如何在php中的mysql查询中获取有用的错误消息？

Question

HTML代码

<form class="col s6 " method="post" enctype="multipart/form-data">

   <div class="input-field col s12">

      <input id="last_name" type="text" name="name" class="validate">

      <label for="last_name">Certificate Name</label>

   </div>

   <button type="submit" id="btnSubmit" name="btnSubmit" class="btn btn-default" style="margin-top:20px;">ADD</button>

</form>

PHP代码

<?php

 include('footer.php'); 

include('conn.php');

if(isset($_POST['btnSubmit']))

{

    $name = mysqli_real_escape_string($conn,$_POST["name"]);

    $sql = "INSERT INTO `isodetail`(`title`) VALUES ('$name')";

    $run = mysqli_query($conn, $sql);

    if($run)

    {

        echo "<script>alert('Certi Added Successfully')</script>";

        echo "<script>window.open('isocerti.php','_self')</script>";

    }

    else

    {

        echo "<script>alert('Something Error!..please try Again..')</script>";

    }

}

 ?>

这会显示一条警告消息Something Error!..please try Again.. ，mysqli_query说明是否有错误？

Answer 1

所以您有正确的错误，现在您可以开始调试它。我的第一个想法是：INSERT isodetail(title, name) VALUES ('{$name}', '')但这只是对数据库结构知识的猜测。