Question

php 填写对应的id修改对应的数据库，但点击修改，数据库没变化，请高人提点！

<?
$conn = mysql_connect("localhost","111","222");
$action = $_POST['action'];
if($action == 'send')
{
$lj_id = $_POST['lj_id'];
$lj_name = $_POST['lj_name'];
$lj_xianghao = $_POST['lj_xianghao'];
$lj_guige = $_POST['lj_guige'];
$lj_pinji = $_POST['lj_pinji'];
$lj_number = $_POST['lj_number'];
mysql_select_db("51ra",$conn);
$sql="update news
set
lj_name='$lj_name',
lj_xianghao='$lj_xianghao',
lj_guige='$lj_guige',
lj_pinji='$lj_pinji',
lj_number='$lj_number'
where
lj_id = '$lj_id' ";
$result = mysql_query($sql,$conn);
}

?>

<div align="center">商品属性修改</div>
<div align="center">
<form method="post" name="shangpinxiugai" action="_goods_dingzhixiugai.php">
<input type="hidden" name="action" value="send">

<dl>
<dd>ID:<input type="text" name="lj_id" /></dd>
<dd>名称:<input type="text" name="lj_name" /></dd>
<dd>型号:<input type="text" name="lj_xianghao" /></dd>
<dd>规格:<input type="text" name="lj_guige" /></dd>
<dd>品级:<input type="text" name="lj_pinji" /></dd>
<dd>数量:<input type="text" name="lj_number" /></dd>
</dl>

<td><input name='bthModify' type='submit' id='bthModify' value='修改' /></td><br>
<td><input type="button" name="Submit2" value="返回" onclick="history.go(-1)" /></td>
</form>
</div>

Answer 1

你运行一下下面代码看看、、、输出什么
<?
$conn = mysql_connect("localhost","111","222");
$action = $_POST['action'];
if($action == 'send')
{
$lj_id = $_POST['lj_id'];
$lj_name = $_POST['lj_name'];
$lj_xianghao = $_POST['lj_xianghao'];
$lj_guige = $_POST['lj_guige'];
$lj_pinji = $_POST['lj_pinji'];
$lj_number = $_POST['lj_number'];
mysql_select_db("51ra",$conn);
$sql="update news
set
lj_name='$lj_name',
lj_xianghao='$lj_xianghao',
lj_guige='$lj_guige',
lj_pinji='$lj_pinji',
lj_number='$lj_number'
where
lj_id = '$lj_id' ";
echo $sql;
exit;

$result = mysql_query($sql,$conn);
}

?>
<div align="center">商品属性修改</div>
<div align="center">
<form method="post" name="shangpinxiugai" action="_goods_dingzhixiugai.php">
<input type="hidden" name="action" value="send">
<dl>
<dd>ID:<input type="text" name="lj_id" /></dd>
<dd>名称:<input type="text" name="lj_name" /></dd>
<dd>型号:<input type="text" name="lj_xianghao" /></dd>
<dd>规格:<input type="text" name="lj_guige" /></dd>
<dd>品级:<input type="text" name="lj_pinji" /></dd>
<dd>数量:<input type="text" name="lj_number" /></dd>
</dl>

<td><input name='bthModify' type='submit' id='bthModify' value='修改' /></td><br>
<td><input type="button" name="Submit2" value="返回" onclick="history.go(-1)" /></td>
</form>
</div>

追问

只加echo $sql;不加exit;如果加exit;的话就没显示，加echo $sql;以后，填写表单出现：update ecs_goods_yudinglj set lj_name='', lj_xianghao='师傅给师傅', lj_guige='', lj_pinji='', lj_number='' where lj_id = '36'
但还是update不了数据库。

追答

商品属性修改

ID:
名称:
型号:
规格:
品级:
数量:

Answer 2

<?
$conn = mysql_connect("localhost","111","222");
$action = $_POST['action'];
if($action == 'send')
{
$lj_id = $_POST['lj_id'];
$lj_name = $_POST['lj_name'];
$lj_xianghao = $_POST['lj_xianghao'];
$lj_guige = $_POST['lj_guige'];
$lj_pinji = $_POST['lj_pinji'];
$lj_number = $_POST['lj_number'];
mysql_select_db("51ra",$conn);
$sql="update news
set
`lj_name`='$lj_name',
`lj_xianghao`='$lj_xianghao',
`lj_guige`='$lj_guige',
`lj_pinji`='$lj_pinji',
`lj_number`='$lj_number'
where
lj_id =$lj_id";
$result = mysql_query($sql,$conn);
}

?>

<div align="center">商品属性修改</div>
<div align="center">
<form method="post" name="shangpinxiugai" action="_goods_dingzhixiugai.php">
<input type="hidden" name="action" value="send">

<dl>
<dd>ID:<input type="text" name="lj_id" /></dd>
<dd>名称:<input type="text" name="lj_name" /></dd>
<dd>型号:<input type="text" name="lj_xianghao" /></dd>
<dd>规格:<input type="text" name="lj_guige" /></dd>
<dd>品级:<input type="text" name="lj_pinji" /></dd>
<dd>数量:<input type="text" name="lj_number" /></dd>
</dl>

<td><input name='bthModify' type='submit' id='bthModify' value='修改' /></td><br>
<td><input type="button" name="Submit2" value="返回" onclick="history.go(-1)" /></td>
</form>
</div>
您试试代码看行不行。