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问PHP MySQL第二次查询失败
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Stack Overflow用户
提问于 2013-04-11 05:09:31
回答 1查看 794关注 0票数 0

我的第二个MySQL查询似乎由于未知的原因而失败。只有在从$dbLink->query()中获得结果时,连接时才不会出现问题;

代码语言:javascript
复制
$photos = getPhotos(getID($_COOKIE[$uname_cookie], $_COOKIE[$passwd_cookie]));

function getID($uName, $passwd) {
    // Just in case, for modularity :P
    include 'misc.php';
    $query = "SELECT ID FROM Users WHERE UserName = '" . $uName . "' AND Password = '" . $passwd . "';";
    $dbLink = new mysqli($host, $user, $pass, $dbname);
    if ($dbLink->connect_errno) {
        echo "Failed to connect to database: " . $dbLink->connect_error;
        return null;
    } else {
        $results = $dbLink->query($query);
        print_r($results->fetch_array());
        $rows = $results->fetch_array();
        return $rows["ID"];
    }
}

function getPhotos($id) {
    // Just in case, for modularity :P
    include 'misc.php';
    $query = "SELECT Location FROM Photos WHERE UserID = " . $id . ";";
    $dbLink = new mysqli($host, $user, $pass, $dbname);
    if ($dbLink->connect_errno) {
        echo "Failed to connect to database: " . $dbLink->connect_error;
        return null;
    } else {
        $results = $dbLink->query($query);
        print_r($results->fetch_array());
        $rows = $results->fetch_array();
        return $rows["ID"];
    }
}

编辑:通过将$dbLink传递给函数IE getPhotos($id,$dbLink)获得它

EN

回答 1

Stack Overflow用户

发布于 2013-04-11 05:21:05

代码语言:javascript
复制
$query = "SELECT Location FROM Photos WHERE UserID = '" . $id . "';";
票数 0
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/15936468

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