for ($i=0; $i<$count; $i++) {
$appid = $chk[$i];
include "dbconnect.php";
$selectquery = mysql_query("SELECT * FROM regform_admin WHERE tid = '$appid'");
$fetch = mysql_fetch_array($selectquery);
$tid = $fetch['tid']; $username = $fetch['username']; $c_month = $fetch['month']; $c_day =$fetch['day']; $c_year = $fetch['year'];
$c_month2 = $fetch['month2']; $c_day2 =$fetch['day2']; $c_year2 = $fetch['year2'];
$pickup = "".$c_month."/".$c_day."/".$c_year."";
$return = "".$c_month2."/".$c_day2."/".$c_year2."";
$pickuploc = "".$fetch['pickupret']." "." ".$fetch['speclocation']."";
$desti = "".$fetch['destination']." "." ".$fetch['location']."";
$vehicle1 = $fetch['vehicle1'];
$datesent = date("n j, Y; G:i"); ;
$rand = rand(98765432,23456789);
include "vehicledbconnect.php";
$vquery = mysql_query("SELECT * FROM vehicletbl WHERE vehicle = '$vehicle1'");
$getvquery = mysql_fetch_array($vquery);
$maxcars = $getvquery['maxcars'];
$carsleft = $getvquery['carsleft'];
if ($carsleft == 0) {
echo '
<script language="JavaScript">
alert("Cannot move reservation to Pending for payment status. No available vehicles left for this reservation.");
</script>';
echo "$vehicle1";
}
嗨,大家好我这里的问题是,如果它插入到数据库查询$车辆没有返回它的值($ vquery =请求mysql_query(“SELECT * FROM vehicletbl WHERE车辆=“$ vehicle1'“);)但是如果它被回显,它会返回它的值。这里的逻辑是,它将选择车辆1中的所有值,其中“车辆”列中的任何值的值将等于$ vehicle1。谢谢您的帮助!变量在MySQL查询
什么都不是通过名称$车辆分配给变量? – Tobias
请显示'$ query'输出的内容 –