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Hello,
Guys, I got a big question here. I have a query where the user can select if they want to see all(Manager and Managers Jr), Only Managers Jr, or Directly select only one.
- All
- Only managers Jr
- John
- Maria
- etc....
The problem is when I select "2 - Only Managers Jr".
Because I have to go on the employee register and find out who the junior managers are (colab_func=3)
, so far so good.
After the previous consultation, where I receive the employee codes, I need to do the consultation itself, where I go in the sales table and look for the sales of these employees. That’s where I’m not getting it.
I’ll put a part of the code here:
$gerente = $_POST["txtgerente"];
if($gerente == 'T'){
$query = "SELECT * FROM venda_dia WHERE venda_dia_aberto='S'";
$tipo = "Todos";
} elseif ($gerente == 'GJR'){
$query_ger = "SELECT colab_id FROM colaborador WHERE colab_funcao in($cod_gerjr)";
$result_ger = mysqli_query($con, $query_ger) or die(mysqli_error());
$row_ger = mysqli_fetch_array($result_ger);
$query = "SELECT * FROM venda_dia WHERE venda_dia_cod_ger in($row_ger) AND venda_dia_aberto='S'";
$tipo = "Gerente Júnior";
} else {
$query = "SELECT * FROM venda_dia WHERE venda_dia_cod_ger='$gerente' AND venda_dia_aberto='S'";
$query_gerente = "SELECT * FROM colaborador WHERE colab_id=$gerente";
$result_gerente = mysqli_query($con, $query_gerente) or die(mysqli_error());
$row_gerente = mysqli_fetch_array($result_gerente);
$tipo = $row_gerente['colab_nome_venda'];
}
?>
<section id="cadastros">
<fieldset id="cadastros"><legend><?php echo "Expositor com dia aberto - $tipo"; ?></legend>
<table id="tab" align="center">
<tr bgcolor="lightblue">
<th width="250px">Expositor</th>
<th width="150px">Dia Faturamento</th>
<th width="250px">Responsável</th>
</tr>
<?php
$result = mysqli_query($con, $query) or die(mysqli_error());
while ($row = mysqli_fetch_array($result)) {
$cod_exp = $row["venda_dia_cod_exp"];
$data = $row["venda_dia_data"];
$cod_colab = $row["venda_dia_cod_colab"];
$cod_colab2 = $row["venda_dia_cod_colab2"];
// ESCREVE UMA LINHA NA TABELA
echo "<tr>";
echo "<td>";
$query_expositor = "SELECT * FROM expositor WHERE expo_id=$cod_exp";
$result_expositor = mysqli_query($con, $query_expositor) or die(mysqli_error());
while ($row_expositor = mysqli_fetch_array($result_expositor)) {
$nome_expo = $row_expositor["expo_nome"];
echo $nome_expo . "</td>";
}
echo "<td id='centro'>" . date('d/m/Y' , strtotime($data)). "</td>";
echo "<td>";
if($cod_colab2 == 0){
$query_colab = "SELECT * FROM colaborador WHERE colab_id=$cod_colab";
$result_colab = mysqli_query($con, $query_colab) or die(mysqli_error());
while ($row_colab = mysqli_fetch_array($result_colab)) {
$colab_nome = $row_colab["colab_nome_venda"];
echo $colab_nome . "</td>";
}
} else {
$query_colab = "SELECT * FROM colaborador WHERE colab_id=$cod_colab2";
$result_colab = mysqli_query($con, $query_colab) or die(mysqli_error());
while ($row_colab = mysqli_fetch_array($result_colab)) {
$colab_nome = $row_colab["colab_nome_venda"];
echo $colab_nome . "</td>";
}
}
echo "</tr>";
}
?>
When executing, gives me the following error:
Notice: Array to string Conversion in C: xampp htdocs Sistema_1_1 est_exp_opened.php on line 65
The line 65:
$query = "SELECT * FROM venda_dia WHERE venda_dia_cod_ger in($row_ger) AND venda_dia_aberto='S'";
Thanks for your help so far...
$row_ger
is an array of results from the first query, you can give aimplode(', ', $row_ger);
or even make a single query replacingWHERE venda_dia_cod_ger in($row_ger)
forWHERE venda_dia_cod_ger in($query_ger)
.– Benilson