在我的错误日志中,我得到了很多PHP消息:PHP通知:未定义的变量:性别名称在$UserData["]下的所有变量。它们是未定义的,因为我试图用这个数组来定义它们。我想知道最好的方法是定义变量,这样我就不会在日志中得到问题了。
if (!$location) $location = "-";
$UserData = array();
$UserData['id'] = $UserInfo['user_id'];
$UserData['username'] = $UserInfo['username'];
$UserData['gender'] = $genderName;
$UserData['age'] = $age;
$UserData['photo'] = $photo;
$UserData['location'] = $location;
$UserData['description'] = $description;
$UserData['isAdminORModerator'] = $typeUser;
$UserData['profile'] = SITE."/".$UserInfo['username'];
$UserData['level'] = $level;
}
return $UserData;
这是所有询问的完整代码。
function commonData($uid)
{
if ($uid)
{
$sql = "
SELECT a.user_id, a.email, a.username, a.displayname, a.level_id, a.photo_id
FROM engine4_users AS a
WHERE a.user_id = ".$uid;
}
$UserInfo = @mysql_fetch_assoc(mysql_query($sql));
if ($UserInfo['user_id'])
{
if ($UserInfo['photo_id'] && $UserInfo['photo_id']!="NULL")
{
$PPhoto = @mysql_fetch_assoc(mysql_query("SELECT a.* FROM engine4_storage_files AS a WHERE a.file_id = ".$UserInfo['photo_id']));
$photo = SOFTLAYER.$PPhoto['storage_path'];
}
else $photo = NO_PHOTO;
$queryMoreProfile = mysql_query("SELECT * FROM engine4_user_fields_values AS a WHERE a.item_id = ".$UserInfo['user_id']);
while ($moreProfile = @mysql_fetch_assoc($queryMoreProfile))
{
//birthday
if ($moreProfile['field_id']==6)
{
$age = getAge($moreProfile['value']);
}
//about
if ($moreProfile['field_id']==13)
{
$description = $moreProfile['value'];
}
//position
if ($moreProfile['field_id']==17)
{
$gender = $moreProfile['value'];
$gendersql = @mysql_fetch_assoc(mysql_query("SELECT a.* FROM engine4_user_fields_options AS a WHERE a.option_id = ".$gender));
$genderName = $gendersql['label'];
}
//location
if ($moreProfile['field_id']==24)
{
$locationNumber = $moreProfile['value'];
$locationsql = @mysql_fetch_assoc(mysql_query("SELECT a.* FROM engine4_user_fields_options AS a WHERE a.option_id = ".$locationNumber));
$location = $locationsql['label'];
}
//level
if ($UserInfo['level_id']==1 or $UserInfo['level_id']==2)
{
$typeUser = '<isAdmin>true</isAdmin>';
$level = 'admin';
}
else if ($UserInfo['level_id']==3)
{
$level = 'moderator';
$typeUser = '<isModerator>true</isModerator>';
}
else if ($UserInfo['level_id']==9 or $UserInfo['level_id']==10 or $UserInfo['level_id']==11 or $UserInfo['level_id']==12 or $UserInfo['level_id']==13 or $UserInfo['level_id']==14)
{
$level = 'premium';
$typeUser = '';
}
else if ($UserInfo['level_id']==8)
{
$level = 'VIP';
$typeUser = '';
}
else
{
$typeUser = '';
$level = 'guest';
}
if (!$location) $location = "-";
$UserData = array();
$UserData['id'] = $UserInfo['user_id'];
$UserData['username'] = $UserInfo['username'];
$UserData['gender'] = $genderName;
$UserData['age'] = $age;
$UserData['photo'] = $photo;
$UserData['location'] = $location;
$UserData['description'] = $description;
$UserData['isAdminORModerator'] = $typeUser;
$UserData['profile'] = SITE."/".$UserInfo['username'];
$UserData['level'] = $level;
}
return $UserData;
}}
你的问题是
$genderName
从未被赋值。
使用if(isset($genderName))
{
...
确保它存在
也可以使用三元操作符
$genderName = isset($genderName) ? $genderName : false;
你必须做
$genderName = "something";
因为否则你不能做
$something = $genderName;
你确定你不是想这么做吗?
$genderName = $UserData['genderName'];
变量插入操作从右插入到左变量
它看起来就像你的代码片段来自一个基于return
关键字的函数。我的猜测是,您已经从函数外部分配给变量$genderName
等。
如果是这种情况,那么注意函数不会将全局变量继承到它们的作用域中。您必须通过使用global
显式请求访问全局变量。如果您不这样做,那么$genderName
引用一个函数作用域的变量,您可能没有分配给它。
例如,在函数的顶部添加:
global $genderName, $age, ...;
您需要为您打算在函数中使用的每个全局变量执行此操作。注意,这个访问是读写的;您可以通过这种方式覆盖存储在全局变量中的值,所以要小心。
你的代码中有一种模式,它不仅仅会引起注意。
这个特殊的注意来自这样一个事实,即变量$genderName
只有在您正在处理的数据集包含moreProfile-field-id为17时才会被定义。只有这样,才会执行从数据库中获取性别名称的查询,变量才会生效。如果不是,变量保持未定义状态,您将得到通知。
实际上,您尝试返回用户的完整记录。如果用户没有保存他的"genderName",返回哪个值?空字符串?空吗?
关于这个模式:你的代码有一种倾向,对失败不是很健壮:
if ($uid)
{
$sql = "
SELECT a.user_id, a.email, a.username, a.displayname, a.level_id, a.photo_id
FROM engine4_users AS a
WHERE a.user_id = ".$uid;
}
$UserInfo = @mysql_fetch_assoc(mysql_query($sql));
$uid
是函数的参数。我希望它存在,否则PHP会抱怨。但是$uid
有很多值不是"真"的,比如0,",array()。如果$uid
被认为是"假的"呢?那么$sql
没有被定义。但它用于查询。并且任何错误都被抑制,因此您无法知道您的查询失败。
顺便说一句:你的SQL是不安全的SQL注入攻击
所以模式是:如果事情运行良好,你就做一些事情,但不注意其他情况下该怎么做。每个if
都有一个else
部分,即使里面没有代码,它也总是在那里。您必须处理else
部分—即使您只是在那里定义相同的变量并分配默认值,如以下所示:
//position
if ($moreProfile['field_id']==17)
{
$gender = $moreProfile['value'];
$gendersql = @mysql_fetch_assoc(mysql_query("SELECT a.* FROM engine4_user_fields_options AS a WHERE a.option_id = ".$gender));
$genderName = $gendersql['label'];
}
else
{
$genderName = "I don't know";
}