使用json_encode时遇到问题？

Question

我在数据库中插入带有json_encode()的数据，现在我只想获取（ select * from <table> ... ）仅数据库的name_unitsin ？ 我要输出此-> salam ＆ mokhles ＆ fadat

以数据库行units ：

[{"name_units":"salam","price_units":"74,554","checkbox_units":["minibar","mobleman"]},
 {"name_units":"mokhles","price_units":"4,851,269","checkbox_units":["mobleman","tv"]},
 {"name_units":"fadat","price_units":"85,642","checkbox_units":["minibar","mobleman","tv"]}]

。

$query_hotel_search = $this->db->query("SELECT * FROM hotel_submits WHERE name LIKE '%$hotel_search%' ORDER BY name asc");
$data = array();
foreach ($query_hotel_search->result() as $row)
{
   $units = json_decode($row->units);
   $data[] = array('name' => $row->name, 'units' =>$units['name_units']); // Line 24
}
echo json_encode($data);

这是上面的代码输出：

遇到PHP错误
严重程度：注意
消息：未定义的索引：name_units
行号：24

[{“ name”：“ Jack”，“ units”：null}]

Answer 1

从json_decode获得的结果是stdClass对象数组，而不是您期望的关联数组。 看来您在数据库的同一单元中有一个JSON字符串数组。

如果要输出，假设这就是数据库表的结构

萨拉姆和mokhles＆fadat

然后试试这个：

foreach( $query_hotel_search->result() as $row ) {
    $units = json_decode( $row->units );
    $names = '';
    foreach( $units as $unit ) {
        $names .= "{$unit->name_units} & ";
    }
}
echo substr( $names, 0, -2 );