<div>
1. Simple, but approximate solution:<br />
select ((today - DOB) / 365.25)::int<br />
<br />
2. Accurate one:<br />
select year(today) - year(DOB) - case when mdy(month(DOB),day(DOB),ye<wbr />ar(today))<wbr /> &gt;&nbsp;today then 1 else 0 end<br />

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1. Simple, but approximate solution:
select ((today - DOB) / 365.25)::int

2. Accurate one:
select year(today) - year(DOB) - case when mdy(month(DOB),day(DOB),year(today)) > today then 1 else 0 end


<div>
running <b>2. Accurate one:</b> in aqua studio I get error:<br />
<br />
&quot;Invalid day in date&quot;<br />
<br />

<pre><code id="code-20-35014658-1">select year(today) - year(mailing_code) - case when mdy(month(mailing_code),day(mailing_code),year(today)) &gt; today then 1 else 0 end
from table_1</code></pre>
<br />
do you know what is wrong?<br />

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running 2. Accurate one: in aqua studio I get error:

"Invalid day in date"



select year(today) - year(mailing_code) - case when mdy(month(mailing_code),day(mailing_code),year(today)) > today then 1 else 0 end
from table_1


do you know what is wrong?


<div>
example DOBs from the mailing_code field:<br />
<b>mailing_code</b><br />
26/03/1941<br />
1/10/1958<br />
21/07/1961
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example DOBs from the mailing_code field:
mailing_code
26/03/1941
1/10/1958
21/07/1961

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<div class="content wysiwyg-content">
The code below is not accurate as it only minuses the year<br />
<br />
<b>year(today)-year(DOB) as AGE</b><br />
<br />
does anybody know a more accurate way of returing a persons age taking into account the day and month of the year as well?<br />
<br />
<br />

</div>
</div>


The code below is not accurate as it only minuses the year

year(today)-year(DOB) as AGE

does anybody know a more accurate way of returing a persons age taking into account the day and month of the year as well?




Informix SQL get age from ddmmyyyy field

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