Question

希望每次看到目标变量时计数器不断增加错误，因为counter = counter + 1

myString = input("Enter a string \n")
target = input("enter the target variable\n")

length = len(myString)

for j in range(length):
    if myString[j] == target:
        counter = counter + 1
print(target, "occured", count, "times")

Answer 1

您需要像这样将counter初始化为0：

counter = 0

myString = input("Enter a string \n")
target = input("enter the target variable\n")
length = len(myString)

for j in range(length):
    if myString[j] == target:
        counter = counter + 1
print(target, "occured", counter, "times")

因此，如果字符串为hello，而目标变量为l。

输出将是： l occured 2 times

Answer 2

解决此问题的Python方法是在字符串target中查找myString变量，然后在找到时counter递增。比较结果应不区分大小写，以获取准确的结果。使用lower()将字符串和目标变量转换为小写。

myString = input("Enter a string \n")
target = input("enter the target variable\n")

count = 0
for j in myString:
    if target.lower() in j.lower():
        count += 1
print(target, "occurred", count, "times")
#Output:
Enter a string 
The combinations of ABC are ABC,BAC,CAB
enter the target variable
b
b occurred 5 times

需要修复（计数器=计数器+ 1）错误

2 个答案: