Question

当我检索两个表时出现错误。我的代码有什么问题？我不知道如何解决这个问题

    <?php
include ('includes/config.php');

$mysqli = new mysqli(DB_SERVER, DB_UNAME,DB_PASSWD,DB_NAME);
if(!$mysqli){
  throw new Exception($mysqli->connect_error, $mysqli->connect_errno);
}

$jqry = $mysqli->prepare("SELECT time FROM table_time ORDER BY time");
if (!$jqry){

  throw new Exception($mysqli->error);

}
$jqry->execute();
$jqry->bind_result($time);
$jqry->store_result();

$times = array();

while ($jqry->fetch()){

        $times[] = $time;

}
$jqry->close();

$gqry = $mysqli->prepare("SELECT table_group.group FROM table_group.table_group ORDER BY group");
if(!$gqry){

  throw new Exception($gqry->error);
}
$gqry->execute();
$gqry->bind_result($group);
$gqry->store_result();

$groups = array();

while ($gqry->fetch()){

  $groups[] = $group;
}


?>

这是我得到的错误：

注意：尝试在第31行的C：\ xampp \ htdocs \ ~Jeremiah \ system5 \ joborder.php中获取非对象的属性

致命错误：C：\ xampp \ htdocs \ ~Jeremiah \ system5 \ joborder.php中未捕获的异常'Exception'：31堆栈跟踪：＃0 {main}抛出

Answer 1

在第31行无效之前

你的SQL。尝试直接对db执行它。

来自手册：

mysqli_prepare() returns a statement object or FALSE if an error occurred.

...那就是为什么我们不在开发环境中压制通知！：）

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1 个答案: