Php SQL联接不返回任何结果_Php_Mysql_Sql

Php SQL联接不返回任何结果

php mysql sql

Php SQL联接不返回任何结果,php,mysql,sql,Php,Mysql,Sql,这是可行的，但我需要更多表中的数据 if(isset($_GET['id']) && $_GET['id'] != null) { $id = $_GET['id']; $sql = "SELECT `maps.name`, `maps.description`, `maps.date`, `maps.mcversion`, `maps.mapid`, `maps.category`, `maps.fo

这是可行的，但我需要更多表中的数据

if(isset($_GET['id']) && $_GET['id'] != null) {

    $id = $_GET['id'];

    $sql = "SELECT 
    `maps.name`,
    `maps.description`,
    `maps.date`,
    `maps.mcversion`,
    `maps.mapid`,
    `maps.category`,
    `maps.format`,
    `users.username`,
    `users.rank`,
    `users.verified`,
    `users.mcusername`,
    COUNT(`views.mapid`) AS `views`,
    COUNT(`likes.mapid`) AS `likes`,
    COUNT(`downloads.mapid`) AS `downloads`,
    COUNT(`subscribes.channelid`) AS `subscribers`
    FROM  `maps` INNER JOIN `users` ON `maps.userid` = `users.id` 
        INNER JOIN `views` ON `maps.mapid` = `views.mapid`
        INNER JOIN `likes` ON `maps.mapid` = `likes.mapid`
        INNER JOIN `downloads` ON `maps.mapid` = `downloads.mapid`
        INNER JOIN `subscribe` ON `mapid.userid` = `subscribe.channelid`
    WHERE `maps.mapid` = '$id'";

    $result = mysqli_query($con,$sql);

    if (mysqli_num_rows($result) > 0)   {
        echo “success”;
    } else {
        header("LOCATION: index.php");
    }

    $sql = "SELECT * FROM `maps` WHERE `id`=$id";
    $result = mysqli_query($con,$sql);

    if (mysqli_num_rows($result) > 0)   {
        viewer($id);
    } else {
        header("LOCATION: index.php");
    }

这个sql连接好吗？为什么它不返回任何结果

使用普通的

$sql=“SELECT*FROM-maps，其中id=$id”一切正常，但我也需要其他表中的数据。
解决方案：
$sql = "SELECT 
    `maps.name`,
    `maps.description`,
    `maps.date`,
    `maps.mcversion`,
    `maps.mapid`,
    `maps.category`,
    `maps.format`,
    `users.username`,
    `users.rank`,
    `users.verified`,
    `users.mcusername`,
    COUNT(`views.mapid`) AS `views`,
    COUNT(`likes.mapid`) AS `likes`,
    COUNT(`downloads.mapid`) AS `downloads`,
    COUNT(`subscribes.channelid`) AS `subscribers`
    FROM  `maps`
        INNER JOIN `users`     ON `maps.userid` = `users.id` 
        INNER JOIN `views`     ON `maps.mapid` = `views.mapid`
        INNER JOIN `likes`     ON `maps.mapid` = `likes.mapid`
        INNER JOIN `downloads` ON `maps.mapid` = `downloads.mapid`
        INNER JOIN `subscribe` ON `mapid.userid` = `subscribe.channelid`
    WHERE `maps.mapid` = '$id'";

谢谢你的帮助
 解决方案：
$sql = "SELECT 
    `maps.name`,
    `maps.description`,
    `maps.date`,
    `maps.mcversion`,
    `maps.mapid`,
    `maps.category`,
    `maps.format`,
    `users.username`,
    `users.rank`,
    `users.verified`,
    `users.mcusername`,
    COUNT(`views.mapid`) AS `views`,
    COUNT(`likes.mapid`) AS `likes`,
    COUNT(`downloads.mapid`) AS `downloads`,
    COUNT(`subscribes.channelid`) AS `subscribers`
    FROM  `maps`
        INNER JOIN `users`     ON `maps.userid` = `users.id` 
        INNER JOIN `views`     ON `maps.mapid` = `views.mapid`
        INNER JOIN `likes`     ON `maps.mapid` = `likes.mapid`
        INNER JOIN `downloads` ON `maps.mapid` = `downloads.mapid`
        INNER JOIN `subscribe` ON `mapid.userid` = `subscribe.channelid`
    WHERE `maps.mapid` = '$id'";

谢谢你的帮助 如果您想保护一个复杂的sql语句，您会如何做？
这个版本可以吗
$sql = "SELECT 
    maps.name,
    maps.description,
    maps.date,
    maps.mcversion,
    maps.mapid,
    maps.category,
    maps.format,
    users.username,
    users.rank,
    users.verified,
    users.mc_username,
    (SELECT COUNT(*) FROM likes WHERE likes.mapid = maps.id) AS likes,
    (SELECT COUNT(*) FROM downloads WHERE downloads.mapid = maps.id) AS downloads,
    (SELECT COUNT(*) FROM subscribe WHERE subscribe.channelid = maps.userid) AS subscribers,
    (SELECT COUNT(*) FROM views WHERE views.mapid = maps.id) AS viewers
    FROM  maps
    INNER JOIN users 
        ON maps.userid = users.id
    WHERE maps.id = '$id'";

数据库连接：
if(isset($_GET['id']) && $_GET['id'] != null) {
  $id = $_GET['id'];

  $stmt = $mysqli->prepare('SELECT id FROM maps WHERE id = ?');
  $stmt->bind_param('i', $id);

  $stmt->execute();

  $result = $stmt->get_result();

  if (mysqli_num_rows($result) == 1)    {
    $row = $result->fetch_assoc();
      $secid = $row["id"];
  } else {
      echo "error2";
  }

