mysql_field_name

(PHP 4, PHP 5)

mysql_field_nameGet the name of the specified field in a result

Avertizare

Această extensie a devenit învechită în PHP 5.5.0 și a fost eliminată în PHP 7.0.0. În locul ei trebuie utilizată extensia MySQLi sau PDO_MySQL. Accesați de asemenea ghidul MySQL: selectarea unei API și FAQ asociat pentru informații suplimentare. Variante alternative pentru această funcție includ:

Descrierea

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

mysql_field_name() returns the name of the specified field index.

Parametri

result

Rezultatul de tip resource, care este evaluat. Acest rezultat provine de la apelul funcției mysql_query ().

field_offset

Deplasamentul numeric al câmpului. field_offset începe de la 0. Dacă field_offset nu există, se generează de asemenea o eroare de nivel E_WARNING.

Valorile întoarse

The name of the specified field index on success sau false în cazul eșecului.

Exemple

Example #1 mysql_field_name() example

<?php
/* The users table consists of three fields:
 *   user_id
 *   username
 *   password.
 */
$link mysql_connect('localhost''mysql_user''mysql_password');
if (!
$link) {
    die(
'Could not connect to MySQL server: ' mysql_error());
}
$dbname 'mydb';
$db_selected mysql_select_db($dbname$link);
if (!
$db_selected) {
    die(
"Could not set $dbname: " mysql_error());
}
$res mysql_query('select * from users'$link);

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

Exemplul de mai sus va afișa:

user_id
password

Note

Notă: Numele câmpurilor întoarse de această funcție sunt sensibile la literele majuscule/minuscule.

Notă:

Pentru compatibilitatea cu versiunile anterioare, următorul pseudonim învechit poate fi utilizat: mysql_fieldname()

A se vedea și

add a note add a note

User Contributed Notes 12 notes

up
13
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 );

?>
up
2
janezr at jcn dot si
18 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"
?>
up
0
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;
up
0
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>';
}
up
-1
matt at iwdt dot net
22 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
up
-2
tiptonentserv at gmail dot com
12 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;
   
}
?>
up
-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);
?>
up
-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
up
-4
bags
13 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.
up
-5
clinnenb at hotmail dot com
18 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++;
    }
}
up
-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>';
}
up
-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.**
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