我有两张表如下: activity_base(保存基本活动信息以及activity_type) activity_email(保存有关电子邮件活动的具体细节)
所有表都有activity_id,引用activity_base。
示例:我正在尝试输出电子邮件活动的数据,该活动根据活动类型从其他表中获取数据。
查询将引用activity_base以获取有关活动的基本信息,然后如果activity_type是电子邮件,我需要它通过运行函数email_activity_details从activity_email表中获取更多数据。
问题是:我无法将$ activity_details显示为未定义的变量。 这是查询: 电子邮件活动信息:
function email_activity_details($activity_id){
global $connection;
$email_activity = "SELECT * FROM activity_email WHERE activity_id ='$activity_id'"
or die("Error: ".mysqli_error($connection));
$query_email_activity = mysqli_query($connection, $email_activity);
return $query_email_activity;
}
活动详情:调用email_activity_details():
function view_full_activity($activity_field){
global $connection;
$contact_id = $_REQUEST['contact_id'];
$activity_id = $_REQUEST['activity_id'];
$get = "SELECT * FROM activity_base WHERE activity_id = '$activity_id' "
or die("Error: ".mysqli_error($connection));
$query = mysqli_query($connection, $get);
//Get activity base information
while ($activity = mysqli_fetch_array($query)){
$activity_related_to_id = $activity ['activity_related_to_id'];
$activity_id = $activity['activity_id'];
$activity_type_id = $activity['activity_type_id'];
}
//Get detailed activity information
//If activity is Email
if ($activity_type_id == "1") {
$email_details = email_activity_details('$activity_id');
while ( $email = mysqli_fetch_assoc($email_details)) {
$activity_details = $email['email_message'];
}
}
switch ($activity_field) {
case 'activity_id':
return $activity_id;
break;
case 'activity_title':
return $activity_title;
break;
default:
# code...
break;
}
}
希望我能够解释清楚。谢谢。
答案 0 :(得分:1)
更改此行
$email_details = email_activity_details('$activity_id');
到这个
$email_details = email_activity_details($activity_id);
说明: 单引号内的美元符号按字面意思处理。
从PHP手册:
特殊字符的变量和转义序列在单引号字符串中出现时不会被展开
http://php.net/manual/en/language.types.string.php#language.types.string.syntax.single