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Top Forums UNIX for Dummies Questions & Answers Cant read in variable on first try Post 85361 by normie on Wednesday 5th of October 2005 01:37:50 AM
Old 10-05-2005
Cant read in variable on first try

Hi,

My problem is :

echo Division
read vDivision

variable1=`cut -c **something****'

echo Do you want to proceed ?
read ans


I cant seem to read in ans on the first try and have to repeatedly enter the return key. If i remove the ` ` statement its ok but i need that line for verification . Any help is appreciated. Thanks.
 

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APC_CAS(3)								 1								APC_CAS(3)

apc_cas - Updates an old value with a new value

SYNOPSIS
bool apc_cas (string $key, int $old, int $new) DESCRIPTION
apc_cas(3) updates an already existing integer value if the $old parameter matches the currently stored value with the value of the $new parameter. PARAMETERS
o $key - The key of the value being updated. o $old - The old value (the value currently stored). o $new - The new value to update to. RETURN VALUES
Returns TRUE on success or FALSE on failure. EXAMPLES
Example #1 apc_cas(3) example <?php apc_store('foobar', 2); echo '$foobar = 2', PHP_EOL; echo '$foobar == 1 ? 2 : 1 = ', (apc_cas('foobar', 1, 2) ? 'ok' : 'fail'), PHP_EOL; echo '$foobar == 2 ? 1 : 2 = ', (apc_cas('foobar', 2, 1) ? 'ok' : 'fail'), PHP_EOL; echo '$foobar = ', apc_fetch('foobar'), PHP_EOL; echo '$f__bar == 1 ? 2 : 1 = ', (apc_cas('f__bar', 1, 2) ? 'ok' : 'fail'), PHP_EOL; apc_store('perfection', 'xyz'); echo '$perfection == 2 ? 1 : 2 = ', (apc_cas('perfection', 2, 1) ? 'ok' : 'epic fail'), PHP_EOL; echo '$foobar = ', apc_fetch('foobar'), PHP_EOL; ?> The above example will output something similar to: $foobar = 2 $foobar == 1 ? 2 : 1 = fail $foobar == 2 ? 1 : 2 = ok $foobar = 1 $f__bar == 1 ? 2 : 1 = fail $perfection == 2 ? 1 : 2 = epic fail $foobar = 1 SEE ALSO
apc_dec(3), apc_store(3). PHP Documentation Group APC_CAS(3)
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