Like playing poker online, other specific websites (deep into the night), chocolate and other candy/food which makes me fat, etc.
Even if i get over an addiction (like answering questions on Yahoo), a new one gets in it's place.
It's bad for my social contacts and a little bit for my work.
Is it possible to break down that pattern? Is it possible without professional help?
2007-05-17 22:37:42 · 7 answers · asked by · 5 in Family & Relationships ➔ Friends
Go down to the bookstore and find a book in the self help section. There are many underlying reasons why people have addictive personalities.
You may be able to help yourself without paying a professional, it all depends on how severe your problem is and if you can fix it yourself. Try reading a few different books till you find something that sounds right for you.
If you can't do this, go see someone, it could help you for the rest of your life.
2007-05-17 23:40:44 · answer #1 · answered by dogcazza 3 · 0⤊ 0⤋
Find a close friend, and have them work it out. If you get an addiction then be able to call and work it out! Go out and play baseketball or a walk to help with the weight, and also a chance for you to get out of the house and computer!
2007-05-18 10:42:39 · answer #2 · answered by fjerseyman 6 · 0⤊ 0⤋
You like me but if you work hard and be in control for some things there is nothing wrong in fun
2007-05-18 05:41:06 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Look up OCD and get familiar with it. Recognizing it and knowing may help you control it. If not seek professional help.
2007-05-18 05:42:36 · answer #4 · answered by Anonymous · 0⤊ 0⤋
It is hard, however it can be done. It is much easier if you enlist someone close to you to help you. If you can't, therapy is a good option.
2007-05-18 05:41:55 · answer #5 · answered by PEGGY S 7 · 0⤊ 0⤋
thats what they want. Be more. prove them wrong by proving them right! above all, be you
2007-05-18 05:42:27 · answer #6 · answered by carl j 2 · 0⤊ 0⤋
try yoga.
it works!
2007-05-18 06:05:39 · answer #7 · answered by Sesshoumaru 1 · 0⤊ 0⤋