MySQL查询不起作用 - 没有错误

For some reason my query isn't working, I'll post the code and output of $_POST.

由于某种原因我的查询不起作用,我将发布$ _POST的代码和输出。

I will also add a screenshot for the structure of the database.

我还将添加数据库结构的屏幕截图。

$_POST: MySQL查询不起作用 - 没有错误

Code:

    ini_set('display_errors',1); 
    error_reporting(E_ALL);
    echo "<pre>";
    var_dump($_POST);
    echo "</pre>";
    $mysqli = new mysqli("localhost", "lunar_casino", "******", "lunar_casino");
    if(isset($_POST['submit'])){
      $error = array();
      if(empty($error)){
        $bonus = $_POST['bonus'];
        $deposit = $_POST['deposit'];
        $offers = $_POST['offers'];
        $link = $_POST['link'];
        $name = $_POST['logo'];
        $q = $mysqli->query("INSERT INTO `lunar_casino`.`casino` VALUES(NULL, '$bonus', '$deposit', '$offers', '$link', '$logo', '$name', NULL)");
        if(!$q){
          echo "<font color='red'><b>There has been an error with our database! Please contact the website administrator!</b></font><br /><br />";
        } else {
          echo "<font color='green'><b>You have successfully added the casino!</b></font><br /><br />";
        }
      } else {
        echo "<font color='red'><b>There were ".count($error)." errors in your form:</b></font><br />";
        foreach($error as $err){
          echo "<b>".$err."</b><br />";
        }
        echo "<br />";
      }
    }

Structure of Database: MySQL查询不起作用 - 没有错误

数据库结构:

If you need more information just let me know!

如果您需要更多信息,请告诉我们!

By the way, the way I know the error is in the query is because I checked if(!$q) and made it display an error message if the query can't be done, and it displays the error message on the page.

顺便说一句,我在查询中知道错误的方式是因为我检查了是否(!$ q)并且如果查询无法完成则显示错误消息,并且它在页面上显示错误消息。

Any ideas why its not working? Also I left out date from the query because I don't know how to add the current date:time into the query.

任何想法为什么它不工作?此外,我从查询中遗漏了日期,因为我不知道如何将当前日期:时间添加到查询中。

If anyone could help with either of these issues please let me know! :)

如果有人可以帮助解决这些问题,请告诉我们! :)

2 个解决方案

#1

I think the problem is that you set NULL the date column. Try NOW() instead of NULL:

我认为问题是你设置NULL日期列。尝试NOW()而不是NULL:

$q = $mysqli->query("INSERT INTO `lunar_casino`.`casino` VALUES(NULL, '$bonus', '$deposit', '$offers', '$link', '$logo', '$name', NOW())");

#2

Your error here is isn't very clear. Here are some debugging options you can try:

你在这里的错误不是很清楚。以下是您可以尝试的一些调试选项:

Name the fields

命名字段

 $q = $mysqli->query("INSERT INTO `lunar_casino`.`casino` VALUES(NULL, '$bonus', '$deposit', '$offers', '$link', '$logo', '$name', NULL)");

 $q = $mysqli->query("INSERT INTO `lunar_casino`.`casino`(field1, field2...) VALUES(NULL, '$bonus', '$deposit', '$offers', '$link', '$logo', '$name', NULL)");

Save your query to a string and print it out. (Gordon Linoff)

将查询保存为字符串并将其打印出来。 (戈登林诺夫)

echo "INSERT INTO `lunar_casino`.`casino` VALUES(NULL, '$bonus', '$deposit', '$offers', '$link', '$logo', '$name', NULL)"

before the query. This can show you what PHP is making for MySQL to feed on.

在查询之前。这可以向您展示PHP为MySQL提供的功能。

Check for errors

检查错误

if ($mysqli->connect_errno) { //check connection error
    printf("Connect failed: %s\n", $mysqli->connect_error); //check query error
    exit();
}

#1