访问嵌套对象数组 [英] Access a nested object array

问题描述

嘿伙计们，我似乎无法正确理解这里的语法，我只是想看看类型是否存在，我试过了

Hey guys I can't seem to get the syntax here right, I'm just trying to see if type exists, I've tried

if($obj->type)    /* do something */
if($obj[0]->type) /* do something */
if($obj->type[0]) /* do something */

关于这个

stdClass::__set_state(
array(
   '0' => 
  stdClass::__set_state(
  array(
     'report_number' => '555',
     'type' => 'citrus',
     'region' => 'Seattle',
     'customer_number' => '9757',
     'customer' => 'The Name',
     'location' => 'West Seattle, WA',
     'shipper' => 'Yamato Transport',
     'po' => '33215',
     'commodity' => 'RARE',
     'label' => 'PRODUCE',
  )),
))

但似乎无法做到正确，我相信这与[0] 是 int 而不是 varchar 但我不知道....

But just can't seem to get it right, I believe it has something to do with [0] being an int instead of varchar but I have no idea....

推荐答案

您问题中的输出是由 var_export 创建的，它将对象的数据表示为 array.但不要被误导，它实际上是一个对象.

The output in your question is created by var_export which represents the data of the object as an array. But don't get mislead by that, it's in fact an object.

通过以下语法访问对象属性:

Object properties are accessed by this syntax:

$obj->property

在您的情况下，属性名为 0，这对于 PHP 来说很难用纯代码处理，因为它不是您可以像这样立即编写的属性名称:

In your case the property is named 0 which is hard for PHP to deal with in plain code as it's not a property name you could write right away like this:

$obj->0

相反，您需要通过将其包含在 {} 中来告诉 PHP 解析器名称是什么:

Instead, you need to tell the PHP parser what the name is by enclosing it into {}:

$obj->{0}

然后您可以进一步遍历以访问 0 的 type 属性:

You can then traverse further on to access the type property of 0:

$obj->{0}->type

希望这有帮助，问题不时出现，这是一个相关的，有更多相关链接:如何访问此对象属性?.

Hope this is helpful, the question comes up from time to time, this is a related one with more related links: How do I access this object property?.

PHP 手册对此有记录，但在 上不是很明显变量变量入口:

The PHP Manual has this documented, but not very obvious on the Variable variables entry:

为了在数组中使用可变变量，您必须解决歧义问题.也就是说，如果您编写 $$a[1] 那么解析器需要知道您是否打算使用 $a[1] 作为变量，或者您是否想要$$a 作为变量，然后是来自该变量的 [1] 索引.解决这种歧义的语法是:${$a[1]} 用于第一种情况，${$a}[1] 用于第二种情况.

In order to use variable variables with arrays, you have to resolve an ambiguity problem. That is, if you write $$a[1] then the parser needs to know if you meant to use $a[1] as a variable, or if you wanted $$a as the variable and then the [1] index from that variable. The syntax for resolving this ambiguity is: ${$a[1]} for the first case and ${$a}[1] for the second.

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