Question

我有一个页面在我改变了一些小的东西后永远加载，我不知道为什么。我使用Chrome的web dev实用程序检查了该页面，它说它无限尝试$_GET某些东西，但我的代码中根本没有这样的请求。我怎样才能尝试调试并找出导致问题的原因？

这是我认为负责的代码块：

<?php
$con = mysql_connect(...);
if (!$con)
{
    die('Could not connect: ' . mysql_error());
}

mysql_select_db(db, $con);

$thisZone = $_SESSION["zoneSelected"];

$newSQL = mysql_query("SELECT id FROM coupons WHERE zone ='$thisZone'");

$numIDs = mysql_num_rows(mysql_query($newSQL));

$newCoups = array();
while($row = mysql_fetch_array($newSQL))
{
    $newCoups[] = $row;
}

//generates the 3 random

$randID1 = mt_rand(1, $numIDs);
do {   
    $randID2 = mt_rand(1, $numIDs);
}
while(in_array($randID2, array($randID1)));
do {   
    $randID3 = mt_rand(1, $numIDs);
}
while(in_array($randID3, array($randID1,$randID2)));

$randArr1 = mysql_fetch_array(mysql_query("SELECT retailerName,savingsDetails,dateExp,qrPicture FROM coupons WHERE id = '$randID1' AND zone = '$thisZone'"));
$randStats1 = mysql_fetch_array(mysql_query("SELECT views,saves,QRScans,prints FROM stats WHERE id = '$randID1'"));
$randArr2 = mysql_fetch_array(mysql_query("SELECT retailerName,savingsDetails,dateExp,qrPicture FROM coupons WHERE id = '$randID2' AND zone = '$thisZone'"));
$randStats2 = mysql_fetch_array(mysql_query("SELECT views,saves,QRScans,prints FROM stats WHERE id = '$randID2'"));
$randArr3 = mysql_fetch_array(mysql_query("SELECT retailerName,savingsDetails,dateExp,qrPicture FROM coupons WHERE id = '$randID3' AND zone = '$thisZone'"));
$randStats3 = mysql_fetch_array(mysql_query("SELECT views,saves,QRScans,prints FROM stats WHERE id = '$randID3'"));

//generates the 3 new

$coupsThisZone = count($newCoups);
$newID1 = $newCoups[($coupsThisZone - 1)];
$newID2 = $newCoups[($coupsThisZone - 2)];
$newID3 = $newCoups[($coupsThisZone - 3)];

$newArr1 = mysql_fetch_array(mysql_query("SELECT retailerName,savingsDetails,dateExp,qrPicture FROM coupons WHERE id = '$newID1' AND zone = '$thisZone'"));
$newStats1 = mysql_fetch_array(mysql_query("SELECT views,saves,QRScans,prints FROM stats WHERE id = '$newID1'"));
$newArr2 = mysql_fetch_array(mysql_query("SELECT retailerName,savingsDetails,dateExp,qrPicture FROM coupons WHERE id = '$newID2' AND zone = '$thisZone'"));
$newStats2 = mysql_fetch_array(mysql_query("SELECT views,saves,QRScans,prints FROM stats WHERE id = '$newID2'"));
$newArr3 = mysql_fetch_array(mysql_query("SELECT retailerName,savingsDetails,dateExp,qrPicture FROM coupons WHERE id = '$newID3' AND zone = '$thisZone'"));
$newStats3 = mysql_fetch_array(mysql_query("SELECT views,saves,QRScans,prints FROM stats WHERE id = '$newID3'"));
?>

Answer 1

do...while的目的是什么，如果你可以用if达到同样的简单效果？ while是一个非常危险的陈述，在大多数情况下总会出现无限循环的情况。

Answer 2

要获得随机x，您可以轻松地将id数组（$ newCoups）洗牌并获取它们的前x个。这将产生正确的结果。在您的代码中，您生成3个随机ID，假设那些id位于该区域中（在某些时候可能会失败）。您可以使用sql来获取新的和随机的，这也是最有效的方式。

随机3：

SELECT c.id, c.retailerName,c.savingsDetails,c.dateExp,c.qrPicture, s.views,s.saves,s.QRScans,s.prints
FROM coupons c
LEFT JOIN stats s ON c.id = s.id 
WHERE c.zone = '$thisZone'
ORDER BY RAND()
LIMIT 3

PHP页面永远加载

5 个答案: