Function that returns mysqli_fetch_array

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Function that returns mysqli_fetch_array

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#1 by (1 votes)
#2 by (0 votes)

0

My scenario is as follows: I have a function called UserSearch ()

function buscaUsuario($conexao)
{

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

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

return $retornar;

}

I also have a file named demand-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 it would enter an infinite loop and crash the browser. And this other way was as follows: in the userSearch () function I created a variable that received mysqli_fetch_array () and returned that variable. So in search-user.php I would call this function and play within a while so the table would be populated until there was no more row inside the array ... And this (as I said up there) made the browser crash , I decided to do it that way that I showed them. However, if I add another field in the select and I command to print within that, the cells of each row in the table are not populated, just the first row, thus: userNameCompleteSequenceNode

I wanted to understand what I'm doing wrong and what would be the best way to do what I'm trying to do ... Thank you folks!

php mysqli

asked by anonymous 28.04.2016 / 21:45

2 answers

Floating Panel My radio input is repeating

score 1 · Answer 1

Dude, you can try doing this:

function buscaUsuario($conexao)
{

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

return mysqli_fetch_array($query)

}

And in the other:

<?php 

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

This makes it much more practical and saves you more than half the lines.

score 0 · Answer 2

I was able 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>