mysql_field_table

(PHP 4, PHP 5)

mysql_field_table — Возвращает название таблицы, которой принадлежит указанное поле

Внимание

Данный модуль устарел начиная с версии PHP 5.5.0, и удалён в PHP 7.0.0. Используйте вместо него MySQLi или PDO_MySQL. Смотрите также инструкцию MySQL: выбор API. Альтернативы для этой функции:

mysqli_fetch_field_direct() [table] или [orgtable]
PDOStatement::getColumnMeta() [table]

Описание

mysql_field_table(resource $result, int $field_offset): string

Возвращает название таблицы, которой принадлежит указанное поле.

Список параметров

result: Обрабатываемый результат запроса. Этот результат может быть получен с помощью функции mysql_query().
field_offset: Числовое смещение поля. field_offset начинается с 0. Если field_offset не существует, генерируется ошибка уровня E_WARNING.

Возвращаемые значения

Имя таблицы в случае успешного выполнения.

Примеры

Пример #1 Пример использования mysql_field_table()

<?php

$query = "SELECT account.*, country.* FROM account, country WHERE country.name = 'Portugal' AND account.country_id = country.id";

// получаем результат из базы данных
$result = mysql_query($query);

// выводит имя таблицы и имя поля
for ($i = 0; $i < mysql_num_fields($result); ++$i) {
    $table = mysql_field_table($result, $i);
    $field = mysql_field_name($result, $i);

    echo  "$table: $field\n";
}

?>

Примечания

Замечание:
Для обратной совместимости может быть использован следующий устаревший псевдоним: mysql_fieldtable()

Смотрите также

mysql_list_tables() - Извлекает список таблиц базы данных MySQL

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Примечания пользователей 6 notes

down

me at thomaskeller dot biz ¶

19 years ago

Beware that if you upgrade to MySQL 5 from any earlier version WITHOUT dumping and reloading your data (just by keeping the binary data in MyISAM table files), you might get weird output on the "table" value for mysql_fetch_field and in this function. Weird means that the table name is randomly set or not.

This behaviour seems to popup only if the SQL query contains a ORDER BY clause. A bug is already reported: 

http://bugs.mysql.com/bug.php?id=14915

To prevent the issue, dump and reload all participating tables in your query or do

CREATE TABLE tmp SELECT * FROM table;
DROP TABLE table;
ALTER TABLE tmp RENAME table;

on each one via commandline client.

down

jorge at rhst dot net ¶

17 years ago

The function below takes a function and returns the col->table mapping as an array.

For example:

$query = “SELECT a.id AS a_id, b.id b_id FROM atable AS a, btable b”

$cols  = queryAlias($query);

print_r($cols);

Returns:

Array
(
    [a] => atable
    [b] => btable
)

I can't promise it's perfect, but this function never hit production cause I ended up using mysqli methods instead.

Enjoy
-Jorge

  /**
   * Takes in a query and returns the alias->table mapping.
   *
   * @param string $query
   * @return array of alias mapping
   */
  
  function queryAlias ( $query ) {
    
    //Make it all lower, we ignore case
    $substr = strtolower($query);
    
    //Remove any subselects
    $substr = preg_replace ( ‘/\(.*\)/’, ”, $substr);
   
    //Remove any special charactors
    $substr = preg_replace ( ‘/[^a-zA-Z0-9_,]/’, ‘ ‘, $substr);
    
    //Remove any white space
    $substr = preg_replace(‘/\s\s+/’, ‘ ‘, $substr);
    
    //Get everything after FROM
    $substr = strtolower(substr($substr, strpos(strtolower($substr),‘ from ‘) + 6));
    
    //Rid of any extra commands
    $substr = preg_replace(
                Array(
                    ‘/ where .*+$/’,
                    ‘/ group by .*+$/’,
                    ‘/ limit .*+$/’ ,
                    ‘/ having .*+$/’ ,
                    ‘/ order by .*+$/’,
                    ‘/ into .*+$/’
                   ), ”, $substr);
    
