Use a label expression with pythonPython parser. Just replaceReplace º with whatever charcharacter you want to use:
def FindLabel ([X]):
s = int([X])
label = ''.join(['º' for i in range(s)])
return label
To get a line break every fifth charachter try:
def FindLabel ( [X] ):
s = int([X])
label = ['●' for i in range(s)]
i = 5
while i < len(label):
label.insert(i, "\r\n")
i += 6
return ''.join(label)
Use a label expression with python parser. Just replace º with whatever char you want to use:
def FindLabel ([X]):
s = int([X])
label = ''.join(['º' for i in range(s)])
return label
To get a line break every fifth charachter try:
def FindLabel ( [X] ):
s = int([X])
label = ['●' for i in range(s)]
i = 5
while i < len(label):
label.insert(i, "\r\n")
i += 6
return ''.join(label)
Use a label expression with Python parser. Replace º with whatever character you want to use:
def FindLabel ([X]):
s = int([X])
label = ''.join(['º' for i in range(s)])
return label
To get a line break every fifth charachter try:
def FindLabel ( [X] ):
s = int([X])
label = ['●' for i in range(s)]
i = 5
while i < len(label):
label.insert(i, "\r\n")
i += 6
return ''.join(label)
You can try code belowUse a label expression with python parser. Just replace º with whatever char you want to use:
def FindLabel ([X]):
s = int([X])
label = ''.join(['º' for i in range(s)])
return label
To get a line break every fifth charachter try:
def FindLabel ( [X] ):
s = int([X])
label = ['º'['●' for i in range(s)]
i = 5
while i < len(label):
label.insert(i, "\r\n")
i += 6
return ''.join(label)
You can try code below. Just replace º with whatever char you want to use:
def FindLabel ([X]):
s = int([X])
label = ''.join(['º' for i in range(s)])
return label
To get a line break every fifth charachter try:
def FindLabel ( [X] ):
s = int([X])
label = ['º' for i in range(s)]
i = 5
while i < len(label):
label.insert(i, "\r\n")
i += 6
return ''.join(label)
Use a label expression with python parser. Just replace º with whatever char you want to use:
def FindLabel ([X]):
s = int([X])
label = ''.join(['º' for i in range(s)])
return label
To get a line break every fifth charachter try:
def FindLabel ( [X] ):
s = int([X])
label = ['●' for i in range(s)]
i = 5
while i < len(label):
label.insert(i, "\r\n")
i += 6
return ''.join(label)
You can try code below. Just replace º with whatever char you want to use:
def FindLabel ([X]):
s = int([X])
label = ''.join(['º' for i in range(s)])
return label
To get a line break every fifth charachter try:
def FindLabel ( [X] ):
s = int([X])
label = ['º' for i in range(s)]
i = 5
while i < len(label):
label.insert(i, "\r\n")
i += 6
return ''.join(label)
You can try code below. Just replace º with whatever char you want to use:
def FindLabel ([X]):
s = int([X])
label = ''.join(['º' for i in range(s)])
return label
You can try code below. Just replace º with whatever char you want to use:
def FindLabel ([X]):
s = int([X])
label = ''.join(['º' for i in range(s)])
return label
To get a line break every fifth charachter try:
def FindLabel ( [X] ):
s = int([X])
label = ['º' for i in range(s)]
i = 5
while i < len(label):
label.insert(i, "\r\n")
i += 6
return ''.join(label)