Question

我已经尝试了几个小时来获得一个解决方案来检查用户名是否存在，但它会不断添加我添加的相同用户名（不输出它存在的错误）。

我目前的职能;

function mysqli_result($res, $row, $field=0) { 
    $res->data_seek($row); 
    $datarow = $res->fetch_array(); 
    return $datarow[$field]; 
} 

function name_exists($name){
    $con = mysqli_connect('localhost','root','','database');
    $name = ($name);
    $query = mysqli_query($con,"SELECT COUNT(id) FROM users WHERE name = '$name'");
    // Count the amount of rows where username = $unsername
    $check = mysqli_num_rows($query);

    return (mysqli_result($query, 0) == 1) ? true:false;
}

我验证用户名和其他内容的完整代码如下所示;

//Validates Username
if(empty($error)) {
    if(name_exists($name)==true){
        $email_error = "Sorry, the username: $name is already Taken.";
        $error = true;    
    }
    if(empty($name)) {
        $error = true;
        $email_error = "You Forgot to Enter Your Username!";
    }
    if(!preg_match('/^\w{5,}$/', $name)){
        $error = true;
        $email_error= "Invalid Username format.";
        // valid username, alphanumeric & longer than or equals 5 chars
    }
}
if(!filter_var($email,FILTER_VALIDATE_EMAIL)) {
    $error = true;
    $email_error = "Please Enter Valid Email ID";

}
if(strlen($password) < 6) {
    $error = true;
    $password_error = "Password must be minimum of 6 characters";
}
if($password != $cpassword) {
    $error = true;
    $cpassword_error = "Password and Confirm Password doesn't match";
}
if (!$error)
{
    // prepare and bind
    $stmt = $con->prepare("INSERT INTO users (name,email,password) VALUES (?, ?, md5(?))");
    $stmt->bind_param("sss", $name, $email, $password);
    //$stmt->execute();
    if( $stmt->execute() == true) {
        $successmsg = "Successfully Registered! <a href='login.php'>Click here to Login</a>";
    } else {
        $errormsg = "Error in registering...Please try again later!";
    }
}

}

数据库结构

id(int8)
name(varchar 30)
email(varchar60)
password(varchar40)

Answer 1

首先，您可以向数据库中的name列添加唯一标识符，以避免重复插入。

其次让查询为

SELECT COUNT(id) 
FROM users 
WHERE name = ' ".$name." ' ";

在单引号内，变量$name被视为字符串$name，而不是被替换为$name

的值

Answer 2

取代：

page = Page.query.filter_by(id=value).one()

使用：

for id in tag_ids:
    tag = Tag.query.filter_by(id=id).one() #wanna avoid this
    page.tags.append(tag)

我测试过，它对我有用

想要检查数据库中是否存在用户名？

2 个答案: