mysqli_query 不向表中插入数据答案

【问题标题】：mysqli_query doesn't insert data into tablemysqli_query 不向表中插入数据
【发布时间】：2018-10-20 15:16:57
【问题描述】：

有人能找到问题吗？它没有给出任何错误，但新行没有出现在数据库中，我不知道问题出在哪里。

  if (isset( $_REQUEST['signupnbtn'] ) ) {
    $age = mysqli_real_escape_string($con,$_REQUEST['ageinput']);
    $discord = mysqli_real_escape_string($con,$_REQUEST['discordinput']);
    $email = mysqli_real_escape_string($con,$_REQUEST['emailinput']);
    $tmp = mysqli_real_escape_string($con,$_REQUEST['tmpinput']);
    $steam = mysqli_real_escape_string($con,$_REQUEST['steaminput']);
    $datum = date("d-m-Y");
    $errorcode = 0;

    $q = "INSERT INTO `admissions` (age, discord, email, tmp, steam, datum)
    VALUES ('$age', '$discord', '$email', '$tmp', '$steam', '$datum')";
    $query2 = "SELECT email FROM `admissions` WHERE email='$email'";
    $sql = mysqli_query($con,$query2);
    $countrows = mysqli_num_rows($sql);
    if($countrows >= 1){
      $errorcode = 1;
    }else {
      $result = mysqli_query($con,$q);
    }
      if ($result) {
        $errorcode = 4;
      }

  }

【问题讨论】：

$countrows 的输出是什么？mysqli_query($con,$q) 行是否被执行？如果是，$result 的输出是什么？你必须调试这个
另外，age 列的类型是什么？您尝试将其视为字符串，但这对于名为 age 的列来说听起来很可疑。
请多描述一下您的问题。它会插入一个空行吗？您确定您的输入名称中没有输入错误吗？
年龄列是一个字符串
它不会向数据库中插入任何内容。

标签： php mysql sql mysqli insert-into

【解决方案1】：

$q = "INSERT INTO admissions (age, discord, email, tmp, steam, datum) VALUES ('$age', '$discord', '$email', '$tmp', '$steam', '$datum')";

【讨论】：

【解决方案2】：

问题已解决，日期应为“Y-m-d”格式，而不是“d-m-Y”，如下所示：

if (isset( $_REQUEST['signupnbtn'] ) ) {
    $age = mysqli_real_escape_string($con,$_REQUEST['ageinput']);
    $discord = mysqli_real_escape_string($con,$_REQUEST['discordinput']);
    $email = mysqli_real_escape_string($con,$_REQUEST['emailinput']);
    $tmp = mysqli_real_escape_string($con,$_REQUEST['tmpinput']);
    $steam = mysqli_real_escape_string($con,$_REQUEST['steaminput']);
    $datum = date("Y-m-d");
    $errorcode = 0;
    $q = "INSERT INTO `admissions` (age, discord, email, tmp, steam, datum)
    VALUES ('$age', '$discord', '$email', '$tmp', '$steam', '$datum')";
    $query2 = "SELECT email FROM `admissions` WHERE email='$email'";
    $sql = mysqli_query($con,$query2);
    $countrows = mysqli_num_rows($sql);
    if($countrows >= 1){
      $errorcode = 1;
    }else {
      $result = mysqli_query($con,$q);
    }
      if ($result) {
        $errorcode = 4;
      }
  }

【讨论】：