PHP sizeof（）返回1错误

Question

This code worked last week, and I don't remember changing this code recently, but for some reason it stopped working today.

I've narrowed down the issue, but I'm confused about it. Basically, $fields = sizeof