Creating a table with a column named ''PASSWORD"

Question

Here is my script and the error I am getting. What am I doing wrong?

create table PHONE_USERS (
 TELEPHONE_NUMBER VARCHAR2(80)  NOT NULL, PRIMARY KEY 
 FIRST_NAME VARCHAR2(80),
 LAST_NAME VARCHAR2(80),
 KEYMAP_LASTNAME CHAR(4), 
"PASSWORD" VARCHAR2(80),
 CONSTRAINT LAST_NAME UNIQUE (TELEPNONE_NUMBER)
 );

Error at Command Line : 7 Column : 31
Error report -
SQL Error: ORA-00904: "TELEPNONE_NUMBER": invalid identifier 00904. 00000 - "%s: invalid identifier"
*Cause:
*Action:

I do beleive that you have to show a little more info/code so that we could get a better picture and see where the problem resides. (We don't steal code) — DaGhostman Dimitrov, Aug 07 '14 at 06:48

score 1 · Accepted Answer · edited Aug 07 '14 at 06:47

1

TELEPNONE_NUMBER is misspelled. Is must be TELEPHONE_NUMBER

create table PHONE_USERS (
 TELEPHONE_NUMBER VARCHAR2(80)  NOT NULL PRIMARY KEY, 
 FIRST_NAME VARCHAR2(80),
 LAST_NAME VARCHAR2(80),
 KEYMAP_LASTNAME CHAR(4), 
"PASSWORD" VARCHAR2(80),
 CONSTRAINT LAST_NAME UNIQUE (TELEPHONE_NUMBER)
 );

edited Aug 07 '14 at 06:47

answered Aug 07 '14 at 06:45

Jens

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@Downvoter: please explaine the downvote! – Jens Aug 07 '14 at 06:51

score 1 · Answer 2 · answered Aug 07 '14 at 06:46

This error occured, because you are placing comma after NOT NULL and TELEPHONE_NUMBER spelling mismatch i think.

Change TELEPNONE_NUMBER to TELEPHONE_NUMBER

create table PHONE_USERS (
 TELEPHONE_NUMBER VARCHAR2(80)  NOT NULL PRIMARY KEY, 
 FIRST_NAME VARCHAR2(80),
 LAST_NAME VARCHAR2(80),
 KEYMAP_LASTNAME CHAR(4), 
"PASSWORD" VARCHAR2(80),
 CONSTRAINT LAST_NAME UNIQUE (TELEPHONE_NUMBER)
 );

Creating a table with a column named ''PASSWORD"

2 Answers2