Well, by the title of the question I could not explain very well what I wanted but here it goes in practice:
I have a select to a table called "posts".
This select wants me to indicate all posts, where the supervisor is the "Test".
After that I make a mysqli_fetch_array
And then I show the $ array ["post"];
However, it only shows me 1 post, and in fact the "Test" supervisor has more than one post.
How can I turn select posts into 1 array?
Thank you.
Gonçalo, based on Gonçalo's answer , but I simplified the loop:
$postos = array();
while($array1 = mysqli_fetch_array($procura1)) $postos[] = $array1['posto'];
This works because when you do variavel[] = 'valor'; the value is added to the end of the array .
Depending on the intended structure, you have an even simpler alternative:
$tudo = mysqli_fetch_all($procura1);
This takes all the results at once, without having to do a loop manually. But in this case, you will get an array of arrays and not values.
If you are running PHP 5.5 or higher, you can use fetch_all like this:
$tudo = mysqli_fetch_all($procura1, MYSQLI_ASSOC); // Pegamos todas as linhas
$postos = array_column($tudo, 'posto'); // extraimos só a coluna 'posto'
The MYSQLI_ASSOC is for the columns to be indexed with names instead of numbers. If we use MYSQLI_NUM it would be necessary array_column($tudo, numero_da_coluna) , which complicates maintenance somewhat if something is changed in SELECT .
More details in the manual:
For this you can make use of repetition loops from php.
Example with While
while($vetor = mysqli_fetch_array($procura1)){
echo $vetor["posto"] . '<br/>';
}
Example with For
$dados = mysqli_fetch_array($procura1)
for($i = 0; $i < count($dados); $i++){
echo $dados[$i]["posto"] . '<br/>';
}
Example with Foreach
$dados = mysqli_fetch_array($procura1)
foreach($dados as $dado){
echo $dado["posto"] . '<br/>';
}
Gonçalo, you can do the following:
$contador = 0;
while($array1 = mysqli_fetch_array($procura1)){
$postos[$contador] = $array1["posto"];
$contador++;
}