ConFoo: Call for paper is now Open

mysql_field_table

(PHP 4, PHP 5)

mysql_field_table指定したフィールドが含まれるテーブルの名前を取得する

警告

この拡張モジュールは PHP 5.5.0 で非推奨になりました。将来のバージョンで削除される予定です。 MySQLi あるいは PDO_MySQL を使うべきです。詳細な情報は MySQL: API の選択それに関連する FAQ を参照ください。 この関数の代替として、これらが使えます。

説明

string mysql_field_table ( resource $result , int $field_offset )

指定したフィールドが含まれるテーブルの名前を返します。

パラメータ

result

評価された結果 リソース。 この結果は、mysql_query() のコールにより得られたものです。

field_offset

数値フィールドオフセット。 field_offset0 から始まります。 field_offset が存在しない場合、 E_WARNING レベルのエラーが発行されます。

返り値

成功した場合にテーブルの名前を返します。

例1 mysql_field_table() の例

<?php

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

// 結果を DB から取得します
$result mysql_query($query);

// テーブル名とフィールド名を一覧表示します
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";
}

?>

注意

注意:

下位互換のために、次の非推奨別名を使用してもいいでしょう。 mysql_fieldtable()

参考

add a note add a note

User Contributed Notes 6 notes

up
0
djafferian at sos dot ri dot gov
2 months 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.
up
0
jorge at rhst dot net
7 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 $out_array;
  }
up
0
spam at blondella dot de
7 years ago
<?php
/*
this function might help in the case described above :-)
*/
function mysql_field_table_resolve_alias($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;
}
?>
up
0
me at thomaskeller dot biz
8 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.
up
0
cptnemo
10 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.
up
-1
mehdi dot haresi at gmail dot com
5 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
To Top