Php 即使值存在,仍尝试获取非对象的属性“月”
我目前正在解码一个JSON文件的一部分,但是得到了一个“尝试获取属性'month'of non object”错误,尽管print\r和var\u dump都显示该值存在 JSON: PHP: 错误:Php 即使值存在,仍尝试获取非对象的属性“月”,php,Php,我目前正在解码一个JSON文件的一部分,但是得到了一个“尝试获取属性'month'of non object”错误,尽管print\r和var\u dump都显示该值存在 JSON: PHP: 错误: Notice: Trying to get property 'month' of non-object in processReviews.php on line 36 Notice: Trying to get property 'year' of non-object in processR
Notice: Trying to get property 'month' of non-object in processReviews.php on line 36
Notice: Trying to get property 'year' of non-object in processReviews.php on line 37
Notice: Trying to get property 'price' of non-object in processReviews.php on line 38
Notice: Trying to get property 'location' of non-object in processReviews.php on line 39
Notice: Trying to get property 'facilities' of non-object in processReviews.php on line 40
Notice: Trying to get property 'view' of non-object in processReviews.php on line 41
Notice: Trying to get property 'unitTypeID' of non-object in processReviews.php on line 42
Notice: Trying to get property 'reviewer' of non-object in processReviews.php on line 43
Notice: Trying to get property 'profileID' of non-object in processReviews.php on line 44
Notice: Trying to get property 'profileType' of non-object in processReviews.php on line 45
Notice: Trying to get property 'unitTypeDescription' of non-object in processReviews.php on line 46
它似乎只有在“review”键之后才会失败,因为其他值肯定存在,并显示在var_dump上:
array(15) {
["reviewID"]=>
int(1)
["markerID"]=>
int(10605)
["userID"]=>
int(1)
["review"]=>
string(4) "Test"
["month"]=>
int(5)
["year"]=>
int(2011)
["price"]=>
int(9)
["location"]=>
int(8)
["facilities"]=>
int(0)
["view"]=>
int(7)
["unitTypeID"]=>
int(3)
["reviewer"]=>
string(5) "admin"
["profileID"]=>
int(3)
["profileType"]=>
string(17) "Middle-age couple"
["unitTypeDescription"]=>
string(25) "Medium Motorhome under 8m"
}
如果有人有任何想法,我们将不胜感激。在您的代码中
$review = (string)$review->review;
$month = (int)$review->month;
因此,在尝试获取$month变量之前,请将变量$review重新分配给其他变量,因此只需将此字段名更改为其他变量,如
$reviewValue = (string)$review->review;
以及代码中的任何其他引用
$review = (string)$review->review;
$month = (int)$review->month;
因此,在尝试获取$month变量之前,请将变量$review重新分配给其他变量,因此只需将此字段名更改为其他变量,如
$reviewValue = (string)$review->review;
还有其他的参考资料不知道我怎么没发现。非常感谢Nigels有时候你被代码迷住了,你不总能看到一些东西。不知道我怎么没发现。非常感谢Nigels有时你会被代码所包围,你不总能看到一些东西。