SQL内部Query不提供输出

/* In $selected m getting data from java code which is successsfully getting stored in $selected. Using $selected in query but cannot get any data after firing the query. */

     $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. */

Query should be like this

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

Change ".$selected."); to '$selected'

Try this... If you use like to search LobName means use %% .Its retrun possible match value kike single character

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

I think it should be the mix of Jocker' and Abdulla Nilam' answer:

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

or make a join

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

• what type of data contained in your $selected = $_GET['selected']; variable?
• how many lines of records your subquery returns?

if your subquery returns more records, you can either limit the number of results:

<?php
  
  /**
    ...
  **/

  $selected = $_GET['selected'];

  $projects = mysql_query("SELECT ProjectName FROM projects WHERE (LobID = (
    SELECT LobID FROM lob WHERE LobName LIKE '%$selected%' LIMIT 1))");

?>

if not, you can do like this:

<?php
      
      /**
        ...
      **/

      $selected = $_GET['selected'];

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

    ?>

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