Re: pg_fetch_array problem

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hmm.. you could try..

for(var i=0; $dados = pg_fetch_array($limite, $i); i++) {
    $nome = $dados["nome"];
     echo "Name: $nome<BR>";
} I believe your error is your not passing pg_fetch_array() the row you wish to return. It *may* or may not work that way. But leaving it open like that is sure to cause an error sooner or later.

Hope this helps
David

angelo.rigo@xxxxxxxxx wrote:

Hi i am trying to build
previous and next links from a table selection, the code below stops working
at the line which says :

while ($dados = pg_fetch_array($limite)) {
   $nome = $dados["nome"];
   echo "Nome: $nome<br>";

the warning says:
Wrong parameter count for pg_fetch_array()

i get this code from a php mysql and try to adapt to my needs, is that any
parameter wrong? wich can be?


<?php

$conn = pg_connect("dbname=sondagem user=postgres");
?>
<?php
$busca = "SELECT nome AS data FROM aprovados ORDER BY nome ASC ";
?>
<?php
$total_reg = "10"; // número de registros por página
?>
<?php
if (!$pagina) {
   $pc = "1";
} else {
   $pc = $pagina;
}
?>
<?php
$inicio = $pc - 1;
$inicio = $inicio * $total_reg;
?>
<?php
$limite = pg_exec("$busca LIMIT $inicio,$total_reg");
$todos = pg_exec("$busca");

$tr = pg_numrows($todos); // verifica o número total de registros
$tp = $tr / $total_reg; // verifica o número total de páginas

// vamos criar a visualização
while ($dados = pg_fetch_array($limite)) {
   $nome = $dados["nome"];
   echo "Nome: $nome<br>";
}

// agora vamos criar os botões "Anterior e próximo"
$anterior = $pc -1;
$proximo = $pc +1;
if ($pc>1) {
   echo " <a href='?pagina=$anterior'><- Anterior</a> ";
}
echo "|";
if ($pc<$tp) {
   echo " <a href='?pagina=$proximo'>Próxima -></a>";
}
?>

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