在PHP中显示SQL查询结果时出错

当我试图在“php”中检索sql查询的结果时，它没有正确显示

我的结果是：

Array ( [0] => 0533 [SUBSTRING(cardnumber, 5, 4)] => 0533 )

而我只是想得到这样的结果

0533 

待展示

我的php文件：

<?php
    $con=mysqli_connect("localhost","root","wickedphat", "bluecard");
    // Check connection
    if (mysqli_connect_errno())
      {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
      }

    $result = mysqli_query($con,"SELECT SUBSTRING(cardnumber, 5, 4) FROM users WHERE username='".($_SESSION['username'])."'");

    while($row = mysqli_fetch_array($result))
      {
      print_r($row) . " " ;
      }

    mysqli_close($con);
    ?>

---

这些结果是正确的，并且与您提供的代码相符。有两点需要注意：

您应该为MySQL函数调用的结果提供一个别名，以便通过PHP更轻松地访问它

您需要在PHP中引用每个结果的标识符，而不是整个结果行

这应表明：

// Gave the result of SUBSTRING(cardnumber, 5, 4) the alias
// of "cardnum"
$result = mysqli_query($con,"SELECT SUBSTRING(cardnumber, 5, 4) as cardnum FROM users WHERE username='".($_SESSION['username'])."'");

while($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
{
    // accessing the value of cardnum by refering to its key in the
    // the array returned by mysqli_fetch_array()
    echo $row['cardnum'] . ' ';
}

---

我相信，如果需要（奇怪的）关联数组键，您需要mysqli\u fetch\u assoc（）

// Gave the result of SUBSTRING(cardnumber, 5, 4) the alias
// of "cardnum"
$result = mysqli_query($con,"SELECT SUBSTRING(cardnumber, 5, 4) as cardnum FROM users WHERE username='".($_SESSION['username'])."'");

while($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
{
    // accessing the value of cardnum by refering to its key in the
    // the array returned by mysqli_fetch_array()
    echo $row['cardnum'] . ' ';
}