mysql_field_name

(PHP 4, PHP 5)

mysql_field_name — Liefert den Namen des angegebenen Feldes aus einem Ergebnis

Warnung

Diese Erweiterung ist seit PHP 5.5.0 als veraltet markiert und wurde in PHP 7.0.0 entfernt. Verwenden Sie stattdessen die Erweiterungen MySQLi oder PDO_MySQL. Weitere Informationen bietet der Ratgeber MySQL: Auswahl einer API. Alternativen zu dieser Funktion umfassen:

mysqli_fetch_field_direct() [name] oder [orgname]
PDOStatement::getColumnMeta() [name]

Beschreibung

mysql_field_name(resource $result, int $field_offset): string|false

mysql_field_name() gibt den Namen des Feldes zurück, das dem angegeben Feldindex entspricht.

Parameter-Liste

result: Das Ergebnis Ressource, das ausgewertet wird. Dieses Ergebnis kommt von einem Aufruf von mysql_query().
field_offset: Der numerische Offset des Feldes. Der field_offset beginnt bei 0. Falls field_offset nicht existiert, wird eine Warnung der Stufe E_WARNING erzeugt.

Rückgabewerte

Gibt den Namen des Feldes zurück, das dem übergebenen Index entspricht. Bei einem Fehler wird false zurückgegeben.

Beispiele

Beispiel #1 mysql_field_name()-Beispiel

<?php
/* Die Benutzertabelle besteht aus 3 Feldern
 *   user_id
 *   username
 *   password.
 */
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
    die('Keine Verbindung zum MySQL-Server: ' . mysql_error());
}
$dbname = 'mydb';
$db_selected = mysql_select_db($dbname, $link);
if (!$db_selected) {
    die("Konnte $dbname nicht auswählen: " . mysql_error());
}
$res = mysql_query('select * from users', $link);

echo mysql_field_name($res, 0) . "\n";
echo mysql_field_name($res, 2);
?>

Das oben gezeigte Beispiel erzeugt folgende Ausgabe:

user_id
password

Anmerkungen

Hinweis: Bei den Spaltennamen, die von dieser Funktion zurückgegeben werden, wird zwischen Groß- und Kleinschreibung unterschieden.

Hinweis:
Für die Abwärtskompatibiliät kann der folgende veraltete Alias verwendet werden: mysql_fieldname()

Siehe auch

mysql_field_type() - Liefert den Typ des spezifizierten Feldes aus einem Ergebnis
mysql_field_len() - Liefert die Länge des angegebenen Feldes

add a note

User Contributed Notes 12 notes

down

anonymous at site dot com ¶

16 years ago


This function is slightly stupid to be honest, why not just make an array of field names... You could consolidate the two of these functions that way and it makes it a lot easier to list them when your script is dynamic.

<?php

    function mysql_field_array( $query ) {
    
        $field = mysql_num_fields( $query );
    
        for ( $i = 0; $i < $field; $i++ ) {
        
            $names[] = mysql_field_name( $query, $i );
        
        }
        
        return $names;
    
    }
    
    // Examples of use
    
    $fields = mysql_field_array( $query );
    
    // Show name of column 3
    
    echo $fields[3];
    
    // Show them all
    
    echo implode( ', ', $fields[3] );
    
     // Count them - easy equivelant to 'mysql_num_fields'
    
    echo count( $fields );

?>

down

janezr at jcn dot si ¶

19 years ago


This is another variant of displaying all columns of a query result, but with a simplified while loop.

