如何检查字符串只包含数字和一个小数点？

Question

My idea is something like this but I dont know the correct code

if (mystring.matches("[0-9.]+")){
  //do something here
}else{
  //do something here
}

I think I'm almost there. The only problem is multiple decimal points can be present in the string. I did look for this answer but I couldn't find how.

Answer 1

If you want to -> make sure it's a number AND has only one decimal <- try this RegEx instead:

if(mystring.matches("^[0-9]*\\.?[0-9]*$")) {
    // Do something
}
else {
    // Do something else
}

This RegEx states:

1. The ^ means the string must start with this.
2. Followed by none or more digits (The * does this).
3. Optionally have a single decimal (The ? does this).
4. Follow by none or more digits (The * does this).
5. And the $ means it must end with this.

Note that bullet point #2 is to catch someone entering ".02" for example.

If that is not valid make the RegEx: "^[0-9]+\\\\.?[0-9]*$"

• Only difference is a + sign. This will force the decimal to be preceded with a digit: 0.02

Answer 2

I think using regexes complicates the answer. A simpler approach is to use indexOf() and substring() :

int index = mystring.indexOf(".");
if(index != -1) {
    // Contains a decimal point
    if (mystring.substring(index + 1).indexOf(".") == -1) {
        // Contains only one decimal points
    } else {
        // Contains more than one decimal point 
    }
}
else {
    // Contains no decimal points 
}

Answer 3

You could use indexOf() and lastIndexOf() :

int first = str.indexOf(".");
if ( (first >= 0) && (first - str.lastIndexOf(".")) == 0) {
    // only one decimal point
}
else {
    // no decimal point or more than one decimal point
}

Answer 4

Simplest

Example:

"123.45".split(".").length();

Answer 5

If you want to check if a number (positive) has one dot and if you want to use regex, you must escape the dot, because the dot means "any char" :-)

see http://docs.oracle.com/javase/6/docs/api/java/util/regex/Pattern.html

Predefined character classes
.   Any character (may or may not match line terminators)
\d  A digit: [0-9]
\D  A non-digit: [^0-9]
\s  A whitespace character: [ \t\n\x0B\f\r]
\S  A non-whitespace character: [^\s]
\w  A word character: [a-zA-Z_0-9]
\W  A non-word character: [^\w]

so you can use something like

System.out.println(s.matches("[0-9]+\\.[0-9]+"));

ps. this will match number such as 01.1 too. I just want to illustrate the \\\\.

Answer 6

int count=0;
For(int i=0;i<mystring.length();i++){    
    if(mystring.charAt(i) == '/.') count++;
}
if(count!=1) return false;

Answer 7

Use the below RegEx its solve your proble

1. allow 2 decimal places ( eg 0.00 to 9.99)

    ^[0-9]{1}[.]{1}[0-9]{2}$ This RegEx states: 1. ^ means the string must start with this. 2. [0-9] accept 0 to 9 digit. 3. {1} number length is one. 4. [.] accept next character dot. 5. [0-9] accept 0 to 9 digit. 6. {2} number length is one. 

2. allow 1 decimal places ( eg 0.0 to 9.9)

    ^[0-9]{1}[.]{1}[0-9]{1}$ This RegEx states: 1. ^ means the string must start with this. 2. [0-9] accept 0 to 9 digit. 3. {1} number length is one. 4. [.] accept next character dot. 5. [0-9] accept 0 to 9 digit. 6. {1} number length is one. 

Answer 8

I create myself to solve exactly question's problem. I'll share you guys the regex:

^(\\d)*(\\.)?([0-9]{1})?$

Take a look at this Online Regex to see work properly

Refer documentation if you wish continue to custom the regex

Documentation