CODEWITHSUNDEEP
php
Sylvia Gerald
Sylvia Gerald

Reputation: 29

Undefined offset: 1 For Loop PHP

The array it is fetching has size 2 , so if we comment out $size it returns 2 , but the for loop prints only the value present at 0th position of table column and not the next at 2th giving the error:

Undefined offset: 1

Code:

$allunstitched="Select p_id from unstitchedproduct";
$resultAllUnstitched= mysqli_query($connection,$allunstitched);
$AllUnstitchedResult= mysqli_fetch_array($resultAllUnstitched);
//$size=count($AllUnstitchedResult);

for ($i=0, $count = count($AllUnstitchedResult); $i < $count; ++$i) {
  print $AllUnstitchedResult[$i];
}

Upvotes: 0

Views: 268

Answers (3)

user1299518
user1299518

Reputation:

Um, i dont think you have a problem in your for-loop; since your query reads ONE field/column it should have 1 iteration (=0 index), so the =1 undefined is correct (in my opinion).

What you should do is a loop that calls repeatedly mysqli_fetch_array to get all rows, like this:

while($AllUnstitchedResult = mysqli_fetch_array($resultAllUnstitched,MYSQLI_NUM)) {
   foreach ($AllUnstitchedResult as $field) {
      print $field;
   }
}

Upvotes: 0

Joni
Joni

Reputation: 111279

You're using mysqli_fetch_array. This function returns the result twice by default: indexed both by number and by column name. Here the result set has only one column, so you get its value twice in the array: first with index 0 and then with a string index. There is no index 1.

To get the results with numerical indexes only, pass MYSQLI_NUM as a second argument to mysqli_fetch_array:

mysqli_fetch_array($resultAllUnstitched, MYSQLI_NUM);

Upvotes: 1

Damien Pirsy
Damien Pirsy

Reputation: 25435

If you start at 0, if count = 5 you'll have 6 elements (0,1,2,3,4,5), beware of that. You need to loop (count-1) times:

 $count = count($AllUnstitchedResult);
 for ($i=0, $i < $count-1; $i++) {
    print $AllUnstitchedResult[$i];
 }

Upvotes: 0

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