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mysql_field_name

(PHP 4, PHP 5)

mysql_field_name結果において指定したフィールド名を取得する

警告

この拡張モジュールは PHP 5.5.0 で非推奨になり、PHP 7.0.0 で削除されました。 MySQLi あるいは PDO_MySQL を使うべきです。詳細な情報は MySQL: API の選択 を参照ください。 この関数の代替として、これらが使えます。

説明

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

mysql_field_name()は、指定したフィールドの 名前を返します。

パラメータ

result

評価された結果 リソース。この結果は、mysql_query() のコールにより得られたものです。

field_offset

数値フィールドオフセット。field_offset0 から始まります。field_offset が存在しない場合、E_WARNING レベルのエラーが発行されます。

戻り値

成功した場合に指定したフィールドの名前を、失敗した場合に false を返します。

例1 mysql_field_name() の例

<?php
/* users テーブルには以下の 3 つのフィールドがある
* 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($res, 0) . "\n";
echo
mysql_field_name($res, 2);
?>

上の例の出力は以下となります。

user_id
password

注意

注意: この関数により返されるフィー ルド名は 大文字小文字を区別 します。

注意:

下位互換のために、次の非推奨別名を使用してもいいでしょう。 mysql_fieldname()

参考

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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