我的var_dump
在变量$results
上显示如下内容
array (size=11)
0 =>
object(stdClass)[69]
public 'Tables_in_database-name-here' => string 'wp_commentmeta' (length=14)
1 =>
object(stdClass)[68]
public 'Tables_in_database-name-here' => string 'wp_comments' (length=11)
2 =>
object(stdClass)[67]
public 'Tables_in_database-name-here' => string 'wp_links' (length=8)
...
10 =>
object(stdClass)[59]
public 'Tables_in_database-name-here' => string 'wp_users' (length=8)
我希望能够从上面的结构中提取表名,并以逗号分隔的形式列出它们,就像这样;
`wp_commentmeta,wp_comments,wp_links....,wp_users`
关于如何从那个结构中提取这些数据,我的大脑冻结了。
我尝试了相当多的选项,因为您可以通过下面的代码跟踪它-每个选项都以错误结束!
foreach ($results as $current_item):
/*
* $current_item is something like this:
*
* object(stdClass)[69]
public 'Tables_in_database-name-here' => string 'wp_commentmeta' (length=14)
*/
# echo $current_item;
# fires an error as "Catchable fatal error: Object of class stdClass could not be converted to string"
# echo $current_item[1];
# fires an error as "Cannot use object of type stdClass as array in..."
# echo array_values($current_item);
# fires an error as "array_values() expects parameter 1 to be array, object given in .."
# echo $current_item['Tables_in_database-name-here'];
#fires an error as "Cannot use object of type stdClass as array in "
how do you get that darn thing? :)
endforeach;
您的主要问题是属性名称Tables_in_database-name-here
不能通过通常的$obj->propertyName
表示。PHP允许您通过$obj->{'invalid-variable-name-but-ok-as-a-property'}
使用字符串属性名。
我只需通过implode()
和array_map()
构建字符串,例如
$string = implode(',', array_map(function($result) {
return $result->{'Tables_in_database-name-here'};
}, $results));
Demo ~ https://eval.in/172247
关于属性名,您还可以将stdclass
对象强制转换为数组。例如…
$associativeArray = (array) $result;
// array(1) { ["Tables_in_database-name-here"] => string(14) "wp_commentmeta" }
$indexedArray = array_values($associativeArray);
// array(1) { [0] => string(14) "wp_commentmeta" }