#1
  1. No Profile Picture
    Junior Member
    Devshed Newbie (0 - 499 posts)

    Join Date
    Feb 2000
    Location
    NY, NY,USA
    Posts
    18
    Rep Power
    0
    I知 having trouble with the line that starts with $dept_name. $dept_name seems to be getting the array index, instead of the value that is in the department_name table. If I type the sql command into the command line the result is correct. I知 trying to get a name, like Accounting, not a number, as I知 getting now. When I just do this: print $row["dept_id"];
    I get the correct numerical value.

    I had this:

    $result = mysql_query ("SELECT * FROM jobs");

    print ("<table>");
    if ($row = mysql_fetch_array($result)) {
    do {
    print ("<tr><td bgcolor=#cccccc>");
    print $row["position_name"];
    print ("</td><td>");
    print ("<b>");
    print $row["dept_id"];
    print ("</b>");
    $dept_name = mysql_query ("SELECT department_name FROM dept_info WHERE dept_id = '$row["dept_id"]'");
    print $dept_name;
    print ("</td></tr>"); }
    while($row = mysql_fetch_array($result)); }
    else {print "Sorry, no records were found!";}

    print ("</table>");
    $result = mysql_query ("SELECT * FROM jobs");

    Someone suggested to change the line to:

    $selectquery = "SELECT department_name FROM dept_info WHERE dept_id = " . $row["dept_id"];
    $dept_name = mysql_query($selectquery);

    But I知 still getting the numeric array index.

  2. #2
  3. No Profile Picture
    Apprentice Deity
    Devshed Loyal (3000 - 3499 posts)

    Join Date
    Jul 1999
    Location
    Niagara Falls (On the wrong side of the gorge)
    Posts
    3,237
    Rep Power
    19
    Read the manual... that's what you are supposed to get. mysql_query returns a result index that you use to in the various mysql_fetch functions to get the results.

    I really don't understand why this is a problem since you did the query from the jobs table correctly.. fetching the result into the array $row.

    Try this:

    list($dept_name)=mysql_fetch_row(mysql_query("select department_name from dept_info where dept_id = '$row[dept_id]'"));

Similar Threads

  1. windows XP chkdsk and file system problems
    By wey97 in forum Windows Help
    Replies: 26
    Last Post: August 2nd, 2005, 11:37 PM
  2. Popup problems from Flash in IE6
    By jmichels in forum Flash Help
    Replies: 19
    Last Post: January 6th, 2004, 03:27 PM
  3. DNS problems
    By banneaux in forum DNS
    Replies: 7
    Last Post: October 10th, 2003, 05:58 AM
  4. Replies: 0
    Last Post: August 26th, 2003, 07:50 PM
  5. Windows problems
    By Known_criminal in forum Windows Help
    Replies: 0
    Last Post: August 23rd, 2003, 07:23 PM

IMN logo majestic logo threadwatch logo seochat tools logo