How to save form image directory in the database

Question

In the system I am developing has a form with a field where the user can attach an image.

After saving the form, the user wants to view the record including the image, so I thought about saving the image in a folder on the system and its directory in the database.

That’s my form page:

<form  method="post" action="update-acidentes.php">
                    <fieldset>
                    <h3 class="subhead" style="color:#39b54a;">Clarificação do problema</h3>
                    <div class="form-field">
                        <h4> O que: </h4>
                        <select style="color:#9e9e9e;" name="OQUE" type="text" id="OQUE" aria-required="true" class="full-width">
                            <option>Clique aqui e selecione...</option>
                            <option>Acidente</option>
                            <option>Quase acidente</option>
                            <option>Trajeto</option>
                            <option>Incêndio</option>
                            <option>Acidente impessoal</option>
                        </select>
                    </div>
                    <h4> Quem </h4>
                    <div class="form-field">
                        <p style="font-size:20px;"> Nome </p>
                        <input style="color:#9e9e9e;" name="NOME" type="text" id="NOME" placeholder="Insira aqui o nome do colaborador" aria-required="true" class="full-width">
                    </div>
                    <div class="form-field">
                        <p style="font-size:20px;"> Descrição do evento </p>
                        <textarea style="color:#9e9e9e;" name="DESCRICAO" type="text" id="DESCRICAO" placeholder="Descreva o ocorrido" aria-required="true" class="full-width"></textarea>
                    </div>
                    <div class="form-field">
                        <p style="font-size:20px;"> Foto </p>
                        <input style="color:#9e9e9e;" name="FOTOA" type="file" id="FOTOA" aria-required="true" class="full-width">
                    </div>
                    <?php require_once('foto-acidente.php'); ?>

                    <div class="form-field">
                        <button type="submit" class="btn btn-primary btn-lg btn-block" style="background-color: #39b54a">Salvar</button>
                    </div>
                    </fieldset>
                </form>

This is the page that saves the form in the bank:

<!--Update de acidentes-->
<?php

    $oque        = $_POST['OQUE'];
    $nome        = $_POST['NOME'];
    $descricao   = $_POST['DESCRICAO'];

    $strcon = mysqli_connect('localhost','root','', 'banco') or die('Erro ao conectar ao banco de dados');
    $sql = "INSERT INTO acidentes (OQUE, NOME, DESCRICAO ) VALUES ('$oque', '$nome', '$descricao')"; 
    mysqli_query($strcon,$sql) or die("Erro ao tentar cadastrar registro" . mysqli_error($strcon));
    mysqli_close($strcon);

    if(isset($_FILES["FOTOA"])){
        $arquivo = $_FILES["FOTOA"];
        //diretorio dos arquivos
        $pasta_dir = "images/arquivosACIDENTES/";
        // Faz o upload da imagem
        $arquivo_nome = $pasta_dir . $arquivo["FOTOA"];
        //salva no banco
        move_uploaded_file($arquivo["FOTOA"], $arquivo_nome);

        $query  = "Insert into acidentes (IMAGEM) values ($arquivo_nome)";
    }

    echo '<script type="text/javascript">
            alert("Salvo com Sucesso !");
            window.history.go(-1);
        </script>';

?>

When I save a new record, the saved message appears successfully and when I open the database really all the columns were saved except the image.

The folder I created to store the images is also empty.

Does anyone know what might be going on?

Answer 1

Tag form form page you need to include attribute enctype and define how "multipart/form-data" which indicates that the form we are using will work with uploading files.

<form  method="post" action="update-acidentes.php" enctype="multipart/form-data">

Several errors in the update-accidents.php page

• Closing the connection before the image Insert in the bank
• The image file name is given by
  $arquivo = $_FILES['FOTOA']['name'];
• The image directory in the database will be
  $arquivo_nome = $pasta_dir . $arquivo;
• The correct upload syntax is
  move_uploaded_file($_FILES["FOTOA"]['tmp_name'], $arquivo_nome);
• In the image index missing involve the variable $arquivo_nome with single quotes
  $query = "Insert into acidentes (IMAGEM) values ('$arquivo_nome')";
• Failed to execute image query
  mysqli_query($strcon,$query) or die("Erro ao tentar cadastrar imagem" . mysqli_error($strcon));

