mysql_field_len
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 );
?>
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 で非推奨になりました。将来のバージョンで削除される予定です。 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 や それに関連する FAQ 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。
- mysqli_fetch_field_direct() [name] あるいは [orgname]
- PDOStatement::getColumnMeta() [name]
説明
string mysql_field_name
( resource
$result
, int $field_offset
)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_offsetは 0 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。
返り値
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
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()
add a note
User Contributed Notes
mysql_field_name - [12 notes]
anonymous at site dot com ¶
5 years ago
janezr at jcn dot si ¶
7 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"
?>
clinnenb at hotmail dot com ¶
7 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;
$i++;
}
}
tiptonentserv at gmail dot com ¶
1 year ago
simple sql to xml converter works with any sql query and returns the name of the table as the root element "row" as each row element and the names of the columns are your children of row. fully tested.
<?php
function sqlToXml($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;
}
?>
bags ¶
2 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.
blackjackdevel at gmail dot com ¶
5 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
matteo.cisilino[no_more]cisilino[spm]com ¶
6 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;
jimharris at blueyonder dot co dot uk ¶
8 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>';
}
colin dot truran at shiftf7 dot com ¶
8 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>';
}
aaronp123 att yahoo dott comm ¶
10 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.**
jason dot chambes at phishie dot net ¶
10 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] </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 ¶
11 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