mysql_fetch_row

(PHP 4, PHP 5)

mysql_fetch_rowOttiene una riga del risultato come un array enumerato

Descrizione

mysql_fetch_row ( resource $risultato ) : array

Restituisce un array che corrisponde ad una riga caricata oppure FALSE se non ci sono più righe.

mysql_fetch_row() carica una riga di dati dal risultato associato all'identificativo specificato. La riga è restituita com un array. Ogni colonna del risultato è memorizzata in un indice dell'array, partendo dall'indice 0.

La susseguente chiamata a mysql_fetch_row() restituisce la successiva riga nell'intervallo del risultato oppure FALSE se non ci sono più righe.

Vedere anche: mysql_fetch_array(), mysql_fetch_object(), mysql_data_seek(), mysql_fetch_lengths() e mysql_result().

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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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