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使用php和mysql获取url的问题
原标题:problem with get url using php and mysql

很抱歉,如果这个问题对一些人来说似乎很容易,但我似乎无法弄清楚。有人告诉我可以来这里,因为这里的人很乐于助人。我对以下代码有问题。当uid被调用到<code>examplepage.php的url中时?uid=5如果我执行page.php,我会得到“一些代码”吗?uid=letters我被重定向到page.php?uid=1。这很好。但是如果用户应该输入<code>page.php?uid=1个字母我得到这个错误。。where子句中的未知列1gh

警告:mysql_fetch_array()要求参数1为resource,在C:wampwwwpage.php的第173行给出布尔值。

只有当用户在get-url-id的末尾输入不需要的字符时,我才会收到这个错误。我该如何防止这种情况发生?我该如何将其重定向到page.php?uid=1。。。请参阅下面的代码

$id = mysql_real_escape_string(@$_GET[ uid ]);

$query = mysql_query("SELECT user.* FROM user WHERE id =  $id ");

if ((mysql_num_rows($query)==0)) {
    header("location:page.php?uid=1");
    die();
}
else {
    while($rows = mysql_fetch_array($query)){
        $foo = $row[ foo ];

        echo "some code";
最佳回答

mysql_query在出现错误时返回false。您应该检查它返回的内容:

$query = "SELECT user.* FROM user WHERE id =  $id "
$result = mysql_query($query);

if ($result === false) {
    die($query. <br/> .mysql_error());
}

然后你就能理解它为什么失败了。我已经将查询添加到die()语句中,因此您也可以手动尝试该查询。

给出的错误是字符串的未知列,这表明您在SQL查询中没有用单引号括起该值,即使问题中的代码是这样做的。

在生产中,您应该采取一切可能的措施来确保错误得到妥善处理。在这种情况下,您可能会想要与未找到用户相同的行为:

header("location:page.php?uid=1");
问题回答

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