Question

我想获得ID为$_GET['id']的“post_title”。为此，我创建了它，但不会回复。

// Informações do Banco de Dados
$db['server'] = '127.0.0.1';
$db['user'] = 'user';
$db['pass'] = 'pass';
$db['database'] = 'databasename';

$connect = mysql_connect($db['server'],$db['user'],$db['pass'],$db['database']) or print (mysql_error());
mysql_select_db($db['database'], $connect);
////////////////////////////////////////////////////////////////////////////////////////

if ($_GET['id'] == NULL){
    print "Não Encontrado!";
    return false;
}
$idPego = $_GET['id'];
echo 'Número: ';
echo $idPego;

$livrotitle = mysql_query("SELECT post_title FROM _posts WHERE id='$_GET['id']' LIMIT 1");
$livro['titulo'] = mysql_result($livrotitle);

Answer 1

替换此

$livrotitle = mysql_query("SELECT post_title FROM _posts WHERE id='$_GET['id']' LIMIT 1");

带

 $livrotitle = mysql_query("SELECT post_title FROM _posts WHERE id='$idPego' LIMIT 1");

注意： 插入查询会抛出错误，因为它有语法错误id='$_GET['id']'

无法在我的数据库中选择一个寄存器

1 个答案: