Extract column data from csv file based on row values
<p>I am trying to use awk/sed to extract specific column data based on row values. My actual files have 15 columns and over 1,000 rows (From a .csv file.) </p>

<p>Simple EXAMPLE: Input; a cdv file with a total of 5 columns and 100 rows. Output; data from column 2 through 5 based on specific row values from column 2. (I have a specific list of the row values I want the operator to filter out. The values are numbers.) </p>

<p>File looks like this:</p>

<pre class="lang-none prettyprint-override"><code>"Date","IdNo","Color","Height","Education"
"06/02/16","7438","Red","54","4"
"06/02/16","7439","Yellow","57","3"
"06/03/16","7500","Red","55","3" 
</code></pre>

<p>Recently Tried in AWK:</p>

<pre><code>#!/usr/bin/awk -f 
#I need to extract a full line when column 2 has a specific 5 digit    value
awk '\
BEGIN { awk -F "," \    
{
    if ( $2 == "19650" ) { \
       {print $1 "," $6} \ 
}
exit }     

chmod u+x PPMDfUN.AWK
</code></pre>

<p>The operator response:</p>

<pre><code>/var/folders/1_/drk_nwld48bb0vfvdm_d9n0h0000gq/T/PPMDfUN-     489939602.998.AWK.command ; exit;
/usr/bin/awk: syntax error at source line 3 source file    /private/var/folders/1_/drk_nwld48bb0vfvdm_d9n0h0000gq/T/PPMDfUN-   489939602.997.AWK
context is
awk &gt;&gt;&gt;  ' &lt;&lt;&lt; 
/usr/bin/awk: bailing out at source line 17
logout
</code></pre>

<p>Output Example: I want full row lines based if column 2 equals 7439 &amp; 7500.</p>

<pre class="lang-none prettyprint-override"><code>“Date","IdNo","Color","Height","Education"
"06/02/16","7439","Yellow","57","3"
"06/03/16","7500","Red","55","3" 
</code></pre>