如何在mysql_fetch_assoc结果中传递post变量

Question

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;


                }

From $insu2_name = $row[$_POST['insu2']] ; I am getting column name, But I want the column Value. output of $insu2_name = alp(column name of database)

Guys help me

Answer 1

Do you know that

select anyRandomSrting from table

will return anyRandomSrting if anyRandomSrting is not a column in the table.

In your case there will not be a column which name in $_POST['insu2'].

Update
Check this query once

"Select $insu2 from tbl_engineering order by  timestamp( `timestamp` ) DESC limit 1"

Please check that i have removed ' from $insu2

Answer 2

Try changing $insu2_name = $row[$_POST['insu2']]; to $insu2_name = $row['insu2'];

Hope this helps.