  $sql = "SELECT 
  maps.name,
  maps.description,
  maps.date,
  maps.mcversion,
  maps.mapid,
  maps.category,
  maps.format,
  users.username,
  users.rank,
  users.verified,
  users.mc_username,
  (SELECT COUNT(*) FROM likes WHERE likes.mapid = maps.id) AS likes,
  (SELECT COUNT(*) FROM downloads WHERE downloads.mapid = maps.id) AS downloads,
  (SELECT COUNT(*) FROM subscribe WHERE subscribe.channelid = maps.userid) AS subscribers,
  (SELECT COUNT(*) FROM views WHERE views.mapid = maps.id) AS viewers
  FROM maps
  INNER JOIN users 
      ON maps.userid = users.id
  WHERE maps.id = '$secid'";

  $result = mysqli_query($con,$sql);

  if (mysqli_num_rows($result) > 0) {
      $row = mysqli_fetch_assoc($result);
      echo $row["name"];
  } else {
      echo "error3";
  }

} else {
    echo "error1";
}

如果您想保护一个复杂的sql语句，您将如何做？
这个版本可以吗
$sql = "SELECT 
    maps.name,
    maps.description,
    maps.date,
    maps.mcversion,
    maps.mapid,
    maps.category,
    maps.format,
    users.username,
    users.rank,
    users.verified,
    users.mc_username,
    (SELECT COUNT(*) FROM likes WHERE likes.mapid = maps.id) AS likes,
    (SELECT COUNT(*) FROM downloads WHERE downloads.mapid = maps.id) AS downloads,
    (SELECT COUNT(*) FROM subscribe WHERE subscribe.channelid = maps.userid) AS subscribers,
    (SELECT COUNT(*) FROM views WHERE views.mapid = maps.id) AS viewers
    FROM  maps
    INNER JOIN users 
        ON maps.userid = users.id
    WHERE maps.id = '$id'";

数据库连接：
if(isset($_GET['id']) && $_GET['id'] != null) {
  $id = $_GET['id'];

  $stmt = $mysqli->prepare('SELECT id FROM maps WHERE id = ?');
  $stmt->bind_param('i', $id);

  $stmt->execute();

  $result = $stmt->get_result();

  if (mysqli_num_rows($result) == 1)    {
    $row = $result->fetch_assoc();
      $secid = $row["id"];
  } else {
      echo "error2";
  }

  $sql = "SELECT 
  maps.name,
  maps.description,
  maps.date,
  maps.mcversion,
  maps.mapid,
  maps.category,
  maps.format,
  users.username,
  users.rank,
  users.verified,
  users.mc_username,
  (SELECT COUNT(*) FROM likes WHERE likes.mapid = maps.id) AS likes,
  (SELECT COUNT(*) FROM downloads WHERE downloads.mapid = maps.id) AS downloads,
  (SELECT COUNT(*) FROM subscribe WHERE subscribe.channelid = maps.userid) AS subscribers,
  (SELECT COUNT(*) FROM views WHERE views.mapid = maps.id) AS viewers
  FROM maps
  INNER JOIN users 
      ON maps.userid = users.id
  WHERE maps.id = '$secid'";

  $result = mysqli_query($con,$sql);

  if (mysqli_num_rows($result) > 0) {
      $row = mysqli_fetch_assoc($result);
      echo $row["name"];
  } else {
      echo "error3";
  }

} else {
    echo "error1";
}

一次删除一个联接，直到得到结果。然后找出您刚才删除的内容找不到预期匹配项的原因。您需要删除背景标记（并确保没有使用任何保留字），或者需要正确使用它们。需要在表名周围设置一个集合，在列名周围设置另一个集合。`maps.name`
不正确。它应该是`maps`.`name`
。对于所有其他类似的实例，情况也是如此。我不能用其他方式来表达。警告：您完全可以使用参数化的预处理语句，而不是手动生成查询。它们由或提供。永远不要相信任何类型的输入，尤其是来自客户端的输入。即使您的查询仅由受信任的用户执行，也要一次删除一个联接，直到得到结果。然后找出您刚才删除的内容找不到预期匹配项的原因。您需要删除背景标记（并确保没有使用任何保留字），或者需要正确使用它们。需要在表名周围设置一个集合，在列名周围设置另一个集合。`maps.name`
不正确。它应该是`maps`.`name`
。对于所有其他类似的实例，情况也是如此。我不能用其他方式来表达。警告：您完全可以使用参数化的预处理语句，而不是手动生成查询。它们由或提供。永远不要相信任何类型的输入，尤其是来自客户端的输入。即使您的查询仅由受信任的用户执行，。警告：您完全可以使用参数化的预处理语句，而不是手动生成查询。它们由或提供。永远不要相信任何类型的输入，尤其是来自客户端的输入。即使您的查询仅由受信任的用户执行，。您是否说此选择查询可能不安全？您能否提供一个类似的sql选择查询，该查询也会执行相同的操作，但处于安全模式？我真的很想学习！：）用占位符替换“$id”
，然后绑定数据。不要在SQL中放入任何变量。哦，我开始理解了，直到现在我才使用了一个简单的函数，从插入到表中的字符串中删除特殊字符。您能告诉我如何绑定它吗？警告：您对参数化预处理语句非常开放，应该真正使用参数化预处理语句，而不是手动构建查询。它们由或提供。永远不要相信任何类型的输入，尤其是来自客户端的输入。即使您的查询仅由受信任的用户执行，。您是否说此选择查询可能不安全？您能否提供一个类似的sql选择查询，该查询也会执行相同的操作，但处于安全模式？我真的很想学习！：）用占位符替换“$id”
，然后绑定数据。不要在SQL中放入任何变量。哦，我开始理解了，直到现在我才使用了一个简单的函数，从插入到表中的字符串中删除特殊字符。你能告诉我怎么装订吗？