mysql_field_name

(PHP 4, PHP 5)

mysql_field_name결과로부터 특정 필드의 이름을 반환

설명

string mysql_field_name ( resource $result , int $field_offset )

mysql_field_name()는 특정 필드 이름을 반환한다.

인수

result

mysql_query() 호출을 통한 결과 resource.

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.

반환값

성공하면 특정 필드 이름을, 실패하면 FALSE를 반환한다.

예제

Example #1 mysql_field_name() 예제

<?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($res0) . "\n";
echo 
mysql_field_name($res2);
?>

위 예제의 출력:

user_id
password

주의

Note: 이 함수가 반환하는 필드 이름은 대소문자를 구별합니다.

Note:

하위 호환을 위하여, 다음의 권장하지 않는 별칭을 사용할 수 있습니다: mysql_fieldname()

참고

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User Contributed Notes 12 notes

up
13
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
2
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"
?>
up
0
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
up
-1
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
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;
   
}
?>
up
-3
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
up
-4
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++;
    }
}
up
-5
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.
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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