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mysql_field_name

(PHP 4, PHP 5)

mysql_field_nameВозвращает название указанной колонки результата запроса

Внимание

Данный модуль устарел начиная с версии PHP 5.5.0, и удалён в PHP 7.0.0. Используйте вместо него MySQLi или PDO_MySQL. Смотрите также инструкцию MySQL: выбор API. Альтернативы для этой функции:

Описание

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

mysql_field_name() возвращает название колонки с указанным индексом.

Список параметров

result

Обрабатываемый результат запроса. Этот результат может быть получен с помощью функции mysql_query().

field_offset

Числовое смещение поля. field_offset начинается с 0. Если field_offset не существует, генерируется ошибка уровня E_WARNING.

Возвращаемые значения

Название поля по указанному индексу в случае успешного выполнения или false в случае ошибки.

Примеры

Пример #1 Пример использования mysql_field_name()

<?php
/* Таблица пользователей состоит из трёх колонок:
* user_id
* username
* password.
*/
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!
$link) {
die(
'Ошибка соединения с MySQL: ' . mysql_error());
}
$dbname = 'mydb';
$db_selected = mysql_select_db($dbname, $link);
if (!
$db_selected) {
die(
"Не удалось выбрать базу $dbname: " . mysql_error());
}
$res = mysql_query('select * from users', $link);

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

Результат выполнения приведённого примера:

user_id
password

Примечания

Замечание: Имена полей, возвращаемые этой функцией являются зависимыми от регистра.

Замечание:

Для обратной совместимости может быть использован следующий устаревший псевдоним: mysql_fieldname()

Смотрите также

  • mysql_field_type() - Возвращает тип указанного поля из результата запроса
  • mysql_field_len() - Возвращает длину указанного поля

add a note

User Contributed Notes 12 notes

up
12
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
3
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
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
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
-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>';
}
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
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
-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
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
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
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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