PHP Notice Undefined offset 1 with array when reading data

Question

I am getting this PHP error   PHP Notice   Undefined offset  1   Here is the PHP code that throws it    file handle   fopen  path   Summary data txt   r      open text file  data   array       create new array map  while   feof  file handle           line of text   fgets  file handle      read in each line      parts   array map  trim   explode       line of text  2            separates line of text by     trim strings for extra space      data  parts 0      parts 1           map the resulting parts into array         results  NAME BEFORE       VALUE AFTER       What does this error mean   What causes this error

User · Answer

How to reproduce the above error in PHP   php gt   yarr   array 3   gt   c   4   gt   d     php gt  echo  yarr 4   d  php gt  echo  yarr 1   PHP Notice   Undefined offset  1 in   usr local lib python2 7 dist-packages phpsh phpsh php 578     eval   d code on line 1   What does that error message mean   It means the php compiler looked for the key 1 and ran the hash against it and didn t find any value associated with it then said Undefined offset  1  How do I make that error go away   Ask the array if the key exists before returning its value like this   php gt  echo array key exists 1   yarr    php gt  echo array key exists 4   yarr   1   If the array does not contain your key  don t ask for its value   Although this solution makes double-work for your program to  check if it s there  and then  go get it      Alternative solution that s faster   If getting a missing key is an exceptional circumstance caused by an error  it s faster to just get the value  as in echo  yarr 1     and catch that offset error and handle it like this  https   stackoverflow com a 5373824 445131

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I just recently had this issue and I didn t even believe it was my mistype   Array  Semester has been set as active    true  Array  Failed to set semester as active    false    And actually it was  I just accidentally typed     rather than

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In your code   parts   array map  trim   explode       line of text  2     You have  quot   quot  as separator  If you use another separator in file  then you will get an  quot Undefined offset  1 quot  but not  quot Undefined offset  0 quot  All information will be in  parts 0  but no information in  parts 1  or  2  etc  Try to echo  part 0   echo  part 1   you will see the information

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Hide php warnings in file  error reporting 0

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This is a  PHP Notice   so you could in theory ignore it   Change php ini   error reporting   E ALL  amp   E NOTICE  amp   E STRICT  amp   E DEPRECATED   To  error reporting   E ALL  amp   E NOTICE   This show all errors  except for notices

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The ideal solution would be as below  You won t miss the values from 0 to n    len count  data   for  i 0  i lt  len  i         echo  data  i     lt br gt

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Update in 2020 in Php7  there is a better way to do this using the Null coalescing operator by just doing the following   data  parts 0      parts 1     null

User · Answer

my quickest solution was to minus 1 to the length of the  array as      len   count  data        for  i 1   i lt   len-1  i           echo  data  i           my offset was always the last value if the count was  140 then it will say offset 140 but after using the minus 1 everything was fine

User · Answer

Change    data  parts 0      parts 1     to  if     isset  parts 1          parts 1    null      data  parts 0      parts 1     or simply    data  parts 0     isset  parts 1      parts 1    null    Not every line of your file has a colon in it and therefore explode on it returns an array of size 1   According to php net possible return values from explode      Returns an array of strings created by splitting the string parameter on boundaries formed by the delimiter       If delimiter is an empty string       explode   will return FALSE  If delimiter contains a value that is not contained in string and a negative limit is used  then an empty array will be returned  otherwise an array containing string will be returned

User · Answer

The output of the error  is because you call an index of the Array that does not exist  for example   arr   Array 1 2 3   echo  arr 3       Error PHP Notice   Undefined offset  1 pointer 3 does not exist  the array only has 3 elements but starts at 0 to 2  not 3

[php] PHP Notice: Undefined offset: 1 with array when reading data

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