Question

我有一个查询，检查数据库中是否存在插入的名字和姓氏，我想检查名字和姓氏或电子邮件，所以如果插入的名字和姓氏存在，那么用户找到了，如果电子邮件只存在于数据库中，那么用户也会找到，该怎么做？

以下是代码：

$query = "SELECT * FROM users WHERE first_name='" . $f_name . "' AND last_name='" . $l_name . "'";
    $result = mysql_query($query);
        $row = mysql_fetch_assoc($result);
if(empty($row)){
    //some code to add the user to the database
}else{
   //user is found
}

所以我希望别的意思是数据库中不存在（名字和姓氏）或（仅电子邮件）

Answer 1

$query = "SELECT * FROM users WHERE (first_name='" . $f_name . "' AND last_name='" . $l_name . "') OR (email = '" . $email . "')";

Answer 2

希望这有帮助，

$query = "SELECT * FROM users WHERE (first_name='" . $f_name . "' AND 
last_name='" . $l_name . "') OR (email = '" . $email . "')";

编辑 -

Barclick Flores Velasquez他的答案对我来说还没有加载，这也是正确的。

Answer 3

只需要使用<?php $query = "SELECT * FROM users WHERE (`first_name` = '".$f_name."' AND `last_name` = '".$l_name."') OR (`email_column_name` = '".$email."') "; $result = mysql_query($query); $row = mysql_fetch_assoc($result); if(empty($row)){ //some code to add the user to the database } else { //user is found }?>条件附加您的查询。

<?php
$mysqli = new mysqli("localhost", "user_name", "password", "db_name");
$stmt = $mysqli->prepare("SELECT * FROM users WHERE (`first_name` = ? AND `last_name` = ?) OR (`email_column_name` = ?) ");
$stmt->bind_param("sss", $f_name, $l_name, $email);
$result = $stmt->execute();

if($result->num_rows > 0){
  //user is found
} else {
  //some code to add the user to the database
}?>

我建议你使用预准备语句来避免SQL注入

面向对象的风格

<?php
$mysqli = mysqli_connect("localhost", "user_name", "password", "db_name");
$stmt = mysqli_prepare($mysqli, "SELECT * FROM users WHERE (`first_name` = ? AND `last_name` = ?) OR (`email_column_name` = ?) ");
mysqli_stmt_bind_param($stmt, "sss", $f_name, $l_name, $email);
$result = mysqli_stmt_execute($stmt);

if(mysqli_num_rows($result) > 0){
  //user is found
} else {
  //some code to add the user to the database
}?>

程序风格

{{1}}

快速链接

如何对查询进行多次检查以选择其中一个为真？

3 个答案: