILES["fileToUpload"]["tmp_name"]));
    $sql="INSERT INTO books (Image) VALUES('$Image')";

This should allow your image to be saved. But remembering the array $_FILES["fileToUpload"]["tmp_name"] should point to an existing file somewhere on your computer.

Putting it all together

There are no errors in your script. Joining everything you posted and creating a test database, I ran and it worked (inserted the image in the database). See the comments:

<?php

//minha conexao com o banco (você já tem isso)
$conn = mysqli_connect('localhost', 'aprendendo', '', 'aprendendo');

//só para depuração
var_dump ( $_FILES );

$target_dir = "uploads/";
$target_file = $target_dir . basename ( $_FILES ["fileToUpload"] ["name"] );
$uploadOk = 1;
$imageFileType = pathinfo ( $target_file, PATHINFO_EXTENSION );
// Check if image file is a actual image or fake image
if (isset ( $_POST ["submit"] )) {
    $check = getimagesize ( $_FILES ["fileToUpload"] ["tmp_name"] );
    if ($check !== false) {
        echo "File is an image - " . $check ["mime"] . ".";
        $uploadOk = 1;
    } else {
        echo "File is not an image.";
        $uploadOk = 0;
    }
}

if (! empty ( $_FILES ["fileToUpload"] ["name"] )) {
    //obtem o conteudo da imagem com file_get_contents
    $Image = $conn->real_escape_string ( file_get_contents ( $_FILES ["fileToUpload"] ["tmp_name"]));
    //criei uma tabela ficticia chamada a, e uma coluna 
    //chamada b (iamgem), substitua pela sua
    $sql = "INSERT INTO a (b) VALUES('$Image')";
    if (mysqli_query ( $conn, $sql )) {
        echo '<h3><font color="red">You have successfully updated </font></h3>';
    } else {
        echo 'Error';
        //se ocorrer algum erro de sintaxe o sql será mostrado na tela
        echo $conn->error;
    }
}