mysql_fetch_row

(PHP 4, PHP 5)

mysql_fetch_row숫자 색인 배열로 결과를 반환

설명

array mysql_fetch_row ( resource $result )

인출된 행과 일치하는 숫자 색인 배열을 반환하고, 앞으로 내부 데이터 포인터를 이동한다.

인수

result

mysql_query() 호출을 통한 결과 resource.

반환값

인출된 행에 대응하는 문자열의 숫자 색인 배열을 반환하거나 더 이상의 행이 없으면 FALSE를 반환한다.

mysql_fetch_row()는 지정된 result 식별자와 연관된 결과로부터 하나의 데이터 행을 인출하며, 배열로 반환한다. 각 결과 컬럼은 0으로 시작하는 배열 오프셋(offset)에 저장이 된다.

예제

Example #1 mysql_fetch_row()를 이용한 하나의 행 인출 예제

<?php
$result 
mysql_query("SELECT id,email FROM people WHERE id = '42'");
if (!
$result) {
    echo 
'Could not run query: ' mysql_error();
    exit;
}
$row mysql_fetch_row($result);

echo 
$row[0]; // 42
echo $row[1]; // the email value
?>

주의

Note: 이 함수는 NULL 필드를 PHP NULL 값으로 설정합니다.

참고

add a note add a note

User Contributed Notes 8 notes

up
-13
a at simongrant dot org
18 years ago
Maybe worth pointing out that all the fields returned by this (and other?) calls are returned with type string. This had me puzzled for quite some time.
up
-17
michael and then an at sign wassupy.com
17 years ago
to print an array, simply use print_r(array name)

like this:
    $myrow = mysql_fetch_row($result);
echo "<pre>";
print_r($myrow);
echo "</pre>";

this will output the array in a readable form, with the index, too. Don't forget the 'pre' tags or the output will be on a single line.
up
-23
larkitetto at gmail dot com
12 years ago
sry :) note now fixed:

<?php
$esi
=mysql_list_tables($db);$ris=mysql_fetch_row($esi);
//example: $db has >= 1 tabs
echo var_dump($ris);
//echoes only array(1). solution:
while($ris=mysql_fetch_row($esi)) echo $ris[0];
/*debug:
$ris=array("1st_tab"); ... $ris=array("n_tab");$ris=false;*/
while ($ris[]=mysql_fetch_row($esi));
//debug:$ris=array(array("1st_tab"), ... array("n_tab"));
echo $ris[n][0];//echo:"n_tab"
echo $ris[0][n];//echo:array | null
?>

hope it helps
up
-28
pepik at gmail dot cz
7 years ago
<?php 
 
require 'prhlavicka.php';
 
pis_hlavicku('Vypis článků');        
   
   require_once
'db.php';
   
$kom = new server();
   
$sql=$kom->query("SELECT autor,nazev,obsah FROM `Clanky_Sadek`");
               while (
$data = mysql_fetch_row($sql)){
                 ECHO
'<br />--AUTOR--<br />'.$data[0].'<br />__NÁZEV ČLÁNKU__<br />'.$data[1].'<br />..OBSAH ČLÁNKU..<br />'.$data[2];     }   
 
  include
'Paticka.html'; ?>
up
-33
m dot s at programmers-online dot net
14 years ago
The following function to read all data out of a mysql-resultset, is may be faster than Rafaels solution:

<?
function mysql_fetch_all($result) {
   while($row=mysql_fetch_array($result)) {
       $return[] = $row;
   }
   return $return;
}
?>
up
-34
ryhan_balboa at yahoo dot com
12 years ago
The following are the basic codes to get a specific row from the mysql db into a $row variable:


$query = "SELECT * FROM table";
$result = mysql_query($query);
$row = mysql_fetch_row($result);

And $row[0], $row[1] ... $row[n] are used to access those field values.

Does anyone know how I can add a new field to $row, so that the field count increases from n to n+1?

I have tried treating $row like an array, and tried array_push function, but didn't work.

Thanks.
up
-36
jhulbert at redf dot com
10 years ago
Creates table from all db info:

<?php
$qry
= "SELECT * FROM exp_member_data";
$res = mysql_query($mem_qry);

function
mysql_fetch_all($res) {
   while(
$row=mysql_fetch_array($res)) {
      
$return[] = $row;
   }
   return
$return;
}

function
create_table($dataArr) {
    echo
"<tr>";
    for(
$j = 0; $j < count($dataArr); $j++) {
        echo
"<td>".$dataArr[$j]."</td>";
    }
    echo
"</tr>";
}

$all = mysql_fetch_all($res);

echo
"<table class='data_table'>";

for(
$i = 0; $i < count($all); $i++) {
   
create_table($all[$i]);
}

echo
"</table>";

?>
up
-49
murapaka dot sateesh at gmail dot com
6 years ago
<?php

$conn
=mysql_connect("localhost","root","");
if(
$conn)
{
    echo(
"Connection Successfully");
   
$seldb=mysql_select_db("test",$conn);
    if(
$seldb)
    {
        echo(
"Database selected successfully");
       
$retrive=mysql_query("select name,number from login1 where name='sateesh'",$conn);
        if(
$retrive)
        {
            echo(
"Successfully data retrived<br>");
           
           
$result=mysql_fetch_row($retrive);
           
                echo (
"Name".$result[0]);
                echo (
"Number".$result[1]);
                   
        }
        else
        {
        echo
"Table not inserted";
        }
        }
        else
        {
            die(
"database not selected");
        }
    }
    else
    {
        die(
"connection faild");   
    }
   
mysql_close($conn);
?>
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