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。