What causes: "Notice: Uninitialized string offset" to appear?

user_113329 0
Viewed 213136

I have a form that users fill out, and on the form there are multiple identical fields, like "project name", "project date", "catagory", etc. Based on how many forms a user is submitting, my goal is to:

  1. loop over the number of forms
  2. create individual SQL insert statements

However, PHP throws me a NOTICE that I don't seem to understand:

Notice:

Notice: Uninitialized string offset: 1 ...dataPasser.php on line 90

PHP

$myQuery = array();

if ($varsCount != 0)
{
  for ($i=0; $i <= $varsCount; $i++)
  {
    $var = "insert into projectData values ('" . $catagory[$i] . "', '" .  $task[$i] . "', '" . $fullText[$i] . "', '" . $dueDate[$i] . "', null, '" . $empId[$i] ."')";
    array_push($myQuery, $var);     
  }
}

There are references to this issue I am having, but they are not exact and I am having trouble deducing where the actual problem stems from. I would greatly appreciate any help in understanding what is causing the array to not initialize properly.

arrays php
5 Answers

This error would occur if any of the following variables were actually strings or null instead of arrays, in which case accessing them with an array syntax $var[$i] would be like trying to access a specific character in a string:

$catagory
$task
$fullText
$dueDate
$empId

In short, everything in your insert query.

Perhaps the $catagory variable is misspelled?

user_812050

It means one of your arrays isn't actually an array.

By the way, your if check is unnecessary. If $varsCount is 0 the for loop won't execute anyway.

cletus_183930

Try to test and initialize your arrays before you use them :

if( !isset($catagory[$i]) ) $catagory[$i] = '' ;
if( !isset($task[$i]) ) $task[$i] = '' ;
if( !isset($fullText[$i]) ) $fullText[$i] = '' ;
if( !isset($dueDate[$i]) ) $dueDate[$i] = '' ;
if( !isset($empId[$i]) ) $empId[$i] = '' ;

If $catagory[$i] doesn't exist, you create (Uninitialized) one ... that's all ; => PHP try to read on your table in the address $i, but at this address, there's nothing, this address doesn't exist => PHP return you a notice, and it put nothing to you string. So you code is not very clean, it takes you some resources that down you server's performance (just a very little).

Take care about your MySQL tables default values

if( !isset($dueDate[$i]) ) $dueDate[$i] = '0000-00-00 00:00:00' ;

or

if( !isset($dueDate[$i]) ) $dueDate[$i] = 'NULL' ;
user_2661350
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