Code with appropriate corrections

<?php

    $oque        = $_POST['OQUE'];
    $nome        = $_POST['NOME'];
    $descricao   = $_POST['DESCRICAO'];
    $arquivo_nome ="";

    $strcon = mysqli_connect('localhost','root','', 'banco') or die('Erro ao conectar ao banco de dados');
    $sql = "INSERT INTO acidentes (OQUE, NOME, DESCRICAO ) VALUES ('$oque', '$nome', '$descricao')"; 
    mysqli_query($strcon,$sql) or die("Erro ao tentar cadastrar registro" . mysqli_error($strcon));

    if(isset($_FILES["FOTOA"])){

        $arquivo = $_FILES['FOTOA']['name'];
        //diretorio dos arquivos
        $pasta_dir = "images/arquivosACIDENTES/";
        // Faz o upload da imagem
        $arquivo_nome = $pasta_dir . $arquivo;
        //salva no banco
        move_uploaded_file($_FILES["FOTOA"]['tmp_name'], $arquivo_nome);

        $query  = "Insert into acidentes (IMAGEM) values ('$arquivo_nome')";
        mysqli_query($strcon,$query) or die("Erro ao tentar cadastrar imagem" . mysqli_error($strcon));
    }

    mysqli_close($strcon);

    echo '<script type="text/javascript">
            alert("Salvo com Sucesso !");
            window.history.go(-1);
        </script>';

?>

Note that this script is saved in the bank as follows:

Perhaps it would be more practical to insert the image in the same line

In this case the update-accidents.php page would be

<?php

    $oque        = $_POST['OQUE'];
    $nome        = $_POST['NOME'];
    $descricao   = $_POST['DESCRICAO'];

    if(isset($_FILES["FOTOA"])){

        $arquivo = $_FILES['FOTOA']['name'];
        //diretorio dos arquivos
        $pasta_dir = "images/arquivosACIDENTES/";
        // Faz o upload da imagem
        $arquivo_nome = $pasta_dir . $arquivo;
        //salva no banco
        move_uploaded_file($_FILES["FOTOA"]['tmp_name'], $arquivo_nome);

    /**************  registro no banco de dados ******************/ 
    $strcon = mysqli_connect('localhost','root','', 'banco') or die('Erro ao conectar ao banco de dados');
    $sql = "INSERT INTO acidentes (OQUE, NOME, DESCRICAO, IMAGEM ) VALUES ('$oque', '$nome', '$descricao', '$arquivo_nome')"; 
    mysqli_query($strcon,$sql) or die("Erro ao tentar cadastrar registro" . mysqli_error($strcon));

    mysqli_close($strcon);
   /**************** fim registro no banco de dados *************/

    }

    echo '<script type="text/javascript">
            alert("Salvo com Sucesso !");
            window.history.go(-1);
        </script>';

?>

If the image is not mandatory, just pass the lines of the record in the database after the closure of the conditional if

Answer 2

good first of all, if you are using a local server, like xampp for example, it is necessary to leave the folder with the images in the same server folder, as it cannot access the files externally.

Second on your form you need to add the enctype tag, as it allows you to pick up the value of the photo path through $_FILES add this in the code

<form  method="post" action="update-acidentes.php" enctype="multipart/form-data">

I hope I’ve helped, anything we’re there

Answer 3

Come on,

1) In the Form tag add the enctype="Multipart/form-data property".
2) Check the field that receives the image in the database, whether it is a char or varchar of considerable size.
3) instead of '/' in the file directory, use the php constant DIRECTORY_SEPARATOR,
ex:
"images". DIRECTORY_SEPARATOR."arquivosACIDENTES". DIRECTORY_SEPARATOR"

Answer 4

The mistake is that you try to use the $arquivo["FOTOA"] as if it were the file name. It is not. This index, nor does it exist. You are probably getting an error of this type:

Notice: Undefined index: FOTOA in ...

It’s right to wear it like this:

    $arquivo_nome = $pasta_dir . $arquivo["name"];
    // envia o arquivo temporário para o diretório
    move_uploaded_file($arquivo["tmp_name"], $arquivo_nome);

Beyond what was quoted by Alvaro Leandro that your form has to look like this:

<form  method="post" action="update-acidentes.php" enctype="multipart/form-data">