mysql_field_seek
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
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
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_offsetは 0 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
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()
add a note
User Contributed Notes
mysql_field_table
mehdi dot haresi at gmail dot com
16-Jul-2009 08:50
jorge at rhst dot net
29-Jul-2007 08:57
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 $out_array;
}
spam at blondella dot de
02-Oct-2006 10:09
<?php
/*
this function might help in the case described above :-)
*/
function mysql_field_table_resolve_alias($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;
}
?>
me at thomaskeller dot biz
23-Nov-2005 06:15
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.
cptnemo
14-Aug-2004 01:18
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.