<?
$query="select * from user";
$result=mysql_query($query);
$numfields = mysql_num_fields($result);

echo "<table>\n<tr>";

for ($i=0; $i < $numfields; $i++) // Header
{ echo '<th>'.mysql_field_name($result, $i).'</th>'; }

echo "</tr>\n";

while ($row = mysql_fetch_row($result)) // Data
{ echo '<tr><td>'.implode($row,'</td><td>')."</td></tr>\n"; }

echo "</table>\n"
?>

down

matt at iwdt dot net ¶

23 years ago


here's one way to print out a row of <th> tags from a table

NOTE: i didn't test this



$result = mysql_query("select * from table");



for ($i = 0; $i < mysql_num_fields($result); $i++) {

    print "<th>".mysql_field_name($result, $i)."</th>\n";

}



post a comment if there's an error

down

-1

matteo.cisilino[no_more]cisilino[spm]com ¶

17 years ago


james, why make so difficult when it's very simple :\

$numberfields = mysql_num_fields($res_gb);

   for ($i=0; $i<$numberfields ; $i++ ) {
       $var = mysql_field_name($res_gb, $i);
       $row_title .= $var;
   }

echo $row_title;

down

-2

jimharris at blueyonder dot co dot uk ¶

19 years ago


The code in the last comment has an obvious mistake in the for loop expression.  The correct expression in the for-loop is $x<$y rather than $x<=$y...

$result = mysql_query($sql,$conn) or die(mysql_error());
$rowcount=mysql_num_rows($result);
$y=mysql_num_fields($result);
for ($x=0; $x<$y; $x++) {
   echo = mysql_field_name($result, $x).'<br>';
}

down

-2

jason dot chambes at phishie dot net ¶

21 years ago


<?
/*
    By simply calling the searchtable() function
    with these variables it will serach the desired
    database and procude a table for each field that
    there is a match.
*/

function searchtable($host,$user,$pass,$database,$tablename,$userquery)
{
    $link   = mysql_connect($host, $user, $pass) or die("Could not connect: " . mysql_error());
    $db     = mysql_select_db($database, $link) or die(mysql_error());
    $fields = mysql_list_fields($database, $tablename, $link);
    $cols   = mysql_num_fields($fields);

    for ($i = 1; $i < $cols; $i++) {
        $allfields[] = mysql_field_name($fields, $i);
    }
    foreach ($allfields as $myfield) {
        $result = mysql_query("SELECT * FROM $tablename WHERE $myfield like '%$userquery%' ");
        if (mysql_num_rows($result) > 0){
            echo "<h3>search <i>$database</i> for <i>$userquery</i>, found match(es) in <i>$myfield</i>: </h3>\n";
            echo "<table border=1 align=\"center\">\n\t<tr>\n";
            for ($i = 1; $i < $cols; $i++) {
                echo "\t\t<th";
                if ($myfield == mysql_field_name($fields, $i)){
                    echo " bgcolor=\"orange\"> ";
                } else {
                    echo ">";
                }
                echo mysql_field_name($fields, $i) . "</th>\n";
            }
            echo "\t</tr>\n";
            $myrow = mysql_fetch_array($result);
            do {
                echo "\t<tr>\n";
                for ($i = 1; $i < $cols; $i++){
                    echo "\t\t<td> $myrow[$i] &nbsp;</td>\n";
                }
                echo "\t</tr>\n";
            } while ($myrow = mysql_fetch_array($result));
            echo "</table>\n";
        }
    }
}

searchtable($host,$user,$pass,$database,$tablename,$userquery);
?>

down

-3

tiptonentserv at gmail dot com ¶

13 years ago


simple sql to xml converter works with any sql query and returns the name of the table as the root element "row" as each row element and the names of the columns are your children of row. fully tested.



<?php

function sqlToXml($host,$user,$pass,$database,$tablename,$query){



    $link   = mysql_connect($host, $user, $pass) or die("Could not connect: " . mysql_error());

    $db     = mysql_select_db($database, $link) or die(mysql_error());

    

    $result = mysql_query($query);

    if(!$result){ die('Invalid query: '.mysql_error()); }

    

    $numOfCols = mysql_num_fields($result);

    $numOfRows = mysql_num_rows($result);

    

    $info = mysql_fetch_assoc($result);

    

    //send headers

    header('Content-type: text/xml');

    header('Pragma: public');        

    header('Cache-control: private');

    header('Expires: -1');

