mysql_fetch_object
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>";
?>
mysql_fetch_row
(PHP 4, PHP 5)
mysql_fetch_row — 結果を添字配列として取得する
説明
array mysql_fetch_row
( resource
$result
)取得された行に対応する配列を返し、内部のデータポインタを前に進めます。
返り値
取得された行に対応する文字列の配列を返します。もう行がない場合は、
FALSE を返します。
mysql_fetch_row()は、指定した結果 ID が指す結果から 1 行分のデータを取得します。 各結果カラムは、オフセット 0 から始まる配列に格納されます。
例
例1 mysql_fetch_row() で行を取得する
<?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]; // 'email' の値
?>
注意
注意: この関数は、 NULL フィールドに PHPの
NULL値を設定します。
参考
- mysql_fetch_array() - 連想配列、添字配列、またはその両方として結果の行を取得する
- mysql_fetch_assoc() - 連想配列として結果の行を取得する
- mysql_fetch_object() - 結果の行をオブジェクトとして取得する
- mysql_data_seek() - 内部的な結果ポインタを移動する
- mysql_fetch_lengths() - 結果における各出力の長さを得る
- mysql_result() - 結果データを得る
add a note
User Contributed Notes
mysql_fetch_row
jhulbert at redf dot com
19-Feb-2010 01:32
larkitetto at gmail dot com
21-Feb-2008 03:29
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
ryhan_balboa at yahoo dot com
16-Nov-2007 09:23
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 $row like an array, and tried array_push function, but didn't work.
Thanks.
m dot s at programmers-online dot net
17-Nov-2005 01:56
The following function to read all data out of a mysql-resultset, is may be faster than Rafaels solution:
<?
function mysql_fetch_all($result) {
while($row=mysql_fetch_array($result)) {
$return[] = $row;
}
return $return;
}
?>
mysql at polyzing dot com
12-Jul-2003 03:05
It is probably worth pointing out that the array elements will actually be of type string, OR NULL if the field is null in the database.
Thus, either use a double equal comparison to look for empty or null
Or, use a triple equal comparison to be able to distinguish the two cases
e.g.
if ($field === '') echo "Empty, not NULL\n";
if ($field === NULL) echo "NULL\n";
if ($field == '') echo "Empty or NULL\n";
michael and then an at sign wassupy.com
08-Apr-2003 12:09
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.
a at simongrant dot org
06-Feb-2002 05:10
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.