if($_POST)
{
$client = $_POST['client'];
$insu1 = $_POST['insu1'];
$insu2 = $_POST['insu2'];
$date = date("dmY");
if($insu2 = 'alp' or $insu2='bppi' or $insu2='cpmp' or $insu2='carp' or $insu2='dsp' or $insu2='eep' or $insu2='earp' or $insu2='mbp')
{
//set @insu2 = $insu2;
$sql2 ="Select '$insu2' from tbl_engineering order by timestamp(`timestamp`) DESC limit 1 ";
$result2=mysql_query($sql2);
//$insu2_name ='';
var_dump($result2);
while($row = mysql_fetch_assoc($result2))
{
echo $insu2;
var_dump($row);
$insu2_name = $row[$_POST['insu2']];
echo $insu2_name;
}
从$insu2_name = $row[$_POST['insu2']]
;我正在获取列名称,但我想要列值。的$insu2_name = alp(column name of database)
如何在mysql_fetch_assoc中传递帖子变量结果
家伙 输出帮我
删除“什么是你的var_dump($结果2)所得到的值; – Anju
尝试更改'$ insu2_name = $ row [$ _ POST ['insu2']];'到'$ insu2_name = $ row ['insu2'];' – Kuya
@ $ result2 = resource(7,mysql result)的@Angel值 –