带有bind的PHP select语句

我试图通过PHP验证一个注册表单，其中包含准备好的语句和绑定参数，如下所示：

$email = $_POST['email'];
$password = $_POST['password']; 
$selectStatement = "SELECT id FROM users WHERE email = ? ";
$stmts = mysqli_prepare($dbc, $selectStatement);
mysqli_stmt_bind_param($stmts,'s', $email);
mysqli_stmt_execute($stmts);
$count = mysqli_stmt_num_rows($stmts);

if (!$stmts){
    die('mysqli error: '.mysqli_error($dbc));
}
while($row = mysqli_stmt_fetch($stmts)){
    if($row != null){   
        echo('User already registered!');
    }
    }else{  
        $insertQuery = "INSERT INTO users(email, password) VALUES(?, ?)";
        $stmtI = mysqli_prepare($dbc, $insertQuery); 
        mysqli_stmt_bind_param($stmtI, "ss", $email, $password);
        mysqli_stmt_execute($stmtI);
        echo('Registration completed!');
}
}
mysqli_stmt_close($stmts);  
mysqli_stmt_close($stmtI);
mysqli_close($dbc); 

---

但是当在数据库中找不到用户时，它应该在else分支上运行，但显然它没有到达那里，甚至在while循环中也没有。我做错了什么

你的代码错了。。。在

}
}else{

只写

}else{

试试这个

$email = $_POST['email'];
$password = $_POST['password'];
$selectStatement = "SELECT id FROM users WHERE email = ? ";
$stmts = mysqli_prepare($dbc, $selectStatement);
mysqli_stmt_bind_param($stmts,'s', $email);
mysqli_stmt_execute($stmts);
$count = mysqli_stmt_num_rows($stmts);

if (!$stmts){
    die('mysqli error: '.mysqli_error($dbc));
 }
while($row = mysqli_stmt_fetch($stmts)){
if($count == 1){
    echo('User already registered!');
}else{
    $insertQuery = "INSERT INTO users(email, password) VALUES(?, ?)";
    $stmtI = mysqli_prepare($dbc, $insertQuery);
    mysqli_stmt_bind_param($stmtI, "ss", $email, $password);
    mysqli_stmt_execute($stmtI);
    echo('Registration completed!');
    }
}
mysqli_stmt_close($stmts);
mysqli_stmt_close($stmtI);
mysqli_close($dbc);

---

While循环仅在从数据库（mysqli_stmt_fetch（$stmts））中找到记录时执行。

您必须像这样更改代码

$email = $_POST['email']; 
$password = $_POST['password']; $selectStatement = "SELECT id FROM users WHERE email = ? "; 
$stmts = mysqli_prepare($dbc, $selectStatement);
mysqli_stmt_bind_param($stmts,'s', $email);
mysqli_stmt_execute($stmts);
$count = mysqli_stmt_num_rows($stmts);
if (!$stmts){ die('mysqli error: '.mysqli_error($dbc));
}
    if($count){
            echo('User already registered!');
    }else{
            $insertQuery = "INSERT INTO users(email, password) VALUES(?, ?)";
            $stmtI = mysqli_prepare($dbc, $insertQuery); 
            mysqli_stmt_bind_param($stmtI, "ss", $email, $password);
            mysqli_stmt_execute($stmtI);
            echo('Registration completed!');
    }

---

对

}

的捕捉很好，我没有注意到！请注意，我将if（$row！=null）{to if（$count==1）{替换为用户不共享电子邮件地址，这是理所当然的。在大多数系统中，无论如何都不应该出现这种情况，但我看到了异常。哦，是的{也是，但这不是问题所在，我以前也尝试过使用计数，但出于某种原因，它总是返回0，即使用户在数据库中。如果没有任何结果，则

while

甚至不会启动，因此无法转到else，它甚至无法到达if！这使得

if

多余同样，如果您在

中输入
，而
中有一个
$row
，即！=null。请不要将密码存储为纯文本。是的，就是这样！我以前也尝试过使用count方法，但出于某种原因，它总是返回0，即使用户在数据库中。请在mysqli_stmt_execute/*store result*/mysqli_stmt_sto之后添加此代码re_result（$stmt）；是的，store_result函数丢失了所有内容！谢谢！