# Php how many three/four/five digit numbers do I have in an array

## Issue

I have an array which has several numbers.
What I want to do is get a loop and run into this array and count how many two-digits numbers, three-digits, four-digits numbers do I have.

I also tried this:

``````\$threeDigits=0;
\$fourDigits=0;
\$fiveDigits=0;
\$sixDigits=0;

\$myArray=(123,1234,12345,123456,1234567,111,222)
for(\$i=1,\$size=count(\$myArray);\$i<=\$size;i++)
{
if(strlen(\$==3)) \$threeDigits++;
else if(strlen(\$i==4)) \$fourDigits++;
else if(strlen(\$i==5)) \$fiveDigits++;
else if(strlen(\$i==6)) \$sixDigits++;
}
echo "There are " .\$fourDigits. " numbers with 4 digits.";
echo "There are " .\$threeDigits. " numbers with 3 digits.";
echo "There are " .\$fiveDigits. " numbers with 5 digits.";
echo "There are " .\$sixDigits. " numbers with 6 digits.";
``````

But somehow it only reads them as one. As you can see, in my array there are three three-digit numbers but when I print it out it says I have only one. What do you think it might be the problem here?

## Solution

Since I couldn’t find any answer that "suited" my code better, I tried and solved this out.

Here’s my code if someone else in the future needs it.

``````<?php
\$threeDigits=0;
\$fourDigits=0;
\$fiveDigits=0;
\$sixDigits=0;

\$myArray=array(123,111,1234,12345,123456);
\$size=count(\$myArray);

foreach(\$myArray as \$el){
if(strlen(\$el) == 3)
{
\$threeDigits++;
}
else if(strlen(\$el)==4)
{
\$fourDigits++;
}
else if(strlen(\$el)==5)
{
\$fiveDigits++;
}
else if (strlen(\$el)==6)
{
\$sixDigits++;
}
}
echo "there are " .\$threeDigits. " three digits numbers -- " ;
echo "there are " .\$fourDigits. " four digits numbers -- " ;
echo "there are " .\$fiveDigits. " five digits numbers -- " ;
echo "there are " .\$sixDigits. " six digits numbers -- " ;
?>
`````` 