从我的sql中选择要在PHP中显示的列

时间：2010-11-01 18:54:57

标签： php mysql

我有一个基本的select *来自我的页面上显示我的表格。我希望该选项只显示某人可从页面中选择的列。这可能吗？

它是一个包含16列的基本表。不是每次都有人想要显示每一列。假设他们只想要显示的列a-e。什么是让他们选择这些列并让它构建查询的最佳方式。

要么过滤掉表头中的dropwons，要么过滤掉。就像你在Excel中添加了一个过滤器一样。

5 个答案:

答案 0 :(得分：3)

另一种方法是将你的'数据表'输出到由JAVASCRIPT修改的东西，使用像jQuery这样的库，你可以使用一个'table plugin'来隐藏/显示列，并重新排序为你请。否则，您将通过后端SQL查询更改进行此排序。

JAVASCRIPT方法让用户可以通过1种格式大量提取隐藏/显示/排序数据。

以下是一些实现此JS方法的插件示例：

享受！

答案 1 :(得分：1)

在不了解您的代码或数据库的任何内容的情况下，一个简单的方法是：

让用户指定列（例如，从列表中选择它们），然后在代码中构造SQL语句，将所选列连接到SELECT列表而不是指定*，然后执行SQL语句。

答案 2 :(得分：0)

这完全取决于你对show的意思。

您可以在select中添加列而不是*，这样只会在返回的列表中提供所选列。

如果这不是您的意思，您需要添加更多详细信息，如问题评论中所述。

答案 3 :(得分：0)

你去（根据需要进行调整）： 

function show_column_selectors($available, $chosen, $labels){
 echo <<<HTML
<form method="POST" action="#">
HTML;
 foreach($available as $column){
  $id = 'show_'.$column;
  $checked = in_array($column, $chosen) ? 'checked="checked"' : '';

  // See how I write the input elements with the same name,
  //                                        ending said name with []?
  // This will make PHP join all their values in an array.
  //                                      (see main() function below)
  echo <<<HTML
<label for="{$id}">
<input id="{$id}" name="columns_to_show[]"
       type="checkbox" {$checked} value="{$column}">
{$labels[$column]}
</label>
HTML;
 }
 echo <<<HTML
 <input type="submit" value="Go!">
</form>
HTML;
}

function do_query($cols){
 // This is boilerplate I hacked just to test this out. Ignore it
 $mysqli = mysqli_connect('server', 'username', 'password', 'database');
 if ($mysqli->connect_error) {
  die('Connect Error (' . $mysqli->connect_errno . ') '.$mysqli->connect_error);
 }

 // Here you build your query from the array $cols,
 //    which contains the columns chosen by the user.
 // See below, in the main() function, how it is obtained.
 $query = "SELECT `".implode("`, `", $cols)."` FROM `table`;";

 // More boilerplate follows
 $result = $mysqli->query($query, MYSQLI_USE_RESULT);
 $rows = array();
 if($result){
  while(($row = $result->fetch_assoc())){
   $rows []= $row;
  }
 }
 mysqli_free_result($result); 
 $mysqli->close();

 return $rows;
}

function show_result($columns, $rows){
 echo "<table><tr>\n";
 foreach($columns as $column){
  echo "<th>{$column}</th>\n";
 }
 echo "<tr>\n";
 foreach($rows as $row){
  echo "<tr>\n";
  foreach($columns as $column){
   echo "<td>{$row[$column]}</td>";
  }
  echo "</tr>\n";
 }
 echo "</table>";
}



function main(){
 // This are the column names as seen by the database
 $AVAILABLE_COLUMNS = array(
  'col1', 'col2', 'col3', 'col4'
 );

 // And these map between the above
    //     and what you want the user to see
 $LABELS = array(
  'col1' => 'Column 1',
  'col2' => 'Column 2',
  'col3' => 'Column 3',
  'col4' => 'Column 4',
 );

 // This is the important part, where you obtain the array
 //                             with user's chosen columns.
 // NOTE: same name as the input elements, but with no []
 if(empty($_POST['columns_to_show'])){
  $chosen_columns = $AVAILABLE_COLUMNS;
 }
 else{
  $chosen_columns = array_intersect($_POST['columns_to_show'], $AVAILABLE_COLUMNS);
 }

 show_column_selectors($AVAILABLE_COLUMNS, $chosen_columns, $LABELS);
 $rows = do_query($chosen_columns);
 show_result($chosen_columns, $rows);
}

main();
?><html>
<head><title>Test</title></head>
<body>
</body>
</html>
function show_column_selectors($available, $chosen, $labels){ echo <<<HTML <form method="POST" action="#"> HTML; foreach($available as $column){ $id = 'show_'.$column; $checked = in_array($column, $chosen) ? 'checked="checked"' : ''; // See how I write the input elements with the same name, // ending said name with []? // This will make PHP join all their values in an array. // (see main() function below) echo <<<HTML <label for="{$id}"> <input id="{$id}" name="columns_to_show[]" type="checkbox" {$checked} value="{$column}"> {$labels[$column]} </label> HTML; } echo <<<HTML <input type="submit" value="Go!"> </form> HTML; } function do_query($cols){ // This is boilerplate I hacked just to test this out. Ignore it $mysqli = mysqli_connect('server', 'username', 'password', 'database'); if ($mysqli->connect_error) { die('Connect Error (' . $mysqli->connect_errno . ') '.$mysqli->connect_error); } // Here you build your query from the array $cols, // which contains the columns chosen by the user. // See below, in the main() function, how it is obtained. $query = "SELECT `".implode("`, `", $cols)."` FROM `table`;"; // More boilerplate follows $result = $mysqli->query($query, MYSQLI_USE_RESULT); $rows = array(); if($result){ while(($row = $result->fetch_assoc())){ $rows []= $row; } } mysqli_free_result($result); $mysqli->close(); return $rows; } function show_result($columns, $rows){ echo "<table><tr>\n"; foreach($columns as $column){ echo "<th>{$column}</th>\n"; } echo "<tr>\n"; foreach($rows as $row){ echo "<tr>\n"; foreach($columns as $column){ echo "<td>{$row[$column]}</td>"; } echo "</tr>\n"; } echo "</table>"; } function main(){ // This are the column names as seen by the database $AVAILABLE_COLUMNS = array( 'col1', 'col2', 'col3', 'col4' ); // And these map between the above // and what you want the user to see $LABELS = array( 'col1' => 'Column 1', 'col2' => 'Column 2', 'col3' => 'Column 3', 'col4' => 'Column 4', ); // This is the important part, where you obtain the array // with user's chosen columns. // NOTE: same name as the input elements, but with no [] if(empty($_POST['columns_to_show'])){ $chosen_columns = $AVAILABLE_COLUMNS; } else{ $chosen_columns = array_intersect($_POST['columns_to_show'], $AVAILABLE_COLUMNS); } show_column_selectors($AVAILABLE_COLUMNS, $chosen_columns, $LABELS); $rows = do_query($chosen_columns); show_result($chosen_columns, $rows); } main(); ?><html> <head><title>Test</title></head> <body> </body> </html>

答案 4 :(得分：0)

我强烈建议您使用YUI Datatable。我经常将它用于类似的应用程序。您可以查询服务器以明确显示哪些列，或者仅返回整个集合（如果不总是需要数据，则效率低下。）

相关问题
最新问题