How to capture the name of a $ _FILE file and print to an echo

Question

How to capture the name of a $ _FILE file and print to an echo

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#1 by (0 votes)

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<form action="upload2.php" method="post" enctype="multipart/form-data">
<input type="file" name="arquivo[]" multiple="multiple"/><br/><br/>
<input type="submit" name="enviar" value="Enviar"/>

upload2.php

<?php
$diretorio = "img/";

if(!is_dir($diretorio)){
echo "pasta $diretorio não existe";
}else{
$arquivo =isset($_FILES['arquivo'])?$_FILES['arquivo'] : FALSE;

for($controle = 0; $controle<count($arquivo['name']);$controle++){
$destino = $diretorio."/".$arquivo['name'][$controle];

    if (move_uploaded_file($arquivo['tmp_name'][$controle],$destino)){
        echo "upload realizado com sucesso<br><br>";

    }
    else{
        echo "erro ao realizar upload";
    }
}
}

php

asked by anonymous 10.06.2018 / 01:42

1 answer

Dynamic interface based on values of an object (Typescript) I can not do Array (in JSON) in Typescript

score 0 · Accepted Answer

The question is that using <input type="file" multiple="multiple" name="upload[]"> PHP will get the array like this:

$_FILES['upload']['name'][n]

Where n is equivalent to the number of files added by <input> . Then to print the names of the files that were selected, you can use for() :

$numero_de_arquivos = count($_FILES['upload']['name']); // conta o número de arquivos adicionados
for($i = 0; $i < $numero_de_arquivos; $i++){
    echo $_FILES['upload']['name'][$i]; // aqui ele pega o nome de cada arquivo e imprime com o echo
    echo "<br>";
}