PHP Doku:: Liefert den Namen eines Feldes in einem Ergebnis - function.mysql-field-name.html

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MySQL Funktionen

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mysql_field_name

(PHP 4, PHP 5)

mysql_field_name Liefert den Namen eines Feldes in einem Ergebnis

Beschreibung

string mysql_field_name ( resource $Ergebnis-Kennung , int $Feldindex )

mysql_field_name() liefert den Namen des Feldes, der dem angegeben Feldindex entspricht. Der Paramater Ergebnis-Kennung muss eine gültige Ergebnis-Kennung sein und Feldindex bestimmt den numerischen Offset des Feldes.

Hinweis:

Der Feldindex beginnt bei 0.

Zur Verdeutlichung: Der Index des dritten Feldes ist demnach 2, der Index des vierten Feldes 3 und so weiter.

Hinweis: Feldnamen, die von dieser Funktion zurückgegeben werden, unterscheiden sich in der Groß-/Kleinschreibung.

Beispiel #1 mysql_field_name() Beispiel

<?php
/* Die Tabelle users enthält drei Felder: 
     user_id
     username
     password
*/
$link mysql_connect('localhost'"mysql_user""mysql_password");
$dbname "mydb";
mysql_select_db($dbname$link)
    or die(
": " mysql_error());
$res mysql_query("select * from users"$link);

echo 
mysql_field_name($res0) . "\n";
echo 
mysql_field_name($res2);
?>
<?php
/* The users table consists of three fields:
 *   user_id
 *   username
 *   password.
 */
$link mysql_connect('localhost''mysql_user''mysql_password');
if (!
$db_selected) {
    die(
'Auswahl der Datenabnk $dbname fehlgeschlagen: ' mysql_error());
}
$dbname 'mydb';
$db_selected mysql_select_db($dbname$link);
if (!
$db_selected) {
    die(
'Auswahl der Datenabnk $dbname fehlgeschlagen: ' mysql_error());
}
$res mysql_query('select * from users'$link);

echo 
mysql_field_name($res0) . "\n";
echo 
mysql_field_name($res2);
?>

Das oben angeführte Beispiel liefert die folgende Ausgabe:

user_id
password

Für Abwärtskompatibilität kann mysql_fieldname() verwendet werden. Diese Funktion ist jedoch veraltet.


11 BenutzerBeiträge:
- Beiträge aktualisieren...
bags
25.07.2010 0:02
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.
anonymous at site dot com
9.03.2008 15:13
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 );

?>
blackjackdevel at gmail dot com
14.11.2007 1:13
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
matteo.cisilino[no_more]cisilino[spm]com
9.01.2007 17:54
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;
janezr at jcn dot si
19.10.2005 16:18
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"
?>
clinnenb at hotmail dot com
5.08.2005 17:19
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++;
    }
}
jimharris at blueyonder dot co dot uk
20.12.2004 15:28
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>';
}
colin dot truran at shiftf7 dot com
17.12.2004 13:44
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>';
}
aaronp123 att yahoo dott comm
21.02.2003 15:27
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.**
jason dot chambes at phishie dot net
21.02.2003 3:07
<?
/*
    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);
?>
matt at iwdt dot net
24.09.2001 3:09
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



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