Unexpected string length when using strlen in php

Question

I am stuck with the issue as currently the result is quite unexpected. I am calculating a hashed keyword length and it is surely giving me an unexpected result.

echo strlen("$2a$08$MphfRBNtQMLuNro5HOtw3Ovu20cLgC0VKjt6w7zrKXfj1bv8tNnNa");

Output - 6

Let me know the reason for this and why it is outputting 6 as a result.

Codepad Link - http://codepad.org/pLARBx6F

try this :echo strlen('$2a$08$MphfRBNtQMLuNro5HOtw3Ovu20cLgC0VKjt6w7zrKXfj1bv8tNnNa'); it takes parameter to a function as php variable and throws notice as variable not defined — Sandip Ghadge
– Sandip Ghadge, Commented Jul 1, 2014 at 11:56
echo or var_dump your string. The reason should become clearer. — cHao
– cHao, Commented Jul 1, 2014 at 11:57

Flosi · Accepted Answer · 2014-07-01 12:07:46Z

5

You must use single quotes '. With the double quotes ("), due to the $ in your string, parts of it get interpreted as variables.

Generally, it's not a bad idea to get accustomed to using single quotes unless you specifically need doubles.

Look at the "variables" contained here. They would be $2a, $08, and $MphfRBNtQM...... The first two couldn't be variables as they start with a number, thus, the 6 characters. The third one indeed could be a proper variable, but since it isn't set, it's empty.

edited Jul 1, 2014 at 12:07

answered Jul 1, 2014 at 11:54

Flosi

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2 Comments

Flosi Over a year ago

Look at the "variables" contained here. They would be $2a, $08, and $MphfRBNtQM...... The first two couldn't be variables as they start with a number, thus, the 6 characters. The third one indeed could be a proper variable, but since it isn't set, it's empty.

Trialcoder Over a year ago

Please add this comment to your answer and I wil accept it..awesome :)

Anik · Accepted Answer · 2014-07-01 11:59:56Z

1

Use the below code to calculate the string length -

echo strlen('$2a$08$MphfRBNtQMLuNro5HOtw3Ovu20cLgC0VKjt6w7zrKXfj1bv8tNnNa');

You need to use single quotes, as at the third occurrence of the symbol $, a alphabet is starting after it and it get treated as a new variable. So before this third occurrence of $ only 6 character were there and you were getting string length as 6

answered Jul 1, 2014 at 11:59

Anik

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Comments

Bora · Accepted Answer · 2014-07-01 11:58:24Z

-2

Try following

<?php

echo strlen('$2a$08$MphfRBNtQMLuNro5HOtw3Ovu20cLgC0VKjt6w7zrKXfj1bv8tNnNa');

?>

If you change your string and remove rest of '$' signs except the first one, then this will work fine because by adding $ it gets a special meaning in PHP.

edited Jul 1, 2014 at 11:58

Bora

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answered Jul 1, 2014 at 11:55

K D

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2 Comments

K D Over a year ago

@Negative Marker, can you please explain the reason?

cHao Over a year ago

Probably because you don't explain what you fixed and why. The little bit of explanation, "adding $ it gets a special meaning in PHP", makes no sense.

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