    $xml = '<?xml version="1.0" encoding="utf-8"?>';

    $xml.= "<{$tablename}>";

    

    if($numOfRows > 0){

        do {

            $xml.= "<row>";

            foreach($info as $column => $value) {

                $xml.= "<{$column}>{$value}</{$column}>";

            }

            $xml.= "</row>";

        }

        while ($info = mysql_fetch_array($result));

    }

    $xml.= "</{$tablename}>";

    

    mysql_free_result($result);    

    return $xml;

    

}

?>

down

-3

blackjackdevel at gmail dot com ¶

16 years ago


Strangely using an aproach like this:
$res=mysql_query("SELECT * FROM `orders`",$conec) or die (mysql_error());

$fields = mysql_num_fields($res);
$out="";
for ($i = 0; $i < $fields; $i++) {
    $fname=mysql_field_name($res, $i);

}

 Outputted the E_Warning:
Warning: mysql_field_name() [function.mysql-field-name]: Field N is invalid for MySQL result index 

 With a lot of different number at N. But expliciting all fields instead of *. Didn't outputted the error.

 It maybe a caracteristic of this mysql database(it is from a open source application) because i never saw this in my own databases. Anyway hope this help if someone face the same strange situation

down

-4

clinnenb at hotmail dot com ¶

19 years ago


The following will create a PHP array, $array, containing the MySQL query results with array indexes of the same name as field names returned by the MySQL query.

while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
    $i=0;
    foreach ($line as $col_value) {
        $field=mysql_field_name($result,$i);
        $array[$field] = $col_value;
        $i++;
    }
}

down

-5

bags ¶

14 years ago


When using aliases, it appears impossible to discover the name of the underlying column.
select `ID` as `anAlias` from `aTable` returns 'anAlias' as the mysql_field_name(). I have tried all the mysql_field_xxx() functions and none return the real column name.

down

-5

colin dot truran at shiftf7 dot com ¶

19 years ago


T simply itterate through all the field names on a result set try using this.

$result = mysql_query($sql,$conn) or die(mysql_error());
$rowcount=mysql_num_rows($result);
$y=mysql_num_fields($result);
for ($x=0; $x<=$y; $x++) {
    echo = mysql_field_name($result, $x).'<br>';
}

This is useful if you have a result set that joins several tables dynamicaly and you are never sure what all the fields will be when you come to display them.

I suggest you place this within a loop through your result rows and include a field flag check  around the echo to only show certain data types like this.

$y=mysql_num_fields($result);
while ($row=mysql_fetch_array($result)) {
  for ($x=0; $x<=$y; $x++) {
    $fieldname=mysql_field_name($result,$x);
    $fieldtype=mysql_field_type($result, $x);
    if ($fieldtype=='string' && $row[$fieldname]!='')    
       echo $row[$fieldname].' , ';
   }
   echo '<br>';
}

down

-5

aaronp123 att yahoo dott comm ¶

21 years ago


You could probably elaborate on this by sending a full sql query to this function...but I titled it simple_query() because it doesn't really allow for joins.  Never the less, if you want to get a quick array full of a single row result set this is painless:

function simple_query($table_name, $key_col, $key_val) {
    // open the db
    $db_link = my_sql_link();
    // query table using key col/val
    $db_rs = mysql_query("SELECT * FROM $table_name WHERE $key_col = $key_val", $db_link);
    $num_fields = mysql_num_fields($db_rs);
    if ($num_fields) {
        // first (and only) row
        $row = mysql_fetch_assoc($db_rs);
        // load up array
        for ($i = 0; $i < $num_fields; $i++) {
            $simple_q[mysql_field_name($db_rs, $i)] = $row[mysql_field_name($db_rs, $i)];
        }
        // and return
        return $simple_q;
    } else {
        // no rows
        return false;
    }
    mysql_free_result($db_rs);
}

**Please note that my_sql_link() is just a function I have to open up a my sql connection.**

add a note