<?PHP
$select = 'SELECT cliente, pedido, data, valor from financial';
$result = mysql_query($select);
$medium = mysql_fetch_row($result);
while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
printf('
<form name="frmFinanceiro" id="frmFinanceiro" action="frmFinanceiro" method="POST">
<div class="row">
<div class="six columns">
<br><br><br>
<input type="text" name="cliente" placeholder="Cliente" value="%s" style="background:#F0E68C; color:black;"/>
<input type="text" name="pedido" placeholder="Pedido" value="%s" style="background:#F0E68C; color:black;"/>
</div>
<div class="six columns">
<br><br><br>
<input type="text" name="data" value="%s" placeholder="Data" style="background:#F0E68C; color:black;"/>
<input type="text" name="Valor" value="%s" placeholder="Valor" style="background:#F0E68C; color:black;"/>
</div>
</div>
<center>
<input type="submit" value="Alterar" class="button success" />
</center>
</form>
', $row[0], $row[1], $row[2], $row[3]);
}
?>
我有這段代碼,它爲什麼會跳第一個結果? 它是隻顯示我的數據庫的ID> 1 :( 我rewrited它3X +和沒有成功行中的錯誤Mysql html/php
刪除'$媒體=和mysql_fetch_row($結果);',基本上要從使用'mysql_fetch_row'您不要使用流中讀取一行,然後在'while'環你讀了第二個等上。 – ddinchev