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;


                }

来自$insu2_name = $row[$_POST['insu2']];我得到列名，但我想要列Value。输出$insu2_name = alp(column name of database)

伙计们帮助我

Answer 1

你知道吗

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

如果 anyRandomSrting 不是表格中的列，

将返回anyRandomSrting。

在您的情况下，$ _POST [＆＃39; insu2＆＃39;]中不会有一个名称列。

<强>更新
检查此查询一次

@PersistenceContext(unitName = "my-pu")
private EntityManager em;
@Override
public void removeUserFromGroup(String username, Group group) {
    Query query = em.createNamedQuery("Group.getByName", Group.class);
    query.setParameter("name", group.getGroupName());
    Group qGroup = (Group) query.getSingleResult();
    // this works
    // Iterator<User> i = qGroup.getUsers().iterator();
    // while (i.hasNext()) {
    // User o = i.next();
    // if (o.getUsername().equals(username)) {
    // System.out.println("eqqq");
    // i.remove();
    // }
    // }
    System.out.println("class: " + qGroup.getUsers().getClass().getName());
    // org.eclipse.persistence.indirection.IndirectList
    qGroup.getUsers().removeIf(u -> u.getUsername().equals(username));// doesn't work
}

请检查我是否删除了＆＃39;来自$ insu2

Answer 2

尝试将$insu2_name = $row[$_POST['insu2']];更改为$insu2_name = $row['insu2'];

希望这有帮助。

如何在mysql_fetch_assoc结果中传递post变量

2 个答案: