2019-05-08 15:14

# 将数字除以月中的日期只返回一位数字

I am trying to calculate what pace our sales team in on for the month, but when I divide our numbers so far (17,305) by the day of the month (08), I am getting the wrong number (2.125).

I've tried converting the date from a string to a number, but I everything I read says php should know how to handle the numbers when it's a string or number.

``````\$dateday = date('d');
\$numberofdays = date('t');

echo \$ztmmoney.' total for month<br>';
echo \$dateday.' day of the month<br>';
\$mavg = \$ztmmoney/\$dateday;
echo \$mavg.' daily average<br>';
echo \$numberofdays.' days in month<br>';
\$pace = \$mavg * \$numberofdays;

echo 'on pace for: '.\$pace;
``````

I should see 17,305/8 = 2,163.125

but this is my output:

17,305 total for month

08 day of the month

2.125 daily average

31 days in month

on pace for: 65.875

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#### 1条回答

• dongshi3605 2年前

You can use this to get rid of the wrong calculation

``````\$ztmmoney = '17,305';
\$ztmmoney = intval(preg_replace('/[^\d.]/', '', \$ztmmoney));
``````
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