简体   繁体   English

使用PHP将任何给定的MySQL SELECT查询显示为HTML表

[英]Display any given MySQL SELECT query as a HTML table using PHP

Sample Query 1 - 示例查询1-

SELECT ID,NAME FROM USERS

Sample Query 2 - 样本查询2-

SELECT Orders.OrderID as ID, Customers.CustomerName as Name, Orders.OrderDate as Date
FROM Orders
INNER JOIN Customers
ON Orders.CustomerID=Customers.CustomerID; 

How can I load title headings to an array in PHP? 如何将标题标题加载到PHP中的数组? I mean ID,NAME in Query 1 and ID,Name,Date in Query 2 我的意思是查询1中的 ID,NAME查询2中的 ID,名称,日期

This is what I'm trying to do : 这就是我想要做的:

I'm creating a PHP function to make HTML table automatically from any given MySQL-Select query 我正在创建一个PHP函数,以通过任何给定的MySQL选择查询自动创建HTML表

This where I'm up to now 我到现在为止

function createTable($query) {
    $sql_link = Connect_MySQLi_DB();// Database Connection
    $sql_link->set_charset("utf8");
    $result = $sql_link->query($query);
    $headings = array('ID','Name','Date');//I need this array to create automatically
    echo '<table>';
    echo '<tr>';
    for ($x = 0; $x <= (count($headings) - 1); $x++) {
        echo '<th>'.$headings[$x].'</th>';
    }
    echo '<tr>';
    while ($row = $result->fetch_object()) {
        echo '<tr>';
        for ($x = 0; $x <= (count($headings) - 1); $x++) {
            echo '<td>' . $row->$headings[$x] . '</td>';
        }
        echo '<tr>';
    }
    echo '</table>';
}

So I need to create that $heading array automatically. 所以我需要自动创建该$ heading数组。 If I can do that, this function can display any MySQL-Select query as a HTML table 如果可以,此函数可以将任何MySQL-Select查询显示为HTML表

If you are using mysqli then fetch_fields seems to do the job. 如果您使用的是mysqli,则fetch_fields似乎可以完成工作。

EDIT: Example: 编辑:示例:

$res = $db->query('SELECT A.id,A.a,A.id + B.id AS idd FROM A natural join B');
var_dump($res->fetch_fields());

Returns information about columns A.id, Aa and about the idd column (both tables have other columns). 返回有关列A.id,Aa和idd列的信息(两个表都有其他列)。 I omitted some fields from the output in order to make it shorter. 为了简化起见,我从输出中省略了一些字段。

array(3) {
  [0]=>
  object(stdClass)#3 (13) {
    ["name"]=>
    string(2) "id"
    ["orgname"]=>
    string(2) "id"
    ["table"]=>
    string(1) "A"
    ["orgtable"]=>
    string(1) "A"
    // more fields here
  }
  [1]=>
  object(stdClass)#4 (13) {
    ["name"]=>
    string(1) "a"
    ["orgname"]=>
    string(1) "a"
    ["table"]=>
    string(1) "A"
    ["orgtable"]=>
    string(1) "A"
    // more fields here
  }
  [2]=>
  object(stdClass)#5 (13) {
    ["name"]=>
    string(3) "idd"
    ["orgname"]=>
    string(0) ""
    ["table"]=>
    string(0) ""
    ["orgtable"]=>
    string(0) ""
    // more fields here
  }
}

Finally this is how I made this function to work (Thanks Hynner for your excellent suggestion), 最后,这就是我使此功能起作用的方式(感谢海因纳的出色建议),

function createTable_from_sql_select_query($query) {
    $sql_link = Connect_MySQLi_DB();// Database connection
    $sql_link->set_charset("utf8");
    $result = $sql_link->query($query);

    // Adding Missing array_column Function for Old PHP Versions (<5.5)
    if (!function_exists('array_column')) {

        function array_column($array, $column) {
            $ret = array();
            foreach ($array as $row)
                $ret[] = $row[$column];
            return $ret;
        }

    }

    $headings = json_decode(json_encode($result->fetch_fields()), true);
    $headings = array_column($headings, 'name');

      $return = '<table>';
      $return .= '<thead><tr>';
      for ($x = 0; $x <= (count($headings) - 1); $x++) {
      $return .= '<th>' . ucwords(str_replace('_', ' ', (strtolower($headings[$x])))) . '</th>';
      }
      $return .= '</tr></thead><tbody>';
      while ($row = $result->fetch_object()) {
      $return .= '<tr>';
      for ($x = 0; $x <= (count($headings) - 1); $x++) {
      $return .= '<td>' . $row->$headings[$x] . '</td>';
      }
      $return .= '</tr>';
      }
      $return .= '</tbody></table>';

      return $return;
}

Now anyone can convert any SQL select query to a HTML table. 现在,任何人都可以将任何SQL选择查询转换为HTML表。

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM