在MySQL数据库，PHP中存储下拉选项 [英] store drop down options in mysql database, PHP

问题描述

当从下拉选择框中选择用户选择的选项时，我需要存储该用户选择的选项，并在以表单提交时将其存储在我的MYSQL数据库中.

I need to store the user selected option when chosen from my drop down select box, and when submitted in my form, store in my MYSQL database.

我已经可以很好地存储和显示信息了，我现在正尝试添加更多内容.我尝试使用相同的方法，但是我无法获取它来存储我想要的东西.

I already have information being stored and displayed fine, i am just trying to add more content now. I have tried using the same method, but i cant get it to store what i want.

select下拉列表在我的表单中看起来像这样:

The select drop down looks like this inside my form:

  <span> Difficulty: </span> 
   <select>
              <option type="text" name="easy" id="easy" value="easy">easy</option>
              <option type="text" name="comfortable" id="comfortable" value="comfortable">comfortable</option>
              <option type="text" name="hard" id="hard" value="hard">hard</option>
              <option type="text" name="veryhard" id="veryhard" value="veryhard">very hard</option>
              <option type="text" name="hardest" id="hardest" value="hardest">Hardest ride in the world!</option>
            </select>
  </label>

并且我正在尝试使用以下方式发送它:

and i am trying to send it using this:

    $difficulty = mysql_real_escape_string($_POST['difficulty']);

使用此查询:

$query = sprintf("INSERT INTO markers (difficulty) VALUES ('%s')", $difficulty);

当我的表设置为INT，20时，允许为空，它将在数据库中存储0. 当我将其设置为VARCHAR(20)时，允许为空，它将存储一个空白字段.

When my table is set up as INT, 20, allow null, it stores 0 in the database. and when i set it up VARCHAR, 20, allow null, it stores a blank field.

我希望它存储用户选择的任何值，即简单，困难，非常困难等.

i want it to store whichever value the user chooses, i.e easy, hard, very hard etc.

非常感谢

推荐答案

$_POST['difficulty']

将不保存值，因为未为SELECT赋予名称".试试

Would not hold a value as the SELECT has not been given a 'name'. Try

<select name="difficulty">

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