    //Remove any JOIN modifiers
    $substr = preg_replace(
                Array(
                    ‘/ left /’,
                    ‘/ right /’,
                    ‘/ inner /’,
                    ‘/ cross /’,
                    ‘/ outer /’,
                    ‘/ natural /’,
                    ‘/ as /’
                   ), ‘ ‘, $substr);
    
    //Replace JOIN statements with commas
    $substr = preg_replace(Array(‘/ join /’, ‘/ straight_join /’), ‘,’, $substr);
    

    $out_array = Array();
    
    //Split by FROM statements
    $st_array = split (‘,’, $substr);
  
    foreach ($st_array as $col) {
      
      $col = preg_replace(Array(‘/ on .*+/’), ”, $col);
      
      $tmp_array = split(‘ ‘, trim($col));
      
      //Oh no, something is wrong, let’s just continue
      if (!isset($tmp_array[0]))
        continue;
        
      $first = $tmp_array[0];
      
      //If the “AS” is set, lets include that, if not, well, guess this table isn’t aliased.
      if (isset($tmp_array[1]))
        $second = $tmp_array[1];
        else  
        $second = $first;
        
      if (strlen($first))
       $out_array[$second] = $first;
      
    }
    
    return $out_array;
  }

down

cptnemo ¶

20 years ago

When trying to find table names for a (My)SQL query containing 'tablename AS alias', mysql_field_table() only returns the alias as specified in the AS clause, and not the tablename.

down

-1

spam at blondella dot de ¶

18 years ago

<?php
/*
this function might help in the case described above :-)
*/
function mysql_field_table_resolve_alias($inQuery,$inResult,$inFieldName) {
   $theNameOrAlias = mysql_field_table($inResult,$inFieldName);
   //check, if AS syntax is being used
   if(ereg(" AS ",$inQuery))  {
      //catch words in query
      $theWords = explode(" ",ereg_replace(",|\n"," ",$inQuery));
      //find the words preceding and following AS
      foreach($theWords as $theIndex => $theWord)  {
         if(trim($theWord) == "AS"
         && isset($theWords[$theIndex-1])
         && isset($theWords[$theIndex+1])
         && $theWords[$theIndex+1] == $theNameOrAlias
         ) {
            $theNameOrAlias = $theWords[$theIndex-1];
            break 1;
         }
      }
   }
   return $theNameOrAlias;
}
?>

down

-1

djafferian at sos dot ri dot gov ¶

10 years ago

This note may apply to anyone who might still be running MySQL 5.0.32 on Debian 4.0.  The mysql_field_table function may return an empty table name if the SELECT query involved uses GROUP BY or ORDER BY and references a view in the FROM clause.  This is caused by MySQL bug 28898, which was fixed in 5.0.46.  I encountered this when I noticed a difference between our production RSS feeds generated on a Debian 5.0.10 server running MySQL 5.0.51a, and the same feed generated on one of our test servers running MySQL 5.0.32 on Debian 4.0.

down

-3

mehdi dot haresi at gmail dot com ¶

15 years ago

For all of you having problems accessing duplicated field names in queries with their table alias i have implemented the following quick solution:

<?php
function mysql_fetch_alias_array($result)
{
    if (!($row = mysql_fetch_array($result)))
    {
        return null;
    }

    $assoc = Array();
    $rowCount = mysql_num_fields($result);
    
    for ($idx = 0; $idx < $rowCount; $idx++)
    {
        $table = mysql_field_table($result, $idx);
        $field = mysql_field_name($result, $idx);
        $assoc["$table.$field"] = $row[$idx];
    }
    
    return $assoc;
}
?>

Lets asume we have 2 tables student and contact each having fID as the index field and want to access both fID fields in php.

The usage of this function will be pretty similar to calling mysql_fetch_array:

<?php
$result = mysql_query("select * from student s inner join contact c on c.fID = s.frContactID");

while ($row = mysql_fetch_alias_array($result))
{
    echo "StudenID: {$row['s.fID']}, ContactID: {$row['c.fID']}";
}
?>

Voila, that's it :)

Please be aware that by using this function, you have to access all fields with their alias name (e.g. s.Name, s.Birhtday) even if they are not duplicated.

If you have questions, just send me a mail.

Best regards,
Mehdi Haresi
die-webdesigner.at

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