Php 如何在数据库中插入随机数不重复

我有一个注册，每个注册的人都有一个唯一的用户id。这个用户id不应该重复，这样每个人都有一个唯一的id。然后用户id被插入到两个不同的表中。但是，每当一个新人在数据库中注册用户id的号码时，总是返回0

这是我的密码：

    $userid = '';
function randomDigits($length){
$numbers = range(0,10000000);
shuffle($numbers);
for($i = 0; $i < $length; $i++){
    global $digits;
    $userid .= $numbers[$i];
}
return $userid;
}

    $query = "INSERT INTO Users (user_id, user_type, name, lastname, email, phone, address, apt, city, state, zip, username, password) VALUES ('$userid','parent','$name', '$lastname', '$email', '$phone', '$address', '$apt', '$city', '$state', '$zip', '$username', '$hash')";
    $result = mysqli_query($query);


    $sql = "INSERT INTO dancers (user_id, name, dancer_name, dancer_middlename, dancer_lastname, dancer_age, dancer_dob, dancer_number, school_code, teacher_name, class_location, date_enrolled, date_comp) VALUES ('$userid', '$name', '$dancer_name', '$dancer_middlename', '$dancer_lastname', '$dancer_age', '$dancer_dob', '$dancer_number', '$schoolcode', '$teacher', '$location', '$date_enrolled', '$date_comp')";


     $result2 = mysqli_query($sql);
     if(!$result = $con->query($query)){
         die('there was an error running query [' . $con->error . ']');
     }else {
       //  header("location: thankyou.html");
     }
    if(!$result2 = $con->query($sql)){
         die('there was an error running query [' . $con->error . ']');
     }else {
         header("location: thankyou.html");
     }
    }

舞者桌

但这应该是：

用户表

舞者桌

用答案编辑2

我在达米安·巴勃罗·冈萨雷斯的回答中使用了D

以下是正确的代码：

$userId = rand(1,9999999);
$check_userId ="select count(*) count from Users where user_id = " . 
$userId;
while ($row['count'] > 0);

$query = "INSERT INTO Users (user_id, user_type, name, lastname, email, phone, address, apt, city, state, zip, username, password) VALUES ('$userid','parent','$name', '$lastname', '$email', '$phone', '$address', '$apt', '$city', '$state', '$zip', '$username', '$hash')";
    $result = mysqli_query($query);


    $sql = "INSERT INTO dancers (user_id, name, dancer_name, dancer_middlename, dancer_lastname, dancer_age, dancer_dob, dancer_number, school_code, teacher_name, class_location, date_enrolled, date_comp) VALUES ('$userid', '$name', '$dancer_name', '$dancer_middlename', '$dancer_lastname', '$dancer_age', '$dancer_dob', '$dancer_number', '$schoolcode', '$teacher', '$location', '$date_enrolled', '$date_comp')";


     $result2 = mysqli_query($sql);
     if(!$result = $con->query($query)){
         die('there was an error running query [' . $con->error . ']');
     }else {
       //  header("location: thankyou.html");
     }
    if(!$result2 = $con->query($sql)){
         die('there was an error running query [' . $con->error . ']');
     }else {
         header("location: thankyou.html");
     }
    }

接近A 在我看来，正如雷伊建议的那样，最好使用一个自动递增字段，它为您涵盖了所有内容，所以您不必担心这些事情

为此，请尝试此MySQL命令。ALTER TABLE Users修改列user\u id INT自动增量。之后，您只需插入其他字段，用户id将始终是唯一的：

$query = "INSERT INTO Users (user_type, name, lastname, email, phone, address, apt, city, state, zip, username, password) VALUES ('parent','$name', '$lastname', '$email', '$phone', '$address', '$apt', '$city', '$state', '$zip', '$username', '$hash')";
$result = mysqli_query($query);

方法B： 插入下一个可用的id：

$query = "INSERT INTO Users (user_id, user_type, name, lastname, email, phone, address, apt, city, state, zip, username, password) SELECT max(user_id)+1,'parent','$name', '$lastname', '$email', '$phone', '$address', '$apt', '$city', '$state', '$zip', '$username', '$hash' from Users";
$result = mysqli_query($query);

看看我做了什么：我使用了插入字段选择，没有值。它将在读取的同时插入id。这要求用户id为数字

方法C 正如我看到的，您使用引号表示用户id，可能是一个字符串字段

1如果用户id为字符串，则使用

基于当前时间（以微秒为单位）获取带前缀的唯一标识符

2然后，为了确定，检查它是否未被使用

do {
    $uniq = uniqid();
    $query ="select count(*) count from Users where user_id = '" . $uniq ."'";
    //execute and do a fetch 
} while ($row['count'] > 0);

do {
    $uniq = rand(1,999999);
    $query ="select count(*) count from Users where user_id = " . $uniq;
    //execute and do a fetch 
} while ($row['count'] > 0);

它将继续尝试，直到id真正唯一为止。这要求user_id是字符串字段，就像varchar一样，因为uniqid返回字母和数字

方法D 与C类似，但为数字

1如果用户_id为整数，则使用

rand-生成一个随机整数

2然后，为了确定，检查它是否未被使用

do {
    $uniq = uniqid();
    $query ="select count(*) count from Users where user_id = '" . $uniq ."'";
    //execute and do a fetch 
} while ($row['count'] > 0);

do {
    $uniq = rand(1,999999);
    $query ="select count(*) count from Users where user_id = " . $uniq;
    //execute and do a fetch 
} while ($row['count'] > 0);

---

它将为您可以使用下面函数的唯一随机数设置。您可以根据需要传递长度

    public static function generateUniqueId($length = 7) {
        $salt = "abcdefghijkmnpqrstuvwxyzABCDEFGHIJKLMNPQRSTUVWXYZ23456789";
        srand(); // start the random generator
        $userId = ""; // set the inital variable
        for ($i = 0; $i < $length; $i++) { // loop and create userId
            $userId .= substr($salt, rand() % strlen($salt), 1);
        }
        return $userId;
    }

---

为什么不使用自动递增id？请查看我的EditA方法不适用于我，因为我在编辑的问题中提到了这个原因。方法B与方法A不起作用的原因相同。因为id是一个整数，所以我使用了方法D，它对我非常有效。因此，我认为方法C也能很好地工作。非常感谢。

do {
    $uniq = rand(1,999999);
    $query ="select count(*) count from Users where user_id = " . $uniq;
    //execute and do a fetch 
} while ($row['count'] > 0);

    public static function generateUniqueId($length = 7) {
        $salt = "abcdefghijkmnpqrstuvwxyzABCDEFGHIJKLMNPQRSTUVWXYZ23456789";
        srand(); // start the random generator
        $userId = ""; // set the inital variable
        for ($i = 0; $i < $length; $i++) { // loop and create userId
            $userId .= substr($salt, rand() % strlen($salt), 1);
        }
        return $userId;
    }