Use ID generated from one FORM in another and UPDATE in bd

Question

Use ID generated from one FORM in another and UPDATE in bd

Asked 9 years, 2 months ago

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1

I recorded this image in the database generated me the ID:20.

Form to save : INICIAL.PHP (Code to save image)

<html>
<head>
    <title></title>
</head>
<body>
    <form enctype="multipart/form-data" method="post">
        <input type="file" name="image" /><br /> 
        <input type="submit" value="Enviar" name="sumit" />
    </form>
    <?php

    if(isset($_POST['sumit']))
    {
        if(getimagesize($_FILES['image']['tmp_name'])== FALSE)
        {
            echo "Selecione uma imagem.";
        }
        else
        {
            $image = addslashes($_FILES['image']['tmp_name']);
            $name = addslashes($_FILES['image']['name']);
            $image = file_get_contents($image);
            $image = base64_encode($image);
            saveimagem($name, $image);
        }
    }
    mostrarimagem();
    function saveimagem($name, $image)
    {

        $host = "localhost";
        $user = "root";
        $pass = "";
        $conexao = mysqli_connect($host, $user, $pass) or die (mysql_error());
        mysqli_select_db($conexao, "teste");
        $sql = "insert into cadastro (foto)  values ('$image')";
        $result = mysqli_query($conexao, $sql);
        if($result)
        {
            echo "<br/>Foi feito o upload" ;
            $id = mysqli_insert_id($conexao);
            echo $id;
        }else
        {
            echo "<br/>Não foi feito o upload";
        }
    }
    function mostrarimagem()
    {
        $host = "localhost";
        $user = "root";
        $pass = "";
        $conexao = mysqli_connect($host, $user, $pass) or die (mysql_error());
        mysqli_select_db($conexao, "teste");
        $sql = "select * from cadastro";
        $result = mysqli_query($conexao, $sql);

        while($row = mysqli_fetch_array($result))
        {
            echo '<img height="300" width="300" src="data:image;base64,'.$row[11].'">';
        }
        mysqli_close($conexao);
    }

    ?>
</body>

Where has the $id = mysqli_insert_id($conexao); is the variable that stored the last record in the database. I need to use it for the update in another form.

WRITE.PHP

<?php
//Irá receber os valores enviados e guardar no banco de dados

<?php
//Irá receber os valores enviados e guardar no banco de dados
if($_POST['enviar']){
    include 'inicial.php';
    $nome = $_POST['NomeAluno'];
    $dataNasc = $_POST['DataNasc'];
    $sexo = $_POST['Sexo'];
    $numcel = $_POST['Celular'];
    $numtel = $_POST['Telefone'];
    $endereco = $_POST['Endereco'];
    $numres = $_POST['NumResid'];
    $uf = $_POST['UF'];
    $rg = $_POST['RG'];
    $prontuario = $_POST['Prontuario'];
    $datavalidade = $_POST['DataValidade'];
    $curso = $_POST['Curso'];
    $semestre = $_POST['Semestre'];
    $periodo = $_POST['Periodo'];
    $email = $_POST['Email'];
    $senha = $_POST['Senha'];

    //"$sql = "SELECT MAX(ID) FROM cadastro";

    // Insere os dados no banco
    $sql = mysqli_query($conexao, "UPDATE cadastro set
                                        nome_aluno = ':$nome',
                                        data_nascimento = ':$dataNasc',
                                        sexo = ':$sexo', 
                                        celular = ':$numcel',
                                        telefone = ':$numtel',
                                        endereco = ':$endereco',
                                        numero = ':$numres',
                                        uf = ':$uf',
                                        rg = ':$rg',
                                        prontuario = ':$prontuario',
                                        data_validade = ':$datavalidade',
                                        curso = ':$curso',
                                        semestre = ':$semestre',
                                        periodo = ':$periodo',
                                        email = ':$email',
                                        senha = ':$senha'
                                  where id = '$id'");

    // Se os dados forem inseridos com sucesso
    if ($sql){
        echo "Você foi cadastrado com sucesso.";
    }
}
?>

I can’t update the table with the other values, without touching the photo that has already been saved. And using the $id variable, I even put the include 'inicial.php';.

I don’t know if it is my Update SQL that is wrong, but when I click save button this error appears: P Notice: Undefined variable: id in C:\wamp\www\ifsp\gravar.php on line 45

1

Those : in front of the variables is what?

– rray

2015/10/01 at 03:07
1

I don’t know why I put it, but it didn’t change anything

– WSS

2015/10/01 at 03:09
Where you make the call from saveimage()? missing to return the $id in this function ;). takes these out : of everything.

– rray

2015/10/01 at 03:10
I just put the include initial.php, with it will not everything that is on the page? I already took

– WSS

2015/10/01 at 03:12
Yes, but you need to call saveimage() pass the values and place at the end of the function return $id;

– rray

2015/10/01 at 03:15
What values? So: include 'inicial.php'
saveimagem($id){
 return $id
 } . Of error yet Parse error: syntax error, unexpected '{' in C:\wamp\www\ifsp\gravar.php on line 22, something I noticed too, that when I click save in the form it redirects me to the page inicial.php, only because of include

– WSS

2015/10/01 at 03:21
in initial.php has more code than the function?

– rray

2015/10/01 at 03:32
Got it, I’ll put it all up there

– WSS

2015/10/01 at 03:37
Dude, you have to see if the other fields accept null, because I realized you add the photo first and then you want to get the generated id to update the data right, you have to see tbm if the id is generated automatically.

