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 — 取得结果中指定字段的字段名
警告
本扩展自 PHP 5.5.0 起已废弃，并在自 PHP 7.0.0 开始被移除。应使用 MySQLi 或 PDO_MySQL 扩展来替换之。参见 MySQL：选择 API 指南来获取更多信息。用以替代本函数的有：
- mysqli_fetch_field_direct() [name] or [orgname]
- PDOStatement::getColumnMeta() [name]
说明
mysql_field_name()
返回指定字段索引的字段名。result
必须是一个合法的结果标识符，field_index
是该字段的数字偏移量。
注意:
field_index从 0 开始。例如，第三个字段的索引值其实是 2，第四个字段的索引值是 3，以此类推。
注意: 此函数返回的字段名大小写敏感。
示例 #1 mysql_field_name() 例子
<?php
/* The users table consists of three fields:
* user_id
* username
* password.
*/
$link = mysql_connect('localhost', "mysql_user", "mysql_password");
$dbname = "mydb";
mysql_select_db($dbname, $link)
or 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);
?>以上例子将产生如下输出：
user_id password
为向下兼容仍然可以使用 mysql_fieldname()，但反对这样做。
参数
-
result -
resource 型的结果集。此结果集来自对 mysql_query() 的调用。
-
field_offset -
数值型字段偏移量。
field_offset从0开始。如果field_offset不存在，则会发出一个E_WARNING级别的错误
返回值
The name of the specified field index on success 或者在失败时返回 false.
示例
示例 #2 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);
?>以上示例会输出：
user_id password
注释
注意: 此函数返回的字段名大小写敏感。
注意:
为了向下兼容，可以使用下列已废弃的别名： mysql_fieldname()
add a note
User Contributed Notes 12 notes
anonymous at site dot com ¶
16 years ago
janezr at jcn dot si ¶
19 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"
?>
matt at iwdt dot net ¶
23 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
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;
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, $x).'<br>';
}
jason dot chambes at phishie dot net ¶
21 years ago
<?
/*
By simply calling the searchtable() function
with these variables it will serach the desired
database and procude a table for each field that
there is a match.
*/
function searchtable($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] </td>\n";
}
echo "\t</tr>\n";
} while ($myrow = mysql_fetch_array($result));
echo "</table>\n";
}
}
}
searchtable($host,$user,$pass,$database,$tablename,$userquery);
?>
tiptonentserv at gmail dot com ¶
13 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;
}
?>
blackjackdevel at gmail dot com ¶
17 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
clinnenb at hotmail dot com ¶
19 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++;
}
}
bags ¶
14 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.
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>';
}
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.**
