0
My mistake is this
Warning: mysql_fetch_array() expects Parameter 1 to be Resource, Boolean Given in C: xampp htdocs BD project proj_details.php on line 25
<?php
$base_dados = "crowdfunding";
$conexao = mysql_connect("localhost","root","") or die(mysql_error());
mysql_select_db($base_dados) or die(mysql_error());
$sql = "SELECT projeto.id, nome, limite, projeto.descricao FROM projeto,(SELECT valor, recompensa.descricao FROM recompensa, projeto WHERE projeto.id = id_projeto)";
$resultado = mysql_query($sql);
?>
<html>
<head>
<title>Projetos Atuais</title>
</head>
<body>
<h1>Detalhes de um Projeto<br /></h1>
<table>
<tr>
<td><strong>ID</strong></td>
<td><strong>Nome</strong></td>
<td><strong>Data Final</strong></td>
<td><strong>Descricao</strong></td>
<td><strong>Valor da Recompensa</strong></td>
<td><strong>Descricao da Recompensa</strong></td>
</tr>
<?php
while($dados = mysql_fetch_array($resultado)){
$id = $dados['id'];
$nome = $dados['nome'];
$data = $dados['limite'];
$descricao_proj = $dados['projeto.descricao'];
$valor = $dados['valor'];
$descricao_rec = $dados['recompensa.descricao'];
echo "<tr>
<td>$id</td>
<td>$nome</td>
<td>$data</td>
<td>$descricao_proj</td>
<td>$valor</td>
<td>$descricao_rec</td>
</tr>";
}
?>
</table><br /><br />
<form action="menu1.php" method="get">
<center><input name="botao" type="submit" value="Voltar"/></center>
</form>
</body>
</html>
The error is clear, the function
mysql_fetch_array()expecting aresourceand is receiving a Boolean. Probably your query is returningfalse.– user28595
Why don’t you use
inner jointo do this query there, ta very strange.– user28595
I took the liberty of re-doing this query for you: SELECT project.id, project.name, project.limit, project.Description, reward.value, reward.Description FROM project INNER JOIN reward ON reward.id_project = project.id
– Hoppy
Bruh, change your duties
mysql_for their dutiesmysqli_, for support reasons. starting with PHP 5.5+ these functions are deprecated and no longer exist in PHP 7– StillBuggin