Question

您好，我有一个网站，它接收一个汽车品牌，并生成一个与数据库对应的车辆ID。尽管有许多解决方案可用于将sql查询结果存储到变量，但在这种情况下它似乎不起作用。

我已经在数据库中尝试过sql语句，它可以使用1个结果输出，这是正确的。

<?php 
session_start();
include "dbconn.php";


$carcat = $_SESSION['selectedcarcat'];
$carbrand = $_POST['carbrand'];
$userid = $_SESSION['loginid'];
$username = $_SESSION['loginname'];
$startdate = $_POST['date1'];
$enddate = $_POST['date2'];
$pick = $_POST['pickuploc'];
$return = $_POST['returnloc'];
$calqty = 0;


    $selcaridsql = "SELECT carid FROM cars WHERE brand='$carbrand' ";

    $caridresult = $dbcnx->query($selcaridsql);

    $caridrow = mysql_fetch_object($caridresult);

    $carid = $caridrow['carid'];

    if (!$caridresult) 
    {
        $errmessage = "Your carid select query failed.";
        echo "<script type='text/javascript'>alert('$errmessage');</script>";
    }


    echo '<br>Debug 1 ';
    echo '<br>The selected qty is '
        .$qtyresult1.'<br />';
    echo '<br>The calculated qty is '
        .$calqty.'<br />';
    echo '<br>The content carid is '
        .$carid.'<br />';
    echo '<br>The content userid is '
        .$userid.'<br />';
    echo '<br>The content start is '
        .$startdate.'<br />';
    echo '<br>The content end is '
        .$enddate.'<br />';
    echo '<br>The content pick is '
        .$pick.'<br />';
    echo '<br>The content return is '
        .$return.'<br />';
        echo '<br>The content carbrand is '
        .$carbrand.'<br />';
?>

错误发生在＆＃34;内容carid为＆＃34;时显示空白结果。通过论坛后，似乎变量$carid没有值，我可能错误地推断。

回声结果：

＆＃13;

Debug 1 
The selected qty is 

The calculated qty is 0

The content carid is 

The content userid is 

The content start is 2016-10-28

The content end is 2016-10-29

The content pick is jurong

The content return is bishan

The content carbrand is Honda

＆＃13;

Answer 1

显然在使用var_dump来调试Chris所述的问题之后，问题被缩小到mysql_fetch_object($caridresult);收集的NULL值。由于数据库连接使用@$dbcnx = new mysqli('localhost','values','values','values');，因此DB类检索数组值是错误的。因此，解决方案将是mysqli_fetch_array。

这些代码可能对尝试调试的人有所帮助：

＆＃13;

<?php // register.php
session_start();
include "dbconn.php";


$carcat = $_SESSION['selectedcarcat'];
$carbrand = $_POST['carbrand'];
$userid = $_SESSION['loginid'];
$username = $_SESSION['loginname'];
$startdate = $_POST['date1'];
$enddate = $_POST['date2'];
$pick = $_POST['pickuploc'];
$return = $_POST['returnloc'];
$calqty = 0;


    $selcaridsql = "SELECT carid FROM cars WHERE brand='$carbrand' ";
    
    echo $selcaridsql."<br>";

    $caridresult = $dbcnx->query($selcaridsql);

    echo "<br>".var_dump($caridresult);

    if ($caridresult->num_rows >0 )
      {
        // if they are in the database register the user id
        echo '<br>Hello more than 1 <br>';  
      }

    else
    {
        echo '<br>Hello less than 1 <br>';
    }


    $caridrow = mysqli_fetch_array($caridresult);

    echo var_dump($caridrow)."<br>"; 

    $carid = $caridrow['carid'];

    echo var_dump($carid)."<br>"; 
    
    if (!$caridresult) 
    {
        $errmessage = "Your carid select query failed.";
        echo "<script type='text/javascript'>alert('$errmessage');</script>";
    }

    echo '<br>Debug 1 ';
    echo '<br>The selected qty is '
        .$qtyresult1.'<br />';
    echo '<br>The calculated qty is '
        .$calqty.'<br />';
    echo '<br>The content carid is '
        .$carid.'<br />';
    echo '<br>The content userid is '
        .$userid.'<br />';
    echo '<br>The content start is '
        .$startdate.'<br />';
    echo '<br>The content end is '
        .$enddate.'<br />';
    echo '<br>The content pick is '
        .$pick.'<br />';
    echo '<br>The content return is '
        .$return.'<br />';
        echo '<br>The content carbrand is '
        .$carbrand.'<br />';
?>

＆＃13;

特别感谢Chris帮助找到错误！

Answer 2

我现在无法对此进行测试。

但是，您已确认$carid没有值（为空）。

您已确认SQL语句在针对数据库运行时返回正确的值。因此，可能的问题在于您使用代码发送到数据库的查询。

您的查询：

$selcaridsql = "SELECT carid FROM cars WHERE brand='$carbrand' ";

我建议您修复查询：

$selcaridsql = "SELECT carid FROM cars WHERE brand='" + $carbrand + "'";

您可以发送字符串文字“$carbrand”，除非您打印出要检查的SQL语句，否则不知道它。

你试过吗？

此外，请检查所有$_POST的值，以确保它们符合您的想法。这对安全也非常重要。

获取那些println语句，让我知道你发现了什么： - ）。

打印出所有内容并确保您的价值符合您的预期。

我99％肯定这就是问题所在。

将单个SQL SELECT语句查询结果分配给PHP变量

2 个答案: