下面是我的代码,在employee表中有一些记录,其pfuidno为null 但它没有在服务器上运行显示我的错误
$sql1="select a.code,a.empname,a.pfuidno from emplmast a ";1
$get1=mysqli_query($conn,$sql1) or die(mysqli_error());
$m_no = mysqli_num_rows($get1);
if ($m_no!=0)
{
while($row1 = mysqli_fetch_array($get1))
{
if ($row1['pfuidno']=='' OR empty($row1['pfuidno']))
{
$m_errmsg='UAN is Blank for Employee Code '.$row1['code'].' Name '.$row1['empname'];
echo "<script language='javascript' type='text/javascript'>alert('$m_errmsg')</script>";
}
}
}
答案 0 :(得分:2)
1.在查询代码行旁边删除1
。
2.在查询中指定列名称时使用_assoc()
。( 不强制 ,但会为您提供更轻的关联数组)
3. OR
需要||
4.需要添加NULL
的检查
代码必须是: -
$sql1="select a.code,a.empname,a.pfuidno from emplmast a "; // remove 1
$get1=mysqli_query($conn,$sql1) or die(mysqli_error());
$m_no = mysqli_num_rows($get1);
if ($m_no > 0){
while($row1 = mysqli_fetch_assoc($get1)){ // use _assoc
if ($row1['pfuidno']=='' || empty($row1['pfuidno']) || $row1['pfuidno'] === NULL){ //check for NULL
$m_errmsg='UAN is Blank for Employee Code '.$row1['code'].' Name '.$row1['empname'];
echo "<script language='javascript' type='text/javascript'>alert('$m_errmsg')</script>";
}
}
}