I'm collecting the birthdates of users in my system, doing so by linking the the user's unique ID to their birthday entry in another table.
我正在收集我的系统中用户的生日日期,通过将用户的唯一ID与另一个表中用户的生日条目相关联来实现这一点。
To prevent users from accidentally/purposefully entering two birthdate entries for their accounts, I'd like to remove to the entry form for birthdays IF the user has already entered a birthday prior.
为了防止用户意外/故意地为他们的帐户输入两个生日条目,如果用户已经提前输入了生日,我想将其删除到生日条目表单中。
For instance:
例如:
$value = mysqli_query("SELECT bd_user_id FROM user_birthdate WHERE
bd_user_id="$user_id";");
From that data, how will I be able to return some form value to determine if whether the user's ID has already been index in user_birthdate or not? (Where $user_id = The current user's ID)
从该数据中,我将如何返回一些表单值,以确定用户的ID是否已经成为user_birthdate中的索引?(其中$user_id =当前用户的ID)
Or perhaps I'm taking the wrong approach here? The logic behind it is what's been getting me.
或者我在这里采用了错误的方法?这背后的逻辑是我一直想要的。
How can I check if whether a value is NOT indexed in a database table?
如何检查数据库表中的值是否被索引?
1 个解决方案
#1
2
You normally query the database as you did
您通常像这样查询数据库
$value = mysqli_query("SELECT bd_user_id FROM user_birthdate WHERE bd_user_id="$user_id";");
Than you use mysqli_num_rows(), and check if it returns 0.
然后使用sqmyli_num_rows()检查它是否返回0。
$num_rows = mysqli_num_rows($value);
if($num_rows > 0){
//exists
}else{
//doesn't exist
}
**Sorry, as Devon said in your case it's mysqli_num_rows not mysql_num_rows.
抱歉,正如德文在你的例子中所说,它是mysqli_num_rows而不是mysql_num_rows。
#1
2
You normally query the database as you did
您通常像这样查询数据库
$value = mysqli_query("SELECT bd_user_id FROM user_birthdate WHERE bd_user_id="$user_id";");
Than you use mysqli_num_rows(), and check if it returns 0.
然后使用sqmyli_num_rows()检查它是否返回0。
$num_rows = mysqli_num_rows($value);
if($num_rows > 0){
//exists
}else{
//doesn't exist
}
**Sorry, as Devon said in your case it's mysqli_num_rows not mysql_num_rows.
抱歉,正如德文在你的例子中所说,它是mysqli_num_rows而不是mysql_num_rows。