在我的數據庫表中插入變量的問題？

Question

<?php 

    $hostname="It is in correctly"; 
    $username="It is in correctly"; 
    $password="It is in correctly"; 
    $dbname="It is in correctly"; 

     $db_conx = mysqli_connect($hostname, $username, $password) OR DIE ("Unable to 
     connect to database! Please try again later."); 

     if(mysqli_connect_errno()){ 
     echo mysqli_connect_error(); 
     exit(); 
     } 

    $select = mysqli_select_db($db_conx,$dbname); 

    $firstname= $_POST["first_name"]; 
    $lastname= $_POST["last_name"]; 
    $email= $_POST["email"]; 
    $password= $_POST["password"]; 
    $gender= $_POST["gender"]; 

    mysqli_query($db_conx,"INSERT INTO users (firstname, lastname, email, password, gender) 
VALUES ($firstname, $lastname, $email, $password, $gender)"); 

mysqli_close($db_conx); 
?> 



     <form method="POST" name="signup" action="php/signup_script.php"> 
<label for="first name"></label> 
<input id="first name" name="first_name" placeholder="First Name" type="text" /> 
<label for="last_name"></label> 
<input id="last name" name="last_name" placeholder="Last Name" type="text" /> 
<br><br><label for="email"> 
</label><input id="email" name="email" placeholder="Email" type="email" /> 

    <label for="password"></label><input id="password" name="password" placeholder="Create Password" type="password" /> 

    <label for="male"><strong>Male</strong></label> 
<input id="male" value="Male" name="gender" type="radio" /> 
<label for="female"><strong>Female</strong> 
</la... <input id="female" value="Female" name="gender" type="radio" /> 

    <label for="submit">"I Agree To Terms And Conditions"</label> 
<input id="submit" value="Submit" type="submit" name="submit"/> 

    </form> 

Answer 1

您忘記將單引號中的插入查詢中的php變量括起來，將其更正。

mysqli_query($db_conx,"INSERT INTO users (firstname, lastname, email, password, gender) 
VALUES ('$firstname', '$lastname', '$email', '$password', '$gender')");