<?php
for ($i = 'a'; $i <= 'z'; $i++)
echo "$i\n";
This snippet gives the following output (newlines are replaced by spaces):
a b c d e f g h i j k l m n o p q r s t u v w x y z aa ab ac ad ae af ag ah ai aj ak al am an ao ap aq ar as at au av aw ax ay az ba bb bc bd be bf bg bh bi bj bk bl bm bn bo bp bq br bs bt bu bv bw bx by bz ca cb cc cd ce cf cg ch ci cj ck cl cm cn co cp cq cr cs ct cu cv cw cx cy cz da db dc dd de df dg dh di dj dk dl dm dn do dp dq dr ds dt du dv dw dx dy dz ea eb ec ed ee ef eg eh ei ej ek el em en eo ep eq er es et eu ev ew ex… on to yz
NOTE: This question was originally posted at StackOverflow.com by Milan Babuškov
- Arthur asked 14 years ago
- last edited 14 years ago
From the docs:
PHP follows Perl’s convention when dealing with arithmetic operations on character variables and not C’s.
For example, in Perl ‘Z’+1 turns into ‘AA’, while in C ‘Z’+1 turns into ‘[‘ ( ord(‘Z’) == 90, ord(‘[‘) == 91 ).
Note that character variables can be incremented but not decremented and even so only plain ASCII characters (a-z and A-Z) are supported.
NOTE: This answer was originally posted at StackOverflow.com by CMS
- Wayne answered 14 years ago
- lol, mind blowing.
NOTE: This comment was originally posted at StackOverflow.com by Madan Sapkota
- in PHP 'z'+1; is 1, not 'aa' but if $a='z'; $a++; then $a=='aa' codepad.org/SnjELFxD
NOTE: This comment was originally posted at StackOverflow.com by codez
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Because once ‘z’ is reached (and this is a valid result within your range, the $i++ increments it to the next value in sequence), the next value will be ‘aa’; and alphabetically, ‘aa’ is < ‘z’, so the comparison is never met
for ($i = 'a'; $i != 'aa'; $i++)
echo "$i\n";
NOTE: This answer was originally posted at StackOverflow.com by Mark Baker
- Amy answered 14 years ago
- mantap..
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Others answers explain the observed behavior of the posted code. Here is one way to do what you want (and it’s cleaner code, IMO):
foreach (range('a', 'z') as $i)
echo "$i\n";
In response to ShreevatsaR’s comment/question about the range function: Yes, it produces the “right endpoint”, i.e. the values passed to the function are in the range. To illustrate, the output I got was:
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
s
t
u
v
w
x
y
z
NOTE: This answer was originally posted at StackOverflow.com by GreenMatt
- Jimmy answered 14 years ago
- last edited 14 years ago
- oke….report
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Wow I really didn’t know about this but its not a big code you can try echo “z” after loop Mark is Absolutely Right I use his method but if you want alternative then this may also you can try
<?php
for($i="a"; $i="y"; $i++)
{
echo "$i\n";
if($i=="z")
{
}
}
echo "z";
?>
NOTE: This answer was originally posted at StackOverflow.com by Mohit Bumb
- Jane answered 13 years ago
- oke keren…
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While the above answers are insightful to what’s going on, and pretty interesting (I didn’t know it would behave like this, and its good to see why.
The easiest fix (although perhaps not the most meaningful) would be just to change the condition to $i != ‘z’
<?php
for ($i = 'a'; $i != 'z'; $i++)
echo "$i\n";
?>
NOTE: This answer was originally posted at StackOverflow.com by jon_darkstar
- Nancy answered 14 years ago
- last edited 14 years ago
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Why not just use range('a','z')?
NOTE: This answer was originally posted at StackOverflow.com by stillstanding
- Dean answered 14 years ago
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Php has the function of looping letters and can exceed beyond single characters, the rest will be done this way: aa ab ac… zz, and so on.
Try this:
<?php
for ($i = 'a'; $i !== 'aa'; $i++)
echo "$i\n";
?>
NOTE: This answer was originally posted at StackOverflow.com by James Dantes
- Tom answered 12 years ago
- last edited 12 years ago
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<?php
$i = 'a';
do {
echo ($j=$i++),"\r\n";
} while (ord($j) < ord($i));
?>
NOTE: This answer was originally posted at StackOverflow.com by Matt H.
- Rose answered 13 years ago
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Other’s already said why PHP doesn’t show what you expect, here’s how you get the result you might want
<?php
for ($i = ord('a'); $i <= ord('z'); $i++)
echo chr($i);
?>
NOTE: This answer was originally posted at StackOverflow.com by Filip Ekberg
- Tim answered 14 years ago
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NOTE: This comment was originally posted at StackOverflow.com by GWW
NOTE: This comment was originally posted at StackOverflow.com by joni
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NOTE: This comment was originally posted at StackOverflow.com by joni
NOTE: This comment was originally posted at StackOverflow.com by Jeffrey