CODEWITHSUNDEEP
php
Sami Al-Subhi
Sami Al-Subhi

Reputation: 4662

Undefined offset with an array

Why can't I use variable which contains a number to specify an array value:

$info(array)
$mySQLHeadings(array)

$infoString(empty String)
$mySQLHeadingString(empty string)

for ($i=0; $i<=count($info) ; $i++){
    if($info[$i] != ""){
        $mySQLHeadingString .= $mySQLHeadings[$i] . ",";
        $infoString .= "'". $info[$i] ."',";
    }
}

PHP says it's an undefined offset $i in the arrays. How can I correct it or do something similar. Thank you so much.

Upvotes: 0

Views: 85

Answers (2)

AlterPHP
AlterPHP

Reputation: 12717

If $info is numerically indexed, you can access elements with $i, but not further than max index !

count() gives you the array length, but max numeric index is (length - 1)

so :

for ($i=0; $i < count($info); $i++) {
    //....
}

Upvotes: 0

matino
matino

Reputation: 17715

You should write for ($i = 0; $i < count($info); $i++). Array indexes start from 0 while count() starts from 1.

Also don't use count() inside for loop - move it before:

$count_info = count($info);
for ($i = 0; $i < $count_info; $i++)

Upvotes: 2

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