undefined offset when using php explode()

匿名 (未验证) 提交于 2019-12-03 01:10:02

问题:

I've written what I thought was a very simple use of the php explode() function to split a name into forename and surname:

// split name into first and last $split = explode(' ', $fullname, 2); $first = $split[0]; $last = $split[1]; 

However, this is throwing up a php error with the message "Undefined offset: 1". The function still seems to work, but I'd like to clear up whatever is causing the error. I've checked the php manual but their examples use the same syntax as above. I think I understand what an undefined offset is, but I can't see why my code is generating the error!

回答1:

this is because your fullname doesn't contain a space. You can use a simple trick to make sure the space is always where

 $split = explode(' ', "$fullname "); 

(note the space inside the quotes)

BTW, you can use list() function to simplify your code

  list($first, $last) = explode(' ', "$fullname "); 


回答2:

BTW, that algorithm won't work all the time. Think about two-word Latina or Italian surnames names like "De Castro", "Dela Cruz", "La Rosa", etc. Split will return 3 instead of 2 words:

Array {   [0] => 'Pedro'   [1] => 'De'   [1] => 'Castro' } 

You'll end up with messages like "Welcome back Ana De" or "Editing Profile of Monsour La".

Same thing will happen for two-word names like "Anne Marie Miller", "William Howard Taft", etc.

Just a tip.



回答3:

This could be due the fact that $fullname did not contain a space character.

This example should fix your problem w/o displaying this notice:

$split = explode(' ', $fullname, 2); $first = @$split[0]; $last = @$split[1]; 

Now if $fullname is "musoNic80" you won't get a notice message.

Note the use of "@" characters.

HTH Elias



回答4:

Presumably, whatever $fullname is doesn't contain a space, so $split is an array containing a single element, so $split[1] refers to an undefined offset.



回答5:

That' strange, it's working correct here. When i try with a string the cat walks and also just the will do and not produce an error. I've outputted it with print_r

What's your $fullname looks like when you get the error?



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