Function that returns mysqli_fetch_array

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My scenario is this: I have a function called searchUsuario()

function buscaUsuario($conexao)
{

$retornar = array();
$query = mysqli_query($conexao, "SELECT user_name FROM usuarios");

while($result = mysqli_fetch_assoc($query))
{
    $retornar[] = $result;
}

return $retornar;

}

Where I also have a file called search-user.php and there is the following code snippet:

                        <?php 

                        $all = buscaUsuario($conexao);
                        $ab = sizeof($all);
                        $i = 0;
                        for ($i = 0; $i < $ab; $i++)
                        {
                         ?>
                         <tr>
                            <td><?=implode($all[$i])?></td>
                         </tr> <?php
                         }
                         ?>

I am returning the query in an array to be displayed in a table. Previously I tried to do it in a more succinct way (at least for me) but I entered an infinite loop and crashed the browser. And this other way was the following: in the function searchUsuario() I created a variable that received mysqli_fetch_array() and returned this variable. Then, in search-user.php I called this function and played inside a while so that the table was populated until there were no more rows inside the array... And that (as I said up there) caused the browser to crash, so I decided to do it the way I showed them. However, if I add one more field in select and have it printed inside it, the cells of each row in the table are not populated, only the first line, so: userName

I wanted to understand what I’m doing wrong and what would be the best way to do what I’m trying to do... Thanks Folks!

php mysqli

  • A foreach does not solve? $all = buscaUsuario($conexao); foreach($all as $item){ echo $item['user_name'] .'<br>';}

  • I had already tried, but it gives the following error: Warning: Illegal string offset 'user_name'

  • So the column name is wrong, that’s all. What’s the correct column name? When in doubt change the echo $item ... for print_r($item) inside the foreach.

  • Nope, that’s right. user_name

  • Puts the result of print_r() in the comment that we have already solved haha

  • The result is: "admin", of which is one of the registered user_name... I believe showed admin first by being ordered alphabetically

  • Not only does admin have more things, the return should be something like this Array ( [chave] => valor )

  • Worse than not brother, returns only "admin" : There in the other file, change mysqli_fetch_assoc to mysqli_fetch_array and print_R returned two admin

  • Aah, inside the foreach, just leave echo $item .'<br>';

  • Again appeared only "admin"

  • So now you’re right to just format in html table, or missed something?

  • Not really, as I have 5 user_name in my database and only one record is being displayed. It is also already formatted in html table D:

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2 answers

1

Dude, you can try doing it this way:

function buscaUsuario($conexao)
{

$query = mysqli_query($conexao, "SELECT user_name FROM usuarios");

return mysqli_fetch_array($query)

}

And the other:

<?php 

$all = buscaUsuario($conexao);
while($all)
{
?>
   <tr>
    <td><?php echo $all["user_name"]; ?></td>
   </tr> 
<?php
}
?>

So it becomes much more practical and you save more than half the lines..

  • This gives an infinite loop, you always return the first line of the query, while will never be false.

  • It gives an infinite loop. That’s how I had done previously and crashed several times the browser...

  • I edited the code, take the function again and test to see

0

I managed to list using this way:

function jQueryArray($sql) {
    $tbl = array();
    $linkConexao = conectaBD();
    $rs = mysqli_query($linkConexao, $sql);

    $i = 0;

    while ($tbl = mysqli_fetch_array($rs)){ $i++;

        $tblReturn[$i] = $tbl;

    }


    desconectaBD($linkConexao);
    return $tblReturn;
}

// Lista na view
$tbl = jQueryArray($sql);   

<table  class="wp-list-table widefat fixed striped posts">
    <thead>
            <tr>
                <td>ID</td>
                <td>Motivo</td>
                <td colspan="2"></td>                
            </tr>
        </thead>

        <?php  foreach ($tbl as $value){   ?>

        <tr>
            <td><?php echo $value['id_motivo'];  ?></td>
            <td><?php echo $value['descricao'];  ?></td>
            <td></td>
        </tr>

        <?php  } ?>

    </table>

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