php mysql查询表单将值发送到其他页面

php mysql查询表单将值发送到其他页面,php,mysql,Php,Mysql,我用mysql查询表单有点麻烦 下面是我的代码示例 <? $query="select * from data where data_id=".$data_id." order by data_id desc"; $result=mysql_query($query); while ($dataresult=mysql_fetch_array($result)) { ?> <form style=

我用mysql查询表单有点麻烦 下面是我的代码示例

    <?
    $query="select * from data where  data_id=".$data_id."  order by data_id desc";
    $result=mysql_query($query);
    while ($dataresult=mysql_fetch_array($result))
            {   
    ?>

   <form style="display:table-row-group" id="form_data" name="form_data" action="view/view_data"  method="get" >
   <input type="hidden" name="ref" id="ref" value="hv" readonly />
   <input type='hidden' name='dataid' id='dataid' value='<? echo"$dataresult[data_id]"?>' readonly />

    <?
    echo" name = $dataresult[name] <br> 
         <button class=buttonimage onclick=$('#form_data').submit() type=button >Pesan</button>";
     ?>

    </form>
    <?
     }
     ?>


您在
name
附近的几个地方缺少了一个quete,并且还更改了
echo
按钮
部分

echo" name =". $dataresult['name'].'<br>'; 
echo "<button class=buttonimage onclick=$('#form_data').submit() type=button >Pesan</button>";

<input type='hidden' name='dataid' id='dataid' value='<? echo"$dataresult[data_id]"?>' readonly />
<input type='hidden' name='dataid' id='dataid' value='<? echo"$dataresult['data_id']"?>' readonly />//Use this 

由于
mysql
已被弃用,请同时使用
mysqli\uuu
函数或
PDO

修改了代码以使其正常工作:

<?php
$query="select * from data where data_id=".$data_id."  order by data_id DESC";
$result=mysql_query($query);
while ($dataresult=mysql_fetch_assoc($result)) // You can use _array also but this one is preferred if using only keys
{   
?>
<form style="display:table-row-group" id="form_data" name="form_data" action="view/view_data"  method="get" >
<input type="hidden" name="ref" id="ref" value="hv" readonly="readonly" />
<input type="hidden" name="dataid" id="dataid" value="<?php echo $dataresult['data_id']; ?>"  readonly="readonly" />
<?php echo "name=".$dataresult['name']; ?> <br />
<button class="buttonimage" onclick="$('#form_data').submit();" type="button">Pesan</button>
</form>
<?php 
 }
 ?>


比桑
<?php
$query="select * from data where data_id=".$data_id."  order by data_id DESC";
$result=mysql_query($query);
while ($dataresult=mysql_fetch_assoc($result)) // You can use _array also but this one is preferred if using only keys
{   
?>
<form style="display:table-row-group" id="form_data" name="form_data" action="view/view_data"  method="get" >
<input type="hidden" name="ref" id="ref" value="hv" readonly="readonly" />
<input type="hidden" name="dataid" id="dataid" value="<?php echo $dataresult['data_id']; ?>"  readonly="readonly" />
<?php echo "name=".$dataresult['name']; ?> <br />
<button class="buttonimage" onclick="$('#form_data').submit();" type="button">Pesan</button>
</form>
<?php 
 }
 ?>