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, PECL mysql:1.0)
mysql_field_name — Get the name of the specified field in a result
Description
string mysql_field_name
( resource $result
, int $field_offset
)
mysql_field_name() returns the name of the specified field index.
Parameters
- result
-
The result resource that is being evaluated. This result comes from a call to mysql_query().
- field_offset
-
The numerical field offset. The field_offset starts at 0. If field_offset does not exist, an error of level E_WARNING is also issued.
Return Values
The name of the specified field index on success, or FALSE on failure.
Examples
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', $link);
echo mysql_field_name($res, 0) . "\n";
echo mysql_field_name($res, 2);
?>
The above example will output:
user_id password
Notes
Note: Field names returned by this function are case-sensitive.
Note: For backward compatibility, the following deprecated alias may be used: mysql_fieldname()
mysql_field_name
anonymous at site dot com
09-Mar-2008 03:13
09-Mar-2008 03:13
blackjackdevel at gmail dot com
14-Nov-2007 01:13
14-Nov-2007 01:13
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
09-Jan-2007 05:54
09-Jan-2007 05:54
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-Oct-2005 04:18
19-Oct-2005 04:18
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
05-Aug-2005 05:19
05-Aug-2005 05:19
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-Dec-2004 03:28
20-Dec-2004 03:28
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-Dec-2004 01:44
17-Dec-2004 01:44
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-Feb-2003 03:27
21-Feb-2003 03:27
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-Feb-2003 03:07
21-Feb-2003 03:07
<?
/*
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] </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-Sep-2001 03:09
24-Sep-2001 03:09
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
