对数组中的每个结果运行php查询

我有两个数据库。Database one有一个上传文档的列表，该列表按标题、id、收件人（按位置）以及文档在数据库中的位置存储。第二个数据库在每次阅读该书时存储详细信息。这将存储id（取自第一个数据库）和登录的用户名

我有两个问题。第一个查找第二个数据库，并将当前用户号的所有ID存储为一个数组。（echo用于测试。）第二个数据库查找第一个数据库，列出所有未读取的文档，但查找与找到的详细信息不同的项目。但是，我一直在通过第一个查询找到的每个id执行第二个查询循环，因此它只搜索第一个编号

这是我的代码：

<?php
$connection = mysql_connect("host","database","password") or die("Couldn't make connection."); 
$db = mysql_select_db("database", $connection) or die("Couldn't select database."); 
$sql = "SELECT *  FROM `READ` WHERE `name` = '$user'";                      
$sql_result = mysql_query($sql,$connection) or die("Couldn't execute query 1."); 

while($row1 = mysql_fetch_array($sql_result))                           
{
   echo "<br>".$row1['item'];
   $ident = $row1["item"];                                              
}

mysqli_close($connection);                                          
?>

<?php
$connection = mysql_connect("host","database","password") or die("Couldn't make connection."); 
    $db = mysql_select_db("database", $connection)  or die("Couldn't select database."); 
$query ="SELECT * FROM UPLOAD WHERE id NOT LIKE '$ident' AND faoGrade LIKE '%$gr%'";       
   //$ident only searching the first number stored
$query_result = mysql_query($query,$connection) or die("Couldn't execute query 2.");  
while($row2 = mysql_fetch_array($query_result))                             
{
   echo "<br>".$row2["title"];      
}

mysqli_close($connection);                                                                      
?>

尝试下面的代码

    <?php
        $connection1 = mysql_connect("host","username","password") or die("Couldn't make connection."); 
        $db = mysql_select_db("database", $connection1) or die("Couldn't select database."); 

        $connection2 = mysql_connect("host","username","password") or die("Couldn't make connection."); 
        $db = mysql_select_db("database", $connection2)  or die("Couldn't select database."); 

        $sql = "SELECT *  FROM `READ` WHERE `name` = '$user'";                      
        $sql_result = mysql_query($sql,$connection1) or die("Couldn't execute query 1."); 

        while($row1 = mysql_fetch_array($sql_result))                           
        {
           echo "<br>".$row1['item'];
           $ident = $row1["item"];

           //$gr is not assigned any value..plz check it
           $query ="SELECT * FROM UPLOAD WHERE id NOT LIKE '$ident' AND faoGrade LIKE '%$gr%'";       
           //$ident only searching the first number stored
           $query_result = mysql_query($query,$connection2) or die("Couldn't execute query 2.");  
           while($row2 = mysql_fetch_array($query_result))                             
           {
              echo "<br>".$row2["title"];      
           }
        }

        mysql_close($connection1);
        mysql_close($connection2);
?>

mysqli\u close
应该记录并完成。我的错误，谢谢你。你需要它们成为独立的数据库吗？它们可能是同一数据库中的独立表吗？我认为OP混淆了“表”和“数据库”！