1
Gentlemen I am trying to collect information from a URL and send to my DB, in the url the information comes all mixed, with the scritp I can already organize them as follows:
[1] => Array
(
[PROTOCOLO] => 1234
[FILA] => Rede
[STATUS] => ABERTO
[DT_ABERTURA] => 17/04/2018 16:10:34
[SERVICO] => Registro de Falhas
[REGIONAL] => EUA
[LOCALIDADE] => IF
[DESCRICAO] => Falha na placa
[INTERROMPEU] =>
[QNT_CLIENTE] => 0
[OBS] => OG: 1234 [ABERTURA]
But I can’t collect this information and release it on DB, I need help. The same at the end of the data organization returns the error Warning: mysqli_query() expects at least 2 Parameters, 1 Given in C: xampp htdocs test.php on line 65.
Line 65 is mysqli_query("INSERT INTO
Follows my code.
<?php
//function dados() {
$url = 'http://pordesemml.arecom.com.br/api/cor/teste.php?usuario=testes&senha=teste';
$contents = file_get_contents($url);
$contents = utf8_encode($contents);
$results = json_decode($contents, true);
echo "<pre>";
print_r($results);
foreach ($results as $key => $value) { //$results é o conteudo que veio daquela url
mysqli_connect (salvaBancoOG($value)); //ele pega cada linha armazena no $value e envia para uma função com o nome salvaBancoOG
}
// }
#Include do arquivo de configurações
function salvaBancoOG($value) {
$data_registro = date_create('now')->format('Y-m-d H:i:s'); //data e hora atual
$data_fim = "0000-00-00 00:00:00";
if ($value[PROTOCOLO] == null)
$value[PROTOCOLO] = " ";
if ($value[FILA] == null)
$value[FILA] = " ";
if ($value[STATUS] == null)
$value[STATUS] = " ";
if ($value[DT_ABERTURA] == null)
$value[DT_ABERTURA] = " ";
if ($value[SERVICO] == null)
$value[SERVICO] = " ";
if ($value[REGIONAL] == null)
$value[REGIONAL] = " ";
if ($value[LOCALIDADE] == null)
$value[LOCALIDADE] = " ";
if ($value[DESCRICAO] == null)
$value[DESCRICAO] = " ";
if ($value[INTERROMPEU] == null)
$value[INTERROMPEU] = " ";
if ($value[QNT_CLIENTE] == null)
$value[QNT_CLIENTE] = " ";
if ($value[OBS] == null)
$value[OBS] = " ";
}
mysqli_query("INSERT INTO 'diario' (`protocolo`, `fila`, `status`, `dt_abertura`, `servico`, `regional`, `localidade`, `descricao`, `massiva`, `qnt_cliente`, `obs`, `id`)
VALUES ('12345678','Tx','Em Aberto','2018-04-17 00:00:00','ula','ula','escrição teste','1','10','deletar','''','''','''','''','NULL');"); //envia para o banco de dados
?>
My connection to the Bank is through a variable:
$conn= mysqli_connect($local_serve, $usuario_serve, $senha_serve, $banco_de_dados) or die ("O servidor não responde!");
Do not put the code image. Edit your question and write the code of a Ctrl+V in your question. Images do not help much in this case
– hugocsl
Hello, Rodrigo! Welcome. As told by @hugocwl, edit your question and enter the code. Unfortunately images are harder to read and the OS and Google search cannot index the code within it.
– Dherik