我正在使用某人编写的PHP类来与BaseCamp API进行交互.
我正在做的特定调用是检索待办事项列表中的项目,这很好.
我的问题是,我不知道如何只访问todo-items返回的对象的属性.这是返回对象的var_dump:
object(stdClass)[6]
public 'completed-count' => string '0' (length=1)
public 'description' => string 'Description String' (length=89)
public 'id' => string '12345' (length=7)
public 'milestone-id' => string '' (length=0)
public 'name' => string 'Error Reports' (length=13)
public 'position' => string '1' (length=1)
public 'private' => string 'false' (length=5)
public 'project-id' => string '58904' (length=7)
public 'tracked' => string 'false' (length=5)
public 'uncompleted-count' => string '1' (length=1)
public 'todo-items' =>
object(stdClass)[3]
public 'todo-item' =>
object(stdClass)[5]
public 'completed' => …Run Code Online (Sandbox Code Playgroud) 我有mysql表,其中包含'operation.date','operation.name'等字符.在将该表数据作为对象获取后,$mysqli->fetch_object()我得到了这个(print_r of row):
stdClass Object
(
[id] => 2
[operation.date] => 2010-12-15
[operation.name] => some_name
)
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我该如何存取权限operation.date,并operation.name和所有其他古怪的命名对象的属性?
所有,
我在键名中有一个带连字符的数组.如何在PHP中提取它的值?它返回0给我,如果我这样访问:
print $testarray->test-key;
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这就是数组的样子
testarray[] = {["test-key"]=2,["hotlink"]=1}
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谢谢