SQL內部查詢不提供輸出

Question

/ *在$ selected中，從Java代碼獲取數據，該數據已成功存儲在$ selected中。 在查詢中使用$ selected，但在觸發查詢后無法獲取任何數據。 * /

     $selected = $_GET['selected'];

     $projects=mysql_query("select ProjectName from projects where LobID =
                ( select LobID from lob where LobName like     ".$selected.");");

 if (!$projects) {
     echo "Could not successfully run query ($projects) from DB: " .mysql_error();
    exit;
}

  /* Have included this query in my php page. All the table names are same.Lob       and projects are two different tables n LobID is primary key in Lob & Foreign key in projects. By executing the above query I am not able to fetch the data in $projects.Instead i am getting mesagecould not successfully run query. please help. */

Answer 1

查詢應該是這樣的

$projects=mysql_query("select ProjectName from projects where LobID =
                     (select LobID from lob where LobName like '$selected')");

更改".$selected."); 到'$selected'

Answer 2

試試這個...如果您喜歡搜索LobName，則表示使用%%。它會重新運行可能的匹配值，例如單個字符

"select ProjectName from projects where LobID =
                ( select LobID from lob where LobName like '%$selected%');"

Answer 3

我認為這應該是喬克（Jocker）和阿卜杜拉·尼拉姆（Abdulla Nilam）的回答：

$projects=mysql_query("select ProjectName from projects  
where LobID = ( select LobID from lob 
where LobName like '%".$selected."%');");

或加入

$projects = mysql_query("SELECT ProjectName FROM projects
JOIN lob ON projects.LobID=lob.LobID
WHERE LobName LIKE '%" . $selected . "%');");

Answer 4

• $selected = $_GET['selected'];包含的數據類型$selected = $_GET['selected']; 變量？
• 您的子查詢返回多少行記錄？

如果您的子查詢返回更多記錄，則可以限制結果數：

 <?php /** ... **/ $selected = $_GET['selected']; $projects = mysql_query("SELECT ProjectName FROM projects WHERE (LobID = ( SELECT LobID FROM lob WHERE LobName LIKE '%$selected%' LIMIT 1))"); ?> 

如果沒有，您可以這樣做：

 <?php /** ... **/ $selected = $_GET['selected']; $projects = mysql_query("SELECT ProjectName FROM projects WHERE (LobID IN ( SELECT LobID FROM lob WHERE LobName LIKE '%$selected%'))"); ?>