我在一个页面写了两个表单
<form enctype="multipart/form-data" method="post" name="upform"> 上传图片:<br/> </td> <td> <input name="upfile" type="file"> <input type="submit" value="上传"><br> 允许上传的文件类型为:<?=implode(', ',$uptypes)?> </td> </form>
在上传完文件后,我要在另一表单填写其他数据并写入数据库中,但我就是获取不到第一个表单上传的图片名,该怎么做呢
回复讨论(解决方案)
考虑写成一个表单
上传的文件名在 $_FILES 中,而不是 $_POST 中
考虑写成一个表单
写成表单 我那个图片预览就没有了,还有上传的值能获取到吗
上传的文件名在 $_FILES 中,而不是 $_POST 中
我取值是正确的,但提交第二个表单文件名就获取不到了,因为我第一个表单是专门上传图片的
第二个表单中没有文件名,当然就取不到
分两次体就的话,要在服务端做缓存
第二个表单中没有文件名,当然就取不到
分两次体就的话,要在服务端做缓存
<form action="" method="post" "> <tr> <td width=80>产品名称</td> <td> <input type="text" name="name" size=20> </td> </tr> <td width=80></td> <td> <input type="submit" name="product" style="height:20px;" value="添加产品"> </td> </tr> </form>
if(isset($_POST[product])){ $db->query("INSERT INTO `p_product` (`pid`,`locate`, `name` " . "VALUES (NULL, '$_FILES[upfile][name]', '$_POST[name]'); $db->Get_admin_msg("img.php","添加成功");
$_FILES[upfile][name]插入没有值啊
第二个表单没有 这个文件上传框,当然就没有值。你用个变量存起来再插入不就行了。
$f_name = $_FILES[upfile][name];
if(isset($_POST[product])){
$db->query("INSERT INTO `p_product` (`pid`,`locate`, `name` " .
"VALUES (NULL, ' $f_name', '$_POST[name]');
$db->Get_admin_msg("img.php","添加成功");

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