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Full Discussion: Replacing $ in variable
Top Forums UNIX for Dummies Questions & Answers Replacing $ in variable Post 302171673 by ashish_uiit on Friday 29th of February 2008 07:07:45 AM
Old 02-29-2008
Quote:
Originally Posted by sank
probably the variable k is not formed correctly. Can you try echo $k and see if you get $DESTDIR/$PKG/$VERSION ? Are you forming the variable k like this : k="\$DESTDIR/\$PKG/\$VERSION" ? if so, the sed command works as expected.
$DESTDIR/$PKG/$VERSION is coming from a read only file to k while reading.
 

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GUPNP_SERVICE_NOTIFY(3) 						 1						   GUPNP_SERVICE_NOTIFY(3)

gupnp_service_notify - Notifies listening clients

SYNOPSIS
bool gupnp_service_notify (resource $service, string $name, int $type, mixed $value) DESCRIPTION
Notifies listening clients that the property have changed to the specified values. PARAMETERS
o $service - A service identifier. o $name - The name of the variable. o $type - The type of the variable. Type can be one of the following values: o GUPNP_TYPE_BOOLEAN - Type of the variable is boolean. o GUPNP_TYPE_INT - Type of the variable is integer. o GUPNP_TYPE_LONG - Type of the variable is long. o GUPNP_TYPE_DOUBLE - Type of the variable is double. o GUPNP_TYPE_FLOAT - Type of the variable is float. o GUPNP_TYPE_STRING - Type of the variable is string. o $value - The value of the variable. RETURN VALUES
Returns TRUE on success or FALSE on failure. ERRORS
/EXCEPTIONS Issues E_WARNING with either not correctly defined type of the variable or the value is not corresponding to the defined type. SEE ALSO
gupnp_service_freeze_notify(3), gupnp_service_thaw_notify(3). PHP Documentation Group GUPNP_SERVICE_NOTIFY(3)
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