Question

所以我目前有一个数据库表，我试图选择所有记录，除了那些包含在我所创建的数组中的记录。作为一些背景背景：

有问题的数据库表的结构是：

server_status:
id int(11)
server_id int(11)
time_checked datetime
status char(1)

将数据导入哈希的PHP脚本如下所示：

$sql2 = "SELECT server_id, time_checked,id from server_status where   time_checked<'$date' order by server_id;";
$result2=$conn->query($sql2);
while($row2 = $result2->fetch_assoc()){

 $server_id = $row2['server_id'];
 $id = $row2['id'];
 $dt = $row2['time_checked'];

 $year = substr($dt,0,4);
 $month = substr($dt,5,2);
 $day = substr($dt,8,2);

 $day = "$year-$month-$day";

 $all[$server_id][$day] = $id;  // ARRAY

}

所以我要做的是创建一个MySQL查询，从数组中读取id（$ id），并从中选择* APART。从查找起来，似乎我将不得不使用'where not'子句，但我不知道如何引用哈希。

进一步澄清：我现在有一个数组，它提取如下所示的数据：

1{
2016-05-05 : 252
2016-05-10 : 406
2016-04-27 : 141
2016-05-04 : 164
2016-05-09 : 263
2016-05-03 : 153
2016-04-26 : 131
2016-04-14 : 1
2016-04-18 : 31
2016-04-21 : 111
2016-04-20 : 61
2016-04-19 : 51
2016-04-15 : 21
2016-04-25 : 121
}
2{
2016-05-10 : 452
2016-05-05 : 198
2016-05-09 : 264
2016-05-04 : 165
2016-04-26 : 132
2016-04-27 : 143
2016-04-25 : 122
2016-04-21 : 112
2016-05-03 : 154
}

我想从这个数组中获取ID（例如154）并选择表中没有任何上述ID的所有内容。我希望这有助于澄清？！

任何帮助将不胜感激！

Answer 1

如果$all是要从中提取不需要的ID的数组，则可能是您提供的代码后需要的内容：

$ids_to_exclude = array();

// iterate through servers
foreach ($all as $server_id => $dates) {
    // iterate through dates of each server
    foreach ($dates as $date => $id) {
        // If a value is not in the array, add it.
        // In case ids don't repeat, you won't need this if
        if (!in_array($id, $ids_to_exclude)) {
             // add $id to the array
             $ids_to_exclude[] = $id;
        }
    }
}

$sql_condition = "where `id` not in (".implode(",",$ids_to_exclude).")";

使用字符串连接编写查询时要小心。阅读SQL Injection以及如何防止它。使用Prepared Statements代替纯连接。

Answer 2

我想你只想要一个NOT IN，对吧？

$sql2 = "SELECT id, server_id, time_checked,id 
         FROM server_status 
         WHERE 
            id NOT IN (".implode(',', $id_array)."
            AND time_checked<'$date' 
         ORDER BY server_id;";

$result2=$conn->query($sql2);

while($row2 = $result2->fetch_assoc()){   
   $server_id = $row2['server_id'];
   $id = $row2['id'];
   $dt = date('Y-m-d', strtotime($row2['time_checked']));
   $all[$server_id][$dt] = $id;  // ARRAY 

}

Answer 3

SOLVE：

$sql2 = "SELECT server_id, time_checked,id from server_status where time_checked<'$date' order by server_id;";
$result2=$conn->query($sql2);
while($row2 = $result2->fetch_assoc()){


 while($row2 = $result2->fetch_assoc()){
    $server_id = $row2['server_id'];
    $id = $row2['id'];
    $dt = date('Y-m-d', strtotime($row2['time_checked']));
    $all[$server_id][$dt] = $id;  // ARRAY
 }
}

 $stack = array();
  $keys = array_keys($all);
  for($i = 0; $i < count($all); $i++) {
      foreach($all[$keys[$i]] as $key => $value) {
        array_push($stack, $value);
      }
  }


$ids = join(',',$stack);
$sql = "SELECT * FROM server_status WHERE time_checked<'$date' AND id NOT IN ($ids)";
$result=$conn->query($sql);
echo "Server status data has been deleted.<br>";

从多维数组中创建另一个数组以仅存储id，并像John Green建议的那样使用NOT IN。

谢谢！

MySQL，PHP：从表中选择*，其中id不在数组中

3 个答案: