php来显示表格;第一列带有mysql列注释,第二列转置了行结果


php to display table; column one with mysql column comments and column two transposed row results

为可能糟糕的标题道歉-最好直观地显示我的目标:

当用户访问SomerL.com/test.php时,我想在页面上显示这样的内容吗?id=101

| Questions      | Answers       |
----------------------------------
| 1(a) First Name| Pedro         |
----------------------------------
| 1(b) Surname   | Millers       |
----------------------------------
| 2(a) Weight    | 150lbs        |
----------------------------------

这将通过查询mysql数据库来实现。第一:

SELECT * FROM Questionnaire WHERE idQuestionnaire=101;

哪个返回:

| idQuestionnaire | FirstName    | Surname    | Weight    |
-----------------------------------------------------------
| 101             | Pedro        | Millers    | 150lbs    |

在我的表设置中,我设置了列注释。即"名字"一栏的注释为"1(a)名字"。要检索所有这些评论,我可以做另一个查询:

SELECT COLUMN_NAMES FROM INFORMATION_SCHEMA.COLUMNS WHERE TABLE_NAME='Questionnaire';

哪个返回:

| COLUMN_COMMENTS |
-------------------
| 1(a) First Name |
-------------------
| 1(b) Surname    |
-------------------
| 2(a) Weight     |
-------------------

我之所以想用数组/循环而不是硬代码来自动化代码,是因为我使用的实际表有400多个字段,字段可能会被修改或添加,所以当这种情况发生时,我不想一直更改php代码。这似乎是一项非常简单的任务,但我一生都找不到相关的文档化解决方案。

我建议您在结果集中使用mysqli fetch_fields()函数,以获取有关结果集中包含的列的信息,而不是在information_SCHEMA中查询视图。

fetch_fields()将适用于仅引用表中列的子集的查询,或返回表达式或返回多个表中的列的查询。

http://php.net/manual/en/mysqli.quickstart.metadata.php

试试这个:

(SELECT 
 "1(a) First Name" as Questions,
 Firstname as Answers
FROM Questionnaire WHERE idQuestionnaire=101)
 union
(SELECT 
 "1(b) Surname" as Questions,
  Surname as Answers
FROM Questionnaire WHERE idQuestionnaire=101)
 union
(SELECT 
 "2(a) Weight" as Questions,
 Weight as Answers
FROM Questionnaire WHERE idQuestionnaire=101);

最后是解决方案。Spencer7593让我上路了,但最终需要的代码如下:

$con=mysqli_connect("localhost","user","password","test");
if (mysqli_connect_errno($con)) {echo "MySQL conn. err:".mysqli_connect_error();}
 $sql = "SELECT column_comment,column_name FROM information_schema.columns  
  WHERE table_name = 'mytablename';";
 $query = mysqli_query($con,$sql) or die(mysql_error());
 $columnArray = array();
 while(($result = mysqli_fetch_array($query, MYSQL_ASSOC))){
 if($result['column_comment'])
  {
  $CurCol = $result['column_comment'];
  }
 else
  {
  $CurCol = "No text stored";
  } 
    $CurName = $result['column_name'];
    $columnArray[$CurName]=$CurCol;
 }
 //echo $columnArray['FirstName']; //test example to return "1(a) First Name"

因此,我循环使用fetch_fields来获得相应的列注释,然后循环使用answers查询的结果来在下一个表列中显示每个答案。