Php mysql查询中的变量

大家好，我这里的问题是，如果将$vehicle插入数据库查询$vquery=mysql_querySELECT*FROM vehicletbl，其中vehicle='$vehicle1'，则$vehicle不会返回其值；但如果它被回响，它将返回其值。这里的逻辑是，它将从VehicleBL中选择所有值，“vehicle”列中任何值的值都将等于$vehicle1。谢谢你的帮助

您的查询没有错误处理。尝试向查询调用添加一些调试：

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";

            }

代码的其余部分很难看，但看起来还行，所以开始看看为什么您没有从查询中得到任何东西

永远不要假设查询成功。

尝试以下方法进行调试：

$result = mysql_query(...) or die(mysql_error());

---

这就是我在sql中查找错误的方法。

Noob程序员？以下是一些需要知道的事情：

$sql = "SELECT * FROM vehicletbl WHERE vehicle = '" . $vehicle1 . "'";
$vquery = mysql_query($sql) or die(mysql_error() . "\n<br>$sql");

安全和安保：

$desti = "".$fetch['destination']." "." ".$fetch['location']."";
// WHY ?? Isn't that easier to do this ?
$desti = $fetch['destination']."  ".$fetch['location'];

---

还有其他的评论，但是首先尝试改进这些观点

没有按名称$vehicle分配任何变量？请显示$query OutputsHanks和yes的内容。代码的其余部分很难看。那是因为我还是一个noob程序员。

// Include the file before the loop ! You're including 20 times your file, but you just need to do it once ! Another thing to know:
include_once("dbconnect.php");

$desti = "".$fetch['destination']." "." ".$fetch['location']."";
// WHY ?? Isn't that easier to do this ?
$desti = $fetch['destination']."  ".$fetch['location'];

// Don't forget to escape your variables before putting it in mysql queries
$appid = mysql_real_escape_string($appid);
$selectquery = mysql_query("SELECT * FROM regform_admin WHERE tid = '$appid'");