Reputation: 1922
So I'm trying to send several iterations of a barcode file to a device. This is the relevant part of the code:
# Start is the first barcode
start = 1234567
# Number is the quantity
number = 3
with open('barcode2.xml', 'rt') as f:
tree = ElementTree.parse(f)
# Iterate over all elements in a tree for the root element
for node in tree.getiterator():
# Looks for the node tag called 'variable', which is the name assigned
# to the accession number value
if node.tag == "variable":
# Iterates over a list whose range is specified by the command
# line argument 'number'
for barcode in range(number):
# The 'A-' prefix and the 'start' argument from the command
# line are assigned to variable 'accession'
accession = "A-" + str(start)
# Start counter is incremented by 1
start += 1
# The node ('variable') text is the accession number.
# The acccession variable is assigned to node text.
node.text = accession
# Writes out to an XML file
tree.write("barcode2.xml")
header = "<?xml version=\"1.0\" standalone=\"no\"?>\n<!DOCTYPE labels SYSTEM \"label.dtd\">\n"
with open("barcode2.xml", "r+") as f:
old = f.read()
f.seek(0)
f.write(header + old)
# Create socket
sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
# Connect to server
host = "xxx.xx.xx.x"
port = 9100
sock.connect((host, port))
# Open XML file and read its contents
target_file = open("barcode2.xml")
barc_file_text = target_file.read()
# Send to printer
sock.sendall(barc_file_text)
# Close connection
sock.close()
This is very much a version one.
The device is failing to receive the files after the first one. Could this be because the port is being reused again too quickly? What's a better way to architect this? Thanks so much for your help.
Upvotes: 0
Views: 1378
Reputation: 61616
target_file = open("barcode2.xml")
barc_file_text = target_file.read()
sock.sendall(barc_file_text)
sock.close()
The socket gets closed, but the file doesn't. The next time through the loop, there is already a lock on the file when you get to the with open...
part.
Solution: Use with open...
here as well. Also, you don't need to baby-step everything; don't give something a name (by assigning it to a variable) if it isn't important.
with open("barcode2.xml", "r") as to_send:
sock.sendall(to_send.read())
sock.close()
Upvotes: 4