如何隐藏错误“试图获取非对象的属性”

I have the following code, which works correctly. The Twitter Friends are listed correctly, however it seems that when the last item is displayed the error "Notice: Trying to get property of non-object" is displayed 4 times.

我有以下代码,它可以正常工作。 Twitter好友列出正确,但似乎当显示最后一项时,错误“通知:试图获取非对象的属性”显示4次。

Since the code works as it should, I would like a way to hide these errors.

由于代码可以正常工作,我想要一种隐藏这些错误的方法。

$connection = getConnectionWithAccessToken($consumerkey, $consumersecret, $accesstoken, $accesstokensecret);
$tweets6 = $connection->get("https://api.twitter.com/1.1/friends/list.json?screen_name=".$twitteruser."&count=".$notweets);
foreach ($tweets6 as $tweet)
{
    for($i = 0; $i < count($tweet); $i++)
    {
        echo $tweet[$i] -> name;
        echo "<br />";
    }
}

5 个解决方案

#1

you can add a checker if the object has a certain property before using its value

如果对象在使用其值之前具有某个属性,则可以添加检查器

if (isset($tweet[$i]->name)) {
    // process

}

#2

Use a simple if condition before print..

打印前使用简单的if条件..

   $connection = getConnectionWithAccessToken($consumerkey, $consumersecret, $accesstoken, $accesstokensecret);
   $tweets6 = $connection->get("https://api.twitter.com/1.1/friends/list.json?screen_name=".$twitteruser."&count=".$notweets);
foreach ($tweets6 as $tweet)
{
   for($i = 0; $i < count($tweet); $i++){  
     if($tweet[$i]){
        echo $tweet[$i] -> name;
        echo "<br />";
     }
   }
}

#3

replace this :

替换这个:

for($i = 0; $i < count($tweet); $i++)

with this :

有了这个 :

for($i = 0; $i < count($tweet) - 1; $i++)

EDIT

for($i = 0; $i < count($tweet); $i++){
    if (isset($tweet[$i]->name)) {
        echo $tweet[$i] -> name;
        echo "<br />";
    }
}

try this

#4

use if empty for prevent that notice.

如果为空则使用以防止该通知。

$connection = getConnectionWithAccessToken($consumerkey, $consumersecret, $accesstoken, $accesstokensecret);
$tweets6 = $connection->get("https://api.twitter.com/1.1/friends/list.json?screen_name=".$twitteruser."&count=".$notweets);
foreach ($tweets6 as $tweet)
{
    for($i = 0; $i < count($tweet); $i++)
    {
        if(empty($tweet[$i]->name)) continue;
        echo $tweet[$i]->name;
        echo "<br />";
    }
}

#5

While the accepted answer will work, PHP has the property_exists() function to do this job and would be more appropriate. It will return true even if the property has a null value, which isset() does not.

虽然接受的答案可行,但PHP有property_exists()函数来完成这项工作,并且更合适。即使属性具有空值,它也将返回true,而isset()则不是。

if (property_exists($tweets[$i], "name")) {
    ....
}

#1