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即时尝试创建更新表单。下面是我的制造年份和车辆状态的HTML。更新值PHP
<div class="form-group">
<label class="control-label" >Year Manufactured:</label>
<select class="form-control" name="yearManufactured" value="<?php if(isset($row['yearManufactured_vehicle'])) { echo $row['yearManufactured_vehicle']; } ?>">
<option>Select</option>
<?php
foreach(range(1950, (int)date("Y")) as $year) {
echo "\t<option value='".$year."'>".$year."</option>\n\r";
}
?>
</select>
</div>
<div class="form-group">
<label>Vehicle Status</label>
<select class="form-control" name="yearManufactured" value="<?php if(isset($row['yearManufactured_vehicle'])) { echo $row['yearManufactured_vehicle']; } ?>">
<option>Select</option>
<?php
foreach(range(1950, (int)date("Y")) as $year) {
echo "\t<option value='".$year."'>".$year."</option>\n\r";
if($row['yearManufactured']==$year){
echo "selected";
}
}
?>
</select>
</div>
在车辆状态中,我用来选择确保插入数据的值以更新形式显示。制造年份也是一样吗?我试图把选择内的选项字段,但它给了我错误。我如何使用制造年份内的选择内部foreach?
请包括错误 – Swellar
,你可以直接呼应'selected',而不是关闭了'php',然后键入它 – Swellar
@Swellar的我编辑的代码,并把选择的。仍然没有变化。我如何解决这个问题? – yuki