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我在 php 页面上使用了以下代码,我正试图让它工作,它运行第一个查询很好(这是一个插入)并将正确的值返回给 $orderid
但它不会返回接下来选择查询的任何内容,如果我取出变量 $myorderid 并输入一个数字,它将返回预期的结果,但是一旦我尝试使用变量,它什么也不返回。
我尝试过使用 mysql_escape_string、sprintf 和现在的 intval 方法,但似乎没有任何效果。
mysql 没有返回错误,只是一个空的结果集。

mysql_query($sSQL) or print(mysql_error());
$orderid = mysql_insert_id();

$myorderid =    $orderid;
$myquery = "SELECT cartOrderID, Sum(pDonationAmt) as amount 
FROM products INNER JOIN cart ON ( cartProdID = pID ) 
WHERE cartOrderID = " . intval($myorderid). " GROUP BY cartOrderID ";

$myresults = mysql_query($myquery);
while ($row = mysql_fetch_assoc($myresults)) 
  { error_log('orderid inserted is:.$orderid.' OrderID in   mysql:'.$row['cartOrderID'].' and the total is: '.$row['amount'], 3, 'inc/atestingdata.txt'); }

感谢您对此的任何帮助!

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2 回答 2

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尝试intval($myorderid)用引号括起来'

 $myquery = "SELECT cartOrderID, Sum(pDonationAmt) as amount 
 FROM products INNER JOIN cart ON ( cartProdID = pID ) 
 WHERE cartOrderID = '" . intval($myorderid). "' GROUP BY cartOrderID ";
于 2012-11-02T19:49:16.353 回答
0

旧的“回声”可以帮助...尝试:

$myquery = "SELECT cartOrderID, Sum(pDonationAmt) as amount 
 FROM products INNER JOIN cart ON (pID = cartProdID) 
 WHERE cartOrderID = $orderid GROUP BY cartOrderID ";
echo $myquery;
于 2012-11-02T19:59:16.803 回答