help with mysql command to update data in a table

ncsuapex · 09-30-2008, 10:30 AM

Here is my setup

database name = database
table name = table
inside this table are several fields:
name
info1
info2
info3

Im trying to replace the data that is in info1, info2 and info3 for a specific name but am not sure how to do it.

Would this work? Just to replace info1, I want to get this working and then I'll edit it to include info2 and info3

update table set info1 = replace(info1,'currentstring','newstring') where name ='specificusername';

EDIT:

Ok so I got bold and tried out this commands:
update table set info1 = replace(info1,'currentstring','newstring') where name ='specificusername';

replacing the values for real database values and it worked. Now. What can I do to have one command to replace all 3 info? values?

CRC123 · 09-30-2008, 11:19 AM

Ok, I don't know what Database you are using, but read this.

In a nutshell:

Code:

UPDATE table_name
SET column1=value, column2=value2,...
WHERE some_column=some_value

Also, you can pre-pend table-names with database names(aka database.table). However, if you're on the sql command line, you can simply use a 'use' statement to specify what database you want the following operations done on:

Code:

use <database name>;

ncsuapex · 09-30-2008, 11:32 AM

I was having a brain fart about which database to change the data in. It hit me after I posted the thread. Now that I have a working command how do I combine the command to include all 3 changes in one command? Ive tried separating them with a , and a ; but the command fails.

play with it long enough and I'll figure it out.

update table set info1 = newstring, info2 = newstring2, info3 = newstring3 where name ='username';