【发布时间】：2017-05-11 18:26:53
【问题描述】：
正如您在这张图片中看到的，我扫描产品的条形码，插入 mysql 表并通过自动刷新将该 mysql 表检索到此页面中。每次我扫描不同的产品时，都会根据 mysql 表添加新行。
使用这些代码我检索 mysql 表并在 html 表中回显：
<table id="dataTable" class="table table-condensed table-bordered" style="margin-top:0px">
<tbody>
<?php
$sat_fatura_no = $_GET['sat_fatura_no'];
$liste_cek = @mysql_query("SELECT * FROM alisfat_listesi WHERE fatura_no = '".$sat_fatura_no."'");
while ($a= mysql_fetch_array($liste_cek)){
$stok_kodu2 = $a['stok_kodu'];
$isim_cek = mysql_query("SELECT * FROM stoklar WHERE sto_RECno = '".$stok_kodu2."'");
while ($isim_al=mysql_fetch_array($isim_cek)){
$urun_isim = $isim_al['sto_isim'];
}
$kdv = $a['kdv'];
$kdv_cek = mysql_query("SELECT * FROM kdvler WHERE kdv_id = '".$kdv."'");
while ($kdv_al = mysql_fetch_array($kdv_cek)){
$kdv_isim = $kdv_al['kdv_isim'];
}
$kdv_cek1 = mysql_query("SELECT * FROM kdvler");
while ($kdv_al1 = mysql_fetch_array($kdv_cek1)){
$kdv_id1 = $kdv_al1['kdv_id'];
$kdv_isim1 = $kdv_al1['kdv_isim'];
}
echo '<tr>
<td width="5%"><a href="satir_sil.php?stok_kodu='.$stok_kodu2.'&sat_fatura_no='.$sat_fatura_no.'"><button class="btn btn-mini btn-danger input1">Sil</button></a></td>
<td width="25%"><input class="input1" type="hidden" name="sth_stok_kod[]" value="'.$stok_kodu2.'" readonly>'.$urun_isim.'</td>
<td width="10%"><input style="text-align:center" class="input1" type="text" name="sth_adet[]" value="" required></td>
<td width="10%">
<select style="margin: 0 !important; padding: 0 !important; width:100%" name="sth_birim[]" required>
<option value="" selected>Seçiniz</option>
<?php $brim2_bul = mysql_query("SELECT * FROM birimler");
while($brim2_cek = mysql_fetch_array($brim2_bul)){
echo "<option ";
echo "value="".$brim2_cek["birim_id"]."">". $brim2_cek["birim_isim"]."</option>";
} ?>
</select>
</td>
<td width="10%"><input class="input1" type="text" id="sth_birim_fiyat" name="sth_birim_fiyat[]" required></td>
<td width="10%">
<select style="margin: 0 !important; padding: 0 !important; width:100%" name="sth_kdv" required>
<option value="<?php echo $stok_vergisi; ?>" selected><?php echo $stok_vergisi; ?></option>
<?php $kdv2_bul = mysql_query("SELECT * FROM kdvler");
while($kdv2_cek = mysql_fetch_array($kdv2_bul)){
echo '<option ';
echo 'value="'.$kdv2_cek['kdv_isim'].'">' .$kdv2_cek['kdv_isim'].'</option>';
} ?>
</select>
</td>
<td width="10%"><input class="input1" type="text" id="sth_iskonto1" name="sth_iskonto1[]" required></td>
<td width="10%"><input class="input1" type="text" id="sth_iskonto2" name="sth_iskonto2[]" required></td>
<td width="10%"><input class="input1" type="text" id="sth_tutar" name="sth_tutar[]" required></td>
</tr>';
} ?>
</tbody>
</table>
我无法修复的唯一部分是那些 SELECT OPTION 部分。通常我可以使用这些选择选项 php 代码检索 mysql 表的每一行，但是当我将它们放在 php echo 中时，通常我无法从数据库中检索任何值。
所以任何人都知道如何让这些选择下拉菜单在 php echo 中保持活跃？
【问题讨论】：
标签： php html mysql drop-down-menu dynamic-tables