如何显示具有多行php的两列

Question

我采用了此示例中给出的代码：

https://www.w3schools.com/php/php_mysql_select.asp

这是来自Phpmyadmin的表格。

表格：帖子

-------------------------------------------------------
|  id  |   a   |   b   |   c   |   d   |   e   |  .... 
-------------------------------------------------------
|  1   |  AA1  |   BB2 |   CC3 |  DD4  |   EE5 |  .... 
-------------------------------------------------------

代码如下（PHP）：

$query = "SELECT * FROM `post` WHERE id ORDER BY id ASC"; 
$result = mysqli_query($db, $query);
while($row = mysqli_fetch_array($result))  { 
    echo $row["a"];  // Out: AA1
    echo $row["b"];  // Out: BB2
    echo $row["c"];  // Out: CC3
    echo $row["d"];  // Out: DD4
    echo $row["e"];  // Out: EE5
    .
    .
    .
}

[当我向数据库中添加一列（phpmyadmin）或导入信息（sql）时，它给出了一条错误消息，其消息如下：

Error:
SQL query: 
MySQL said: Documentation
#1118 - Row size too large (> 8126). Changing some columns to TEXT or BLOB may help. 
In current row format, BLOB prefix of 0 bytes is stored inline.

我用上面的代码更改了表，如下所示：（$ data）

----------------------------
|  id  |  name   |  value  |
----------------------------
|  1   |    a    |   AA1   |
----------------------------
|  2   |    b    |   BB2   |
----------------------------
|  3   |    c    |   CC3   |
----------------------------
|  4   |    d    |   DD4   |
----------------------------
|  5   |    e    |   EE5   |
----------------------------
.
.
.

现在修改的代码如下：

$query = "SELECT * FROM `post` (`name`,`value`) WHERE id ORDER BY id ASC"; 
$result = mysqli_query($db, $query);
$data = [];
while($row = mysqli_fetch_array($result))  { 
    echo $data[$row['value']]['a'];  // out : AA1
    echo $data[$row['value']]['b'];  // out : BB2
    echo $data[$row['value']]['c'];  // out : CC3
    echo $data[$row['value']]['d'];  // out : DD4
    echo $data[$row['value']]['e'];  // out : EE5
    .
    .
    .
}

当我刷新页面并显示错误消息时，其消息如下：

Notice: Undefined index: AA1
Notice: Undefined index: BB2
Notice: Undefined index: CC3
Notice: Undefined index: DD4
Notice: Undefined index: EE5

可能是什么原因，我在做什么错？

Answer 1

change：$query = "SELECT * FROM post(name,value ) WHERE id ORDER BY id ASC";

收件人：

$query = "SELECT * FROM `post` WHERE id = id  ORDER BY id ASC"; 
 $result = mysqli_query($db, $query);
 $data = array();
     while($row = mysqli_fetch_array($result))  {
       $data[] = $row;
     }

我打赌您不知道如何将参数传递给where子句，并且您可能不需要查询。因此，尝试这样：

$query = "SELECT * FROM `post` ORDER BY id ASC"; 
$result = mysqli_query($db, $query);
$data = array();
     while($row = mysqli_fetch_array($result))  {
       $data[] = $row;
     }