我有一个带有MYSQL的PHP代码,其中代码获取数据库选择并在表格中显示结果。
我需要的是能够检查所选变量是否为空。
如果它是空的我不想在表格中显示该字段。
$sql = $wpdb->prepare("select i.siteID
, i.siteNAME
, i.equipmentTYPE
, c.latitude
, c.longitude
, c.height
, o.ownerNAME
, o.ownerCONTACT
, x.companyNAME
, y.subcontractorCOMPANY
, y.subcontractorNAME
, y.subcontractorCONTACT
from site_info i
LEFT
JOIN owner_info o
on i.ownerID = o.ownerID
LEFT
JOIN company_info x
on i.companyID = x.companyID
LEFT
JOIN subcontractor_info y
on i.subcontractorID = y.subcontractorID
LEFT JOIN site_coordinates2 c
on i.siteID=c.siteID
where
i.siteNAME = %s
AND
o.ownerID = %d
AND
x.companyID = %d
",$site_name,$owner_name,$company_name);
$query_submit =$wpdb->get_results($sql, OBJECT);
echo "<br>";
echo "<br>";
// table that will dsiplay the results based on the user's selection //
echo "<table class='t1' width='30%'> ";
echo "<tr>";
echo "<th>Site Name</th>";
echo "<th>Owner Name</th>";
echo "<th>Company Name</th>";
//echo "<th>Subcontractor Name</th>";
echo "<th>Site ID</th>";
echo "<th>Equipment Type</th>";
echo "<th> Lattitude</th>";
echo "<th>Longitude </th>";
echo "<th> Height</th>";
echo "<th> Owner Contact</th>";
echo "<th> Sub Contact</th>";
echo "<th> Sub company Name</th>";
echo "</tr>";
foreach ($query_submit as $obj) {
echo "<tr>";
echo "<td>".$obj->siteNAME."</td>";
echo "<td>".$obj->ownerNAME."</td>";
echo "<td>".$obj->companyNAME."</td>";
if(!$obj->subcontractorNAME){
echo "<th>Subcontractor Name</th>";
echo "<td>".$obj->subcontractorNAME."</td>";
}
else{
}
echo "<td>".$obj->siteID."</td>";
echo "<td>".$obj->equipmentTYPE."</td>";
echo "<td>".$obj->latitude."</td>";
echo "<td>".$obj->longitude."</td>";
echo "<td>".$obj->height."</td>";
echo "<td>".$obj->ownerCONTACT."</td>";
echo "<td>".$obj->subcontractorCONTACT."</td>";
echo "<td>".$obj->subcontractorCOMPANY."</td>";
echo "</tr>";
}