如果值不在表中，则显示标准消息

问题描述：

I want to be able to display a fallback statement if an entry in my database table doesn't exist.

On a user profile page I would like to display a phone number. If a user hasn't entered a phone number I would like to display a message to say 'no number has been provided.'

At present, I have been able to display nothing if the value doesn't exist, but this isn't ideal

How would I amend the current code I am using (below) to achieve this?

<?php if(!empty($profile['profile_phonenumber'])){ ?>
<?php echo $profile['profile_phonenumber'] ?>
<?php } ?>

如果数据库表中的条目不存在，我希望能够显示回退语句。 p>

在用户个人资料页面上，我想显示一个电话号码。如果用户没有输入电话号码，我想显示一条消息，说“没有提供号码”。 p>

目前，如果该值不存在，我就无法显示任何内容，但这并不理想 p>

我将如何修改我正在使用（下面）实现的当前代码？ p>

 ＆lt;？php if（！empty（$ profile ['profile_phonenumber']））{？＆gt; \  n＆lt;？php echo $ profile ['profile_phonenumber']？＆gt; 
＆lt;？php}？＆gt; 
  code>  pre> 
  div>

答

You can do it several way: Through php:

echo  (!empty($profile['profile_phonenumber'])) ?
     $profile['profile_phonenumber'] : 'No phone provided';

if you are using mysql: then in select query:

select if(length(profile_phonenumber)>0,profile_phonenumber,'profile_phonenumber') as profile_phonenumber,<other column name>  from <tablename> where <your_condition>

答

Use this code with else clause:

if (!empty($profile['profile_phonenumber'])) {
    echo $profile['profile_phonenumber'];
} else {
    echo 'No phone provided';
}

答

Use Ternary operator. Try this code:

<?php 
  echo (!empty($profile['profile_phonenumber'])) ? $profile['profile_phonenumber'] : "no number has been provided"; 
?>

如果值不在表中，则显示标准消息

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