Manually getting the date in filename with awk

fix.sh

#!/bin/bash

CURRENT_DIR="$( cd "$(dirname "${BASH_SOURCE[0]}")" && pwd )"
cd "$CURRENT_DIR"

workpath="$1"

cd "${workpath}"
for image in *.jpeg; do
    date_format=$(echo "${image}" | awk '{gsub(/-/, ":", $3); print $3}')
    time_format=$(echo "${image}" | awk '{print $5}')

    twelve_h_format=$(echo "${image}" | awk '{sub(/M.+/, "M", $6); print $6}')
    hour=$(echo "${time_format}" | awk -F "." '{print $1}')

    if [[ "${twelve_h_format}" == "PM" ]]; then
        hour=$((hour + 12))
    fi

    hour=$(printf "%02d" "${hour}")
    minute_and_second=$(echo "${time_format}" | awk -F "." '{print $2":"$3}')
    time_format="${hour}:${minute_and_second}"

    full_datetime="${date_format} ${time_format}"

    "$CURRENT_DIR/exiftool.exe" -overwrite_original "-AllDates=${full_datetime}" "${image}"
done