– Romario Pires

2015/10/01 at 03:46
Yes ID is auto_increment and when the photo is saved the fields are to accept null. : s

– WSS

2015/10/01 at 03:51
Has been answered then!

– Romario Pires

2015/10/01 at 03:59
Similar: http://answall.com/questions/89986/pega-o-id-da-%C3%Baltima-line-in-database

– Edilson

2015/10/01 at 11:43
@Edilson, no guy are not similar. This first didn’t even know how to get the ID. Now I don’t know use it

– WSS

2015/10/01 at 17:34
It’s basically the same thing. And another thing, even though you showed me what the problem is, it would help more if you said what you really want, or what you want the whole script to do.

– Edilson

2015/10/01 at 17:55
Sorry. I’m not used to this forum yet. : x

– WSS

2015/10/01 at 18:37

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2 answers

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by rray • **66,288** points · Answer 1 · 2015-10-01T03:50:44+00:00

First make the function return the id inserted with a return $id.

function saveimagem($name, $image) {

//codigo omitido

   if($result){
      echo "<br/>Foi feito o upload" ;
      $id = mysqli_insert_id($conexao);
      echo $id;
      return $id;   //<--- linha adicionada
   }else{
      echo "<br/>Não foi feito o upload";
   }
   return false;
}

Then in write.php take the id of saveimagem() with an assignment, then pass the variable in update

write php.

else {
    $image = addslashes($_FILES['image']['tmp_name']);
    $name = addslashes($_FILES['image']['name']);
    $image = file_get_contents($image);
    $image = base64_encode($image);
    $id_imagem = saveimagem($name, $image); //<--- linha nova atribuição
}

Dai lá non update, pass $id_imagem.

where id = $id_imagem");

by Romario Pires • **831** points · Answer 2 · 2015-10-01T03:52:02+00:00

Look do the following, let’s make your code simpler!

Create a new file

It can be called a connection

php connection.

    $host = "localhost";
    $user = "root";
    $pass = "";
    $conexao = mysqli_connect($host, $user, $pass) or die (mysql_error());
  //Conectar

Inicial.PHP

 <?php session_start(); ?>
<html>
 <head>
    <title></title>
 </head>
 <body>
    <form enctype="multipart/form-data" method="post">
        <input type="file" name="image" /><br /> 
        <input type="submit" value="Enviar" name="sumit" />
    </form>
    <?php

    if(isset($_POST['sumit']))
    {
        if(getimagesize($_FILES['image']['tmp_name'])== FALSE)
        {
            echo "Selecione uma imagem.";
        }
        else
        {
            $image = addslashes($_FILES['image']['tmp_name']);
            $name = addslashes($_FILES['image']['name']);
            $image = file_get_contents($image);
            $image = base64_encode($image);
            include_once 'conexao.php';
            saveimagem($name, $image);
        }
    }

    mostrarimagem();

    function saveimagem($name, $image)
    {
        mysqli_select_db($conexao, "teste");
        $sql = "insert into cadastro (foto)  values ('$image')";
        $result = mysqli_query($conexao, $sql);
        if($result)
        {
            echo "<br/>Foi feito o upload" ;
            $_SESSION['id_temp'] = mysqli_insert_id($conexao);
            echo $id;
        }else
        {
            echo "<br/>Não foi feito o upload";
        }
    }

    function mostrarimagem()
    {
        mysqli_select_db($conexao, "teste");
        $sql = "select * from cadastro";
        $result = mysqli_query($conexao, $sql);

        while($row = mysqli_fetch_array($result))
        {
            echo '<img height="300" width="300" src="data:image;base64,'.$row[11].'">';
        }
        mysqli_close($conexao);
    }

    ?>
</body>
</html>

Write.php

<?php
session_start();
//Irá receber os valores enviados e guardar no banco de dados
if($_POST['enviar']){
    include_once 'conexao.php';
    $nome = $_POST['NomeAluno'];
    $dataNasc = $_POST['DataNasc'];
    $sexo = $_POST['Sexo'];
    $numcel = $_POST['Celular'];
    $numtel = $_POST['Telefone'];
    $endereco = $_POST['Endereco'];
    $numres = $_POST['NumResid'];
    $uf = $_POST['UF'];
    $rg = $_POST['RG'];
    $prontuario = $_POST['Prontuario'];
    $datavalidade = $_POST['DataValidade'];
    $curso = $_POST['Curso'];
    $semestre = $_POST['Semestre'];
    $periodo = $_POST['Periodo'];
    $email = $_POST['Email'];
    $senha = $_POST['Senha'];

    // Insere os dados no banco
    $sql = mysqli_query($conexao, "UPDATE cadastro set
                                        nome_aluno = '$nome',
                                        data_nascimento = '$dataNasc',
                                        sexo = '$sexo', 
                                        celular = '$numcel',
                                        telefone = '$numtel',
                                        endereco = '$endereco',
                                        numero = '$numres',
                                        uf = '$uf',
                                        rg = '$rg',
                                        prontuario = '$prontuario',
                                        data_validade = '$datavalidade',
                                        curso = '$curso',
                                        semestre = '$semestre',
                                        periodo = '$periodo',
                                        email = '$email',
                                        senha = '$senha'
                                  where id =".$_SESSION['id_temp']);

    // Se os dados forem inseridos com sucesso
    if ($sql){
        echo "Você foi cadastrado com sucesso.";
        if(isset($_SESSION['id_temp'])){
            unset($_SESSION['id_temp']);
        }else{
            echo "<script>alert('ocorreu um erro, pois a session para o id não foi criada')</script>";
        }
    }