PHP string undefined offset error

Question

I have one .txt file and one .php file

.txt:

15118553;1315239266;0;1;EURUSD;1.4111;0;0;1315239282;1.4109;0;0;-27;0;

.php:

<?php
$text = file_get_contents('test.txt', true);
$sir = explode(";", $text);
$d=count($sir);

for($i=0; $i<=$d ; $i++) 
    echo "string = ". $i ."-------". $sir[$i]. "<br>" ;
?>

in localhost :

string = 0-------15118553
string = 1-------1315239266
string = 2-------0
string = 3-------1
string = 4-------EURUSD
string = 5-------1.4111
string = 6-------0
string = 7-------0
string = 8-------1315239282
string = 9-------1.4109
string = 10-------0
string = 11-------0
string = 12--------27
string = 13-------0
string = 14-------

Notice: Undefined offset: 15 in C:\xampp\htdocs\test\index.php on line 7
string = 15-------

Why do they display 15 strings? In the txt file there are only 13 strings. How do I fix this error?

Tip: do not use for to iterate arrays. Use foreach.

georg
– georg

2014-08-08 09:11:03 +00:00
Commented Aug 8, 2014 at 9:11 — georg
– georg, Commented Aug 8, 2014 at 9:11

k102 · Accepted Answer · 2014-08-08 09:04:47Z

2

Why "Undefined offset":

Change it to:

for($i=0; $i < $d ; $i++) Because you are starting to count elements from zero, not from one.

Why 14:

Because your input string ends with the delimiter, so after exploding you get jne extra string.

You can either cycle to $d - 1, or use an if condition to check if this string is an empty string.

answered Aug 8, 2014 at 9:04

k102

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1 Comment

Aleksei Matiushkin Over a year ago

Trimming the value of $text is needed as well to avoid printing empty string @pos 14.

Awlad Liton · Accepted Answer · 2014-08-08 09:17:17Z

1

That error means that you do not have offset '15' in $sir array.

You should iterate until $count - 1.

But as you are a beginner for skipping this types of issues for array always use foreach.

Try like this:

 foreach($str as $strValue) {
    echo "string = ". $strValue. "<br>" ;
}

answered Aug 8, 2014 at 9:17

Awlad Liton

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Comments

Abhishek Goel · Accepted Answer · 2014-08-08 09:08:00Z

0

for($i=0; $i<$d ; $i++)

loop will run from 0 to n-1

answered Aug 8, 2014 at 9:08

Abhishek Goel

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Comments

ChoiBedal · Accepted Answer · 2014-08-08 09:23:45Z

Because array start at the index 0.

So, when you count lines by yourself, you start : "1, 2, 3, 4,..." when an array starts "0, 1, 2, 3 ...". So you have a 15th line that appears, but crashes because $d[15] = null;

Moreover, you wrote "$i<=$d". And $d[14] = "", because it is between ;and the end of the line. You can echo "Count : ".$d, and see Count : 14.

To understand better :

Line[x] | Array[x] | count($sir) | Array <= $d // $d = count($sir);

   1          0          14          TRUE
   2          1          14          TRUE
   3          2          14          TRUE
   4          3          14          TRUE
   5          4          14          TRUE
   6          5          14          TRUE
   7          6          14          TRUE
   8          7          14          TRUE
   9          8          14          TRUE
   10         9          14          TRUE
   11         10         14          TRUE
   12         11         14          TRUE
   13         12         14          TRUE
   14         13         14          TRUE
   15         14         14          TRUE  // --> Array[14] == "". PHP prints empty string, but doesn't crash.
   16         15         14          FALSE // --> Array[15] == null. PHP crashes.

That's why $i (which equals Array[x] here) stacks until $i = 15 before crashing.

You might want to write :

for($i = 0;$i < $d-1;i++)
{
// do something;
}

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