MySQL选择查询无法正常工作。

Question

<?php
$hostname='localhost';
$user='root';
$password='';
$connect=mysqli_connect($hostname,$user,$password,'a_railway');

if(!$connect)
 {
   die('Could not connect');
  }

$rs="central";
$rd="central";
$s="A";
$d="B";

$dist=call_dist($rs,$rd,$s,$d,$connect);
echo $dist;

function call_dist($rs,$rd,$s,$d,$connect)
{
    $src_dist="SELECT Distance FROM $rs WHERE Station=$s ";
    $dest_dist="SELECT Distance FROM $rs WHERE Station=$d "; 
    $src_dist=mysqli_query($connect,$src_dist);
    $dest_dist=mysqli_query($connect,$dest_dist);
     $dist=abs($src_dist-$dest_dist);
    return $dist;
}
?>

为什么我的Sql查询不起作用？ 数据库没有问题。 但查询不会执行。 我想在此程序中找到通过减去点A和B的值得出的距离。

Answer 1

如果station是varchar，则需要在字符串前后加上引号。

$src_dist="SELECT Distance FROM $rs WHERE Station='$s' ";

但是，你不应该永远像一个字符串插入值。 请使用准备好的语句！

阅读此参考资料，它将使您的代码和工作变得更加轻松： http : //php.net/manual/en/mysqli.quickstart.prepared-statements.php

编辑您也可以在sql中进行数学运算：

$dist_query = mysqli_query("SELECT ABS(s.Distance - d.Distance) as dist FROM
                ( SELECT Distance FROM $rs WHERE Station='$s' ) as s,
                ( SELECT Distance FROM $rs WHERE Station='$d' ) as d");
$dist_result = $dist_query->fetch_assoc();
return $dist_result[0]['dist'];