new names and new script

docker-deploy

@@ -0,0 +1,56 @@
+#!/bin/bash -e
+
+while test $# -gt 0; do
+    case "$1" in
+	(-h|--help) cat <<EOF
+$0 [OPTIONS] FILES…
+
+OPTIONS
+
+  -h, --help           show this help
+
+FILES…
+
+  List of stack names or yaml files to deploy. There must be yaml
+  files with the same name. The extension .yaml may be given or not.
+
+DESCRIPTION
+
+  Deploys stack from yaml files. The stack name is identical to the
+  file name, but wihout path and without .yaml extendsion.
+
+EXAMPLS
+
+  The following calls ado the same and deploy a local yaml file:
+
+    $0 yaml-file-name
+    $0 yaml-file-name.yaml
+    $0 ./yaml-file-name.yaml
+
+  Deploy three files from three sources:
+
+    $0 /path/to/file1/first.yaml /path/to/file2/second.yaml third
+
+EOF
+		    exit;;
+	(*) break;;
+    esac
+    if test $# -lt 1; then
+	echo "error: missing argument, try $0 --help" 1>&2
+	exit 1
+    fi
+    shift
+done
+
+for f in $*; do
+    f=${f%.yaml}
+    if ! test -e ${f}.yaml; then
+	echo "ERROR: no file ${f}.yaml" 1>&2
+	exit 1
+    fi
+    echo "... deploying $f"
+    for d in $(sed -n 's/^ *source: *//p' ${f}.yaml); do
+        test -e $d || mkdir -p $d
+    done
+    docker stack deploy --compose-file ${f}.yaml ${##*/f}
+done