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)