在PHP中,可以在对象上完成"var_dump"而不包括设置为其属性的对象的转储吗？

Question

PHP的"var_dump"函数以递归方式输出对象的属性.我想知道是否有一种"转储"对象的方法,但不要在原始对象中转储递归对象.

原始转储:

object(Class_Name)#1 (3) {
  ["label":protected]=>
  string(16) "My Label"
  ["name":protected]=>
  string(16) "name"
  ["object":protected]=>
  object(Class_Name)#2 (2) {
    ["id":protected]=>
    NULL
    ["classes":protected]=>
    array(0) {
    }
  }
}

通缉转储:

object(Class_Name)#1 (3) {
  ["label":protected]=>
  string(16) "My Label"
  ["name":protected]=>
  string(16) "name"
  ["object":protected]=>
  object(Class_Name)#2 (2) { ... }
}

Answer 1

你可以写自己的

/**
 * Schows all visible Vars of a Object, only in the first Level.
 * To get private or protected we need to call the class
 * in the Context of the Object ($this)
 * @param object $obj The Object
 * @param string|null $newLineCharacter The New Line Character (If null, it is based on \n for CLI or <br/> on web
 * @return void
 */
function firstLevelVarDump($obj, $newLineCharacter = null) {
    //Decide which new Line Character we use (Based on Loïc suggestion)
    if ($newLineCharacter === null) {
        $newLineCharacter = php_sapi_name() == 'cli' ? PHP_EOL : '<br/>';
    }
    //Get all visible Items
    $data = get_object_vars($obj);

    //Loop through each Item
    foreach ($data as $key => $item) {
        //Display Key + Type
        echo $key . ' => ' . gettype($item);

        //Extract Details, beased on the Type
        if (is_string($item)) {
            echo '(' . strlen($item) . ') "' . $item . '"';
        } elseif (is_bool($item)) {
            echo '(' . ($item ? 'true' : 'false') . ')';
        } elseif (is_integer($item) || is_float($item)) {
            echo '(' . $item . ')';
        } elseif (is_object($item)) {
            echo '(' . get_class($item) . ')';
        }

        //Line Break
        echo $newLineCharacter;
    }
}