I need to convert numbers to have .00 after them, but only if the number is an integer, or it has just 1 number after the decimal point, like so:
我需要在它们之后将数字转换为.00,但仅当数字是整数时,或者它在小数点后只有1个数字,如下所示:
1.4 = 1.40
45 = 45.00
34.77 = 34.77
What reg exp to use for this simple case?
reg exp用于这个简单的案例?
5 个解决方案
#1
1
number_format($number, 2, '.', '');
Read more at PHP.net. You don't need to determine if a number is an integer or not -- as long as it's a number, it will be formatted to two decimal places.
在PHP.net上阅读更多内容。您不需要确定数字是否为整数 - 只要它是一个数字,它将被格式化为两个小数位。
If you'd like the thousands separator, change the last parameter to ','
.
如果您想要千位分隔符,请将最后一个参数更改为“,”。
#2
#3
1
Check out PHP's built-in function number_format
查看PHP的内置函数number_format
You can pass it a variable and it'll format it to the correct decimal places
您可以将变量传递给它,并将其格式化为正确的小数位
$number = 20;
if (is_int($number)) {
$number = number_format($number, 2, '.', '');
}
#5
0
$num = 0.00638835;
$avg = sscanf($num,"%f")[0] /100;
echo sprintf("%.10f", $avg);
result 0.0000638835
#1
1
number_format($number, 2, '.', '');
Read more at PHP.net. You don't need to determine if a number is an integer or not -- as long as it's a number, it will be formatted to two decimal places.
在PHP.net上阅读更多内容。您不需要确定数字是否为整数 - 只要它是一个数字,它将被格式化为两个小数位。
If you'd like the thousands separator, change the last parameter to ','
.
如果您想要千位分隔符,请将最后一个参数更改为“,”。
#2
2
You can also use printf or sprintf
您也可以使用printf或sprintf
printf("%01.2f", '34.77');
$formatted_num = sprintf("%01.2f", '34.77');
#3
1
Check out PHP's built-in function number_format
查看PHP的内置函数number_format
You can pass it a variable and it'll format it to the correct decimal places
您可以将变量传递给它,并将其格式化为正确的小数位
$number = 20;
if (is_int($number)) {
$number = number_format($number, 2, '.', '');
}
#4
#5
0
$num = 0.00638835;
$avg = sscanf($num,"%f")[0] /100;
echo sprintf("%.10f", $avg);
result 0.0000638835