mysql_fetch_array 有时会失败

发布于 2024-11-02 09:24:30 字数 819 浏览 2 评论 0原文

我正在尝试实现与在线支付框架的连接。

其中一个文件给我带来了一些麻烦，因为有时代码可以工作，有时则不能...我不明白为什么...

这就是代码失败的地方...

$sql = "select transidmerchant,totalamount from nsiapay where     transidmerchant='".$order_number."'and trxstatus='Verified'";
$result = mysql_query($sql);
**$checkout = mysql_fetch_array($result);**
echo "sql : ".$sql;
$hasil=$checkout['transidmerchant'];
echo "hasil: ".$hasil;
$amount=$checkout['totalamount'];
echo "amount: ".$amount;
    // Custom Field
if (!$hasil) {
  echo 'Stop1';
} else {
    if ($status=="Success") {}
}

这只是代码的一部分，但是我认为您尝试查看问题就足够了...它在粗体行上失败，$checkout = mysql_fetch_array($result); 奇怪的是，“echo sql”有效，并且它显示了正确的值，但是当我将它们放在数组中时，有时变量会被传递，有时则不会......所以，当到达 < code>if (!$hasil) 它会失败，因为该值为空...但有时它会起作用...

对可能发生的情况有什么想法吗？

谢路易斯

原文

I'm trying to implement the connection to a online payment framework.

One of the files is giving me some trouble, because sometimes the code works, sometimes it doesn't... And I can't understand why...

Here's where the code is failing...

$sql = "select transidmerchant,totalamount from nsiapay where     transidmerchant='".$order_number."'and trxstatus='Verified'";
$result = mysql_query($sql);
**$checkout = mysql_fetch_array($result);**
echo "sql : ".$sql;
$hasil=$checkout['transidmerchant'];
echo "hasil: ".$hasil;
$amount=$checkout['totalamount'];
echo "amount: ".$amount;
    // Custom Field
if (!$hasil) {
  echo 'Stop1';
} else {
    if ($status=="Success") {}
}

It's just part of the code but I think it's enough for you to try to see the problem... It fails on the bold line, $checkout = mysql_fetch_array($result);
The weird thing is that the "echo sql" works, and it shows the right values, but then when I put them on the array, sometimes the variables are passed, sometimes they're not... And so, when getting to if (!$hasil) it fails because the value is empty... but sometimes it works...

Any ideas on what might be happen?

Thans
Luis

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￡冰雨忧蓝° 2024-11-09 09:24:30

失败的唯一原因是查询不返回任何内容。

正确的方法是检查是否有返回的东西：

$sql = "select transidmerchant,totalamount from nsiapay where     transidmerchant='".$order_number."'and trxstatus='Verified'";
$result = mysql_query($sql);
if($checkout = mysql_fetch_array($result)){
    $hasil = $checkout['transidmerchant'];
    echo "hasil: ".$hasil;
    $amount=$checkout['totalamount'];
    echo "amount: ".$amount;
        // Custom Field
    if (!$hasil) {
      echo 'Stop1';
    } else {
        if ($status=="Success") {}
    }
}else{
    echo "Empty query result";
}

The only way this fail is when query doesn't return anything.

The correct way would be to check if there is something returned:

$sql = "select transidmerchant,totalamount from nsiapay where     transidmerchant='".$order_number."'and trxstatus='Verified'";
$result = mysql_query($sql);
if($checkout = mysql_fetch_array($result)){
    $hasil = $checkout['transidmerchant'];
    echo "hasil: ".$hasil;
    $amount=$checkout['totalamount'];
    echo "amount: ".$amount;
        // Custom Field
    if (!$hasil) {
      echo 'Stop1';
    } else {
        if ($status=="Success") {}
    }
}else{
    echo "Empty query result";
}

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