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    mysql_field_table

    (PHP 4, PHP 5)

    mysql_field_table — Obtiene el nombre de la tabla en la que está el campo especificado

    Advertencia

    Esta extensión fue declarada obsoleta en PHP 5.5.0 y eliminada en PHP 7.0.0. En su lugar debería utilzarse las extensiones MySQLi o PDO_MySQL. Véase también la guía MySQL: elegir una API. Las alternativas a esta función son:

    Descripción

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

    Devuelve el nombre de la tabla en la que está el campo especificado.

    Parámetros

    result

    El resultado resource que está siendo evaluado. Este resultado proviene de una llamada a mysql_query().

    field_offset

    El número del campo a buscar. El valor de field_offset comienza en 0. Si field_offset no existe, un error de nivel E_WARNING es emitido.

    Valores devueltos

    El nombre de la tabla en caso de éxito.

    Ejemplos

    Ejemplo #1 Un ejemplo de mysql_field_table()

    <?php

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

    // obtener el resultado desde la BD
    $resultado = mysql_query($consulta);

    // Lista el nombre de la tabla y luego el nombre del campo
    for ($i = 0; $i < mysql_num_fields($resultado); ++$i) {
    $tabla = mysql_field_table($resultado, $i);
    $campo = mysql_field_name($resultado, $i);

    echo     "$tabla: $campo\n";
    }

    ?>

    Notas

    Nota:

    Por razones de compatibilidad con versiones anteriores, los siguientes alias obsoletos podrían usarse: mysql_fieldtable()

    Ver también

    Found A Problem?

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    User Contributed Notes 6 notes

    up
    1
    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.
    up
    0
    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 $out_array;
    }
    up
    0
    cptnemo
    20 years ago
    When trying to find table names for a (My)SQL query containing 'tablename AS alias', mysql_field_table() only returns the alias as specified in the AS clause, and not the tablename.
    up
    -1
    spam at blondella dot de
    18 years ago
    <?php
    /*
    this function might help in the case described above :-)
    */
    function mysql_field_table_resolve_alias($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;
    }
    ?>
    up
    -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.
    up
    -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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