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#1
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I’m 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’m trying to get a name, like Accounting, not a number, as I’m 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’m still getting the numeric array index. |
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#2
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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]'")); |
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