Question

我从早上开始整整一天都在没有解决方案的情况下处理这个问题。

我有这些数据来自数据库。我使用PHP PDO连接。

Array (
 [0] => stdClass Object ( [testing_id] => 4 [testing_name] => please [testing_location] => kjnkdsnkdnskjndkjsndjknskdnsk )
 [1] => stdClass Object ( [testing_id] => 3 [testing_name] => please [testing_location] => jknds ndns )
 [2] => stdClass Object ( [testing_id] => 2 [testing_name] => please [testing_location] => be done to me ) )

我想重命名对象中的键，而不是将testing_id重命名为id，将testing_name重命名为name等。

我已经写了如下所示的函数数量

function remove_keys($arr, $table) {
  $object = new stdClass();
  foreach ($arr as $key => $val) {
    $x = (array) $val;
    foreach ($x as $key2 => $value) {
      $new_key = str_replace($table, '', $key2);
      $object->$new_key = $value;
    }
  }
  return $object;
}

和这个

function replaceKey(&$array,$table) {    
  $x = array();
  foreach($array as $k => $v){           
    $new_key = str_replace($table, '', $k);
    array_push($x, $new_key);
  } 
  $array = array_combine($x, $array);
  return $array;
}

在所有情况下，我只获得一个对象结果，而不是重命名整个对象

stdClass Object ( [id] => 2 [name] => please [location] => be done to me )

如何重命名对象中的每个索引并重命名完整对象？请帮忙

我需要输出像这样

Array ( 
  [0] => stdClass Object ( [id] => 4 [name] => please [location] => kjnkdsnkdnskjndkjsndjknskdnsk )
  [1] => stdClass Object ( [id] => 3 [name] => please [location] => jknds ndns ) 
  [2] => stdClass Object ( [id] => 2 [name] => please [location] => be done to me ) )

我在这里搜索过没有任何类似的解决方案

Answer 1

您正在覆盖foreach中的对象值：

$object->$new_key = $value;

$object始终是同一个变量，尝试这样的事情：

function remove_keys($arr, $table) {
  $temp_array = array();
  foreach ($arr as $key => $val) {
    $object = new stdClass();
    $x = (array) $val;
    foreach ($x as $key2 => $value) {
      $new_key = str_replace($table, '', $key2);
      $object->$new_key = $value;
    }
    $temp_array[] = $object;
  }
  return $temp_array;
}

这将为您提供一系列对象。

在php中重命名对象键

1 个答案: