嘿家伙 我试图开发一个先进的使用PHP的消费者网站,我试图获取数据提交到另一个页面时卡住了某处。我想要做的就是获取存储在数组中的复选框值,并由用户检查。我迫切需要你的帮助,请在这里是代码:我不能提交表格
$sql2="SELECT * FROM alinanfis WHERE alinanfis.fis_id='".$fis."'" ;
$resultFis=mysql_query($sql2);
if(mysql_num_rows($resultFis)>0)
{
print "<form method='POST' action='deletionResult.php'>";
print "<table>";
print "<tr>";
print "<th style='background: transparent;'></th>";
print "<th>Fis No</th>";
print "<th>isim</th>";
print "<th>soyisim</th>";
print "<th >Tarih</th>";
print "<th>Fis Tipi</th>";
print "<th>Nerede</th>";
print "<th>Litre</th>";
print "<th>Tutar</th>";
print "</tr>";
while($rowAlinan=mysql_fetch_array($resultFis))
{
$sqlFisTipi="SELECT * FROM atype WHERE a_id='".$rowAlinan['a_id']."'" ;
$resultFisTipi=mysql_query($sqlFisTipi);
$rowFisTipi=mysql_fetch_array($resultFisTipi);
$sqlNerede="SELECT * FROM isyeri WHERE i_id='".$rowAlinan['nerde']."'" ;
$resultNerede=mysql_query($sqlNerede);
$rowNerede=mysql_fetch_array($resultNerede);
$sqlMID="SELECT * FROM musteri WHERE m_id='".$rowAlinan['m_id']."'" ;
$resultMID1=mysql_query($sqlMID);
$rowMID1=mysql_fetch_array($resultMID1);
print "<tr>";
print "<td><input name='checkBox[]' type='checkbox' value='".$rowAlinan['fis_id']."' />
php</td>";
print "<td>".$rowAlinan['fis_id']."</td>";
print "<td>".$rowMID1['m_name']."</td>";
print "<td>".$rowMID1['m_lastName']."</td>";
print "<td>".$rowAlinan['alinan_tarih']."</td>";
print "<td>".$rowFisTipi['a_name']."</td>";
print "<td>".$rowNerede['i_name']."</td>";
print "<td>".$rowAlinan['litre']."</td>";
print "<td>".$rowAlinan['tutar']."</td>";
print "</tr>";
}
print '<div class="form_settings">';
print "<input class='submit' type='submit' name= 'send' value='Send'>";
/
print '</div>';
print "</table>";
print "</form>";
}//end of if(num_rows>0)
else
echo '*no such receipt found!!';
mysql_close($con);
这里是整个代码:http://pastebin.com/SBiiSBHu。
问题是什么? – 2011-03-04 21:54:40
你得到了什么错误? – 2011-03-04 21:57:32
伙计们,当我点击提交按钮时,它什么都不做,并重新定向页面本身,但我希望它提交值到deleseResult.php页面 – 2011-03-04 22:01:23