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Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

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mysql_获取_数组(mysql)_Mysql_Arrays_Fetch - Fatal编程技术网

mysql_获取_数组(mysql)

mysql_获取_数组(mysql),mysql,arrays,fetch,Mysql,Arrays,Fetch,我是初学者,我有一个错误 Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Program Files\VertrigoServ\www\index.php on line 35 而代码 $sresult = mysql_query("SELECT code, location FROM banners"); while ($row_s = mysql_fetch

我是初学者,我有一个错误

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Program Files\VertrigoServ\www\index.php on line 35
而代码

$sresult = mysql_query("SELECT code, location FROM banners");
while ($row_s = mysql_fetch_array($sresult))
{
    $banner[$row_s["location"]]=$row_s["code"];
}

这个查询有问题。 尝试:

试试这个

$sresult = mysql_query("SELECT code, location FROM banners");
if (!$result) {
    die('Invalid query: ' . mysql_error());
}
while ($row_s = mysql_fetch_array($sresult))
{
    $banner[$row_s["location"]]=$row_s["code"];
}
检查错误是什么。

或die()
:-(:-)(-我一直不明白为什么人们一直在使用这个…甚至一开始就开始使用它…同意die(),我个人从来没有在我自己的代码中使用过它,但这是在这里写作时首先想到的,奇怪:)
$sresult = mysql_query("SELECT code, location FROM banners");
if (!$result) {
    die('Invalid query: ' . mysql_error());
}
while ($row_s = mysql_fetch_array($sresult))
{
    $banner[$row_s["location"]]=$row_s["code"];
}