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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。

mysqli_fetch_field_direct() [name] あるいは [orgname]
PDOStatement::getColumnMeta() [name]

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_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。

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

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

mysql_field_type() - 結果において指定したフィールドの型を取得する
mysql_field_len() - 指定したフィールドの長さを返す

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User Contributed Notes 12 notes

down

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 );

?>

down

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"
?>

down

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

down

-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;

down

-2

jimharris at blueyonder dot co dot uk ¶

20 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>';
}

down

-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);
?>

down

-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;

    

}

?>

down

-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

down

-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++;
    }
}

down

-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.

down

-5

colin dot truran at shiftf7 dot com ¶

20 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>';
}

down

-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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mysql_field_name

説明

パラメータ

戻り値

例

注意

参考

Found A Problem?

User Contributed Notes 12 notes