[英]select the name as selected in dropdown
从上帝表中查询选择上帝
$sth =$dbh->prepare("SELECT god_id,god_name_ml,god_name_en,image,info_ml,info_en,details_ml,
details_en,rounds_ml,rounds_en,mantra_ml,mantra_en,display_order FROM god");
$sth->execute();
从神表中查询选择神,这里god_id是神表中的外键
$stmt = $dbh->prepare("SELECT deity_id,god_id,deity_name_ml,deity_name_en,info_ml,info_en,details_ml,
details_en,mantra_ml,mantra_en,display_order FROM deity
WHERE deity_id = :deity_id");
$stmt->bindValue(':deity_id',$deity_id,PDO::PARAM_INT);
$stmt->execute();
$result = $stmt->fetchAll();
$temp_array=$result[0];
$god_id=$temp_array['god_id'];
在上帝表的下拉菜单中显示上帝的显示代码
<?php while ($row = $sth->fetch(PDO::FETCH_ASSOC)) { ?>
<option value="<?php echo $row['god_id'];?>">
<?php echo $row['god_name_en']; ?>
</option>
<?php } ?>
</option>
</select>
我需要的是,在下拉列表中,神表中存在的God表中给定的god_id使其成为选定表。 只选择我需要的ID
只需检查两个ID是否相同,请尝试以下操作:
<?php while ($row = $sth->fetch(PDO::FETCH_ASSOC)) { ?>
<option <?php if($row['god_id'] == $god_id) { echo "selected='selected'"; } ?> value="<?php echo $row['god_id'];?>"><?php echo $row['god_name_en']; ?></option>
<?php } ?>
</option>
</select>
尝试这个:
<?php while ($row = $sth->fetch(PDO::FETCH_ASSOC)) { ?>
<option <?php if($row['id'] == $stmt->god_id) { echo "selected='selected'"; } ?> value="<?php echo $row['god_id'];?>"><?php echo $row['god_name_en']; ?></option>
<?php } ?>
</option>
</select>
在编写每个选项时,请比较God表中的God_I是否与当前选项的god_ID相匹配。 如果确实存在,则在该值之后附加文本“ selected”。
所以选择的选项看起来像
<option value='1' selected>foo</option>
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