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. [EMAIL PROTECTED]), 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
[EMAIL PROTECTED]
