<select>在这个循环中表现得很奇怪

Question

认为如果我只是向您展示代码可能是最好的.

while ($row = mysql_fetch_array($enheder)) {
echo"<tr>
<td>&nbsp;" .$row['Model'].
"<td>&nbsp;" .$row['SN'].
"<td>&nbsp;" .$row['Softwarever'].
"<td>&nbsp;" .$row['KobtDato'].
"<td>&nbsp;" .$row['Saelger'].
"<td>&nbsp;" .$row['ServiceDato'];

$dato = mysql_query("SELECT * FROM Servicetxt WHERE Servicenum IN ($where);")
or die(mysql_error());

?><form>
<select name="dato"><?
while ($datorow = mysql_fetch_array($dato)) {
    if ($datorow['Servicenum'] == $row['ID1']) {
        ?>
        <option value="something">something</option>
        <?
    }
?></select>
</form><?

}

echo "<td>&nbsp;" .$row['CalDato'].
"<td>&nbsp;" .$row['KundeID'].
"<td>&nbsp;" .$row['ID1'].
"<td>&nbsp;" .$row['KundeRefNo'];


}

抱歉,我无法发布截图,我只需要解释它.如果有两个以上的选项,它会很奇怪 - 不是将选项放在下拉列表中,而是只打印空下拉列表下方的所有选项.如果只有一个选项,它的效果非常好.但为什么？这可能是一个简单的愚蠢的错误,但这是早上05.46,我现在已经很久没睡觉了:D请帮忙.

Answer 1

你没有关闭if条件也没有关闭表格单元格

mysql不推荐使用mysqli或pdo

<select name="dato"><?
while ($datorow = mysql_fetch_array($dato)) {
    if ($datorow['Servicenum'] == $row['ID1']) {
        ?>
        <option value="something">something</option>
        <?
    }// here was the mistake
}
?></select>