php mysql查询表单将值发送到其他页面
我用mysql查询表单有点麻烦 下面是我的代码示例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=
<?
$query="select * from data where data_id=".$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
}
?>