Re: SELECT statement

Ron, Thanks for your help, but my problem still there even when I made the following changes: $sql = "SELECT * FROM \"Remodel\" WHERE Email=\"".$find."\""; By the way, you're correct when saying that "WO" and "Email" are column names and '"$find" is what the user entered when searching for a record. At 03:39 PM 4/4/2001 -0700, you wrote:

At 03:24 PM 4/4/2001 -0700, Julio Cuz, Jr. wrote:

Every time I run the following code, I get this error if I use a NUMBER for the 'Email' case: "Warning: Unable to jump to row 0 on PostgreSQL result index 2 in /html/rccd/remodel/display2.php on line 35"

The problem is that your query is equivalent to going: SELECT * FROM "Remodel" WHERE "WO" LIKE 1; // or some other numerical value The "WO" is a string because it is in quotes so you are saying: WHERE string = integer Ain't gonna happen so the result is always empty. What you want probably is (assuming WO is a column name): $sql = "SELECT * FROM \"Remodel\" WHERE WO=\"".$find."\""; It is equals since you are using an integer instead of a string (like is for strings).

or, if I use a string (i.e. jcuz@rccd.cc.ca.us), I get this error message: "Warning: PostgreSQL query failed: ERROR: Attribute 'jcuz' not found in /html/rccd/remodel/display2.php on line 33"

Same problem. "EMAIL" means the string "EMAIL" not the column called "EMAIL". Cheers, Ron -----------------------------------------------------------------------------

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Julio Cuz, Jr. Riverside Community College jcuz@rccd.cc.ca.us