douxun1407 2015-06-01 18:23
浏览 47
已采纳

mysqli_query没有按预期工作

I have the php:

$con=mysqli_connect("127.0.0.1","foo","bar","quaz");
if($con){
    $sql = "INSERT INTO `virtual_users` (`domain_id`, `password` , `email`) 
    VALUES ('2', ENCRYPT('".$password."', CONCAT('\$6\$', SUBSTRING(SHA(RAND()), -16))), '".$user."@".$domain."');";

    $query = mysqli_query($con, $sql);

    if(!$query){
        echo $sql;
    }else{
        echo "Success adding new email user!";
    }
}

For some reason when I run this query it always returns $sql. This means that the connection is fine but the query is not.

When I then run the the echo of $sql directly on the mysql database it works perfectly!! I have no idea what is going wrong! Any ideas?

  • 写回答

1条回答 默认 最新

  • dsbo44836129 2015-06-01 18:26
    关注

    That's because you should be outputting the reason for the failure, not the query that caused the failure:

    $result = mysqli_query($con, $sql) or die(mysqli_error($con));
                                      ^^^^^^^^^^^^^^^^^^^^^^^^^^^
    

    There's exactly ONE way for a query to succeed, and a near infinite number of ways for it to fail. Just having "valid" sql means nothing.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

报告相同问题?

悬赏问题

  • ¥15 用hfss做微带贴片阵列天线的时候分析设置有问题
  • ¥50 我撰写的python爬虫爬不了 要爬的网址有反爬机制
  • ¥15 Centos / PETSc / PETGEM
  • ¥15 centos7.9 IPv6端口telnet和端口监控问题
  • ¥120 计算机网络的新校区组网设计
  • ¥20 完全没有学习过GAN,看了CSDN的一篇文章,里面有代码但是完全不知道如何操作
  • ¥15 使用ue5插件narrative时如何切换关卡也保存叙事任务记录
  • ¥20 海浪数据 南海地区海况数据,波浪数据
  • ¥20 软件测试决策法疑问求解答
  • ¥15 win11 23H2删除推荐的项目,支持注册表等