使用php插入mysql数据库时出错

Question

尝试使用函数插入db时出现错误！ 我首先尝试从我的“ included connect.php”中传递$connection变量，但是它让我出错，无法传递对象或类似的东西。 现在，我在函数中重复了连接部分，但仍然无法添加用户！

现在不显示错误，而是跳转到else ...例如

 <?php
  function enterToDb($user, $pass, $email, $gender){
    // take those and insert into users

    $dbname = "mydb";
    $connection = mysqli_connect("localhost", "root", "root", $dbname);
    //query
    $query= "insert into users";
    $query.=" (`username`, `password`, `email`)";
    $query.= " values ('".$user."','".$pass."','".$email."','".$gender."')";
    if(mysqli_query($connection, $query)){
      ?>
          <div class="alert alert-success" role="alert">Congratulations! you're a member of E-Bridge community!
          </br>

          </div>

        <?php
    }else{
      ?>
      <div class="alert alert-danger" role="alert">Error signing you up! Please try again! 
      </div>
      <?php
    }



  }

如上所示是该函数，它进入该函数但直接跳转到else

这是我提到的功能

<?php
 echo"in sign up pages";
    $yes=0;
    if(isset($_POST['submit'])){
      // get all the info from the form
      $username= $_POST['username'];
      $password= $_POST['password'];
      $email= $_POST['email'];
      $gender=$_POST["radioGroup"];
      // check if the user already exists in the db
      $query= "select * from users where ";
      $query.= "username =\"".$username."\"";
      $query.= " or email=\"". $email."\"";
      //echo $query;
      $select= mysqli_query($connection, $query);
      if(mysqli_num_rows($select)>=1){ // checking if rows exist
        // if so
        ?>
          <div class="alert alert-danger" role="alert">Credential already exists, please try another name or email.</div>
        <?php

        display();



      }else{
         //echo $username." ".$password." ".$email." ".$gender; 
        enterToDb($username, $password, $email, $gender);

      //display();
      }


      // if not there > error
      // else = exitst >> start session and direct to topics or profile ;)



    }else{


      display();

    }

 ?>

Answer 1

$query.=" (`username`, `password`, `email`)";
$query.= " values ('".$user."','".$pass."','".$email."','".$gender."')";

3个字段名称，将插入4个值。 我认为您会发现需要在字段中添加性别。