Question

我一直收到错误“未定义的变量：myquery - 查询为空”我已经尝试了许多不同的事情几个小时但仍然无法让它工作。我是非常新的PHP，发现它令人困惑。所以请详细解释或链接我有用的东西，以便我能更好地理解它。这是代码：

 <?php
include_once("scripts/connect_db.php");
$totals = "";
$rating = "";
$sql = mysql_query("SELECT `ratings` FROM `blog_posts` WHERE `id`='1'");
$result = mysql_query($myquery) or die(mysql_error());
while($row = mysql_fetch_array($sql)){
$myNums = $row["ratings"];
    $kaboom = explode(",", $myNums);
    $result = array_count_values($kaboom);
    foreach($result as $key => $value){
        if ($value =="1"){
            $howMany = "person";
            }else{
                $howMany = "people";
            }
            if($key ==""){
                $pic = "images/starsNorm.png";
            }
            else if($key == "1"){
                $stars = "star";
                $pic = "images/1lit.png";
            }else if($key == "2"){
                $stars = "stars";
                $pic = "images/2lit.png";
            }else if($key == "3"){
                $stars = "stars";
                $pic = "images/3lit.png";
            }else if($key == "4"){
                $stars = "stars";
                $pic = "images/4lit.png";
            }else if($key == "5"){
                $stars = "stars";
                $pic = "images/5lit.png";
            }
$totals .= '<p class="small" style="color:#32CD32;">' . $key . ' ' . $stars . ': <img src="' . $pic . '" alt="stars" />
        ' . $value . ' ' .$howMany . '</p>';
    }
    $count = count($kaboom);
    $sum = array_sum($kaboom);
    $avg = $sum / $count;
    $roundit = floor($avg);

    if($roundit == 0) {
        $rating = '<p class="small" style="color:#32CD32;">This ... has not yet been rated. You can be first!</p>';
    }else if ($count == 1) {
        $rating = '<p class="small" style="color:#32CD32;">Current Article Rating: ' . $roundit . '/5 stars <img id="myStars" src="images/starsNorm.png"
        alt="stars"/></p>';
    }else if($count > 1) {
        $rating = '<p class="small" style="color:#32CD32;">Current Article Rating: ' . $roundit . '/5 stars <img id="myStars" src="images/starsNorm.png"
        alt="stars"/></p>';
    }else{
        $rating = "sorry there is an error in the system... please try refreshing the page";
    }
}
?>

Answer 1

此行无效：

$result = mysql_query($myquery) or die(mysql_error());

您没有在任何地方定义$ myquery，这正是错误消息告诉您的内容。

我猜它上面的那一行应该只是一个字符串然后那是你要传入的变量。

E.g。

$sql = "SELECT `ratings` FROM `blog_posts` WHERE `id`='1'";
$result = mysql_query( $sql ) or die(mysql_error());

Answer 2

你没有定义$ myquery变量。你有$ sql。试试这个：

$ result = mysql_query（$ sql）或die（mysql_error（））;

Answer 3

试试这个

$result = mysql_query("SELECT `ratings` FROM `blog_posts` WHERE `id`='1'");

while ($row = mysql_fetch_array($result)) {  

}

Answer 4

如果你确定连接，查询，获取代码是正确的那么可能是mysql_connect正在连接但没有看到你的数据库。运行此代码以确保找到您的数据库

mysql_connect('localhost') or die ("Connect error");

$res = mysql_query("SHOW DATABASES");
while ($row = mysql_fetch_row($res)) {
    echo $row[0], '<br/>';
}

如果找不到，那么您需要修改数据库的权限。例如，进入相关数据库的“权限”选项卡，将“用户名”设置为“任何用户”并检查适用的权限。

未定义的变量：myquery - 查询为空

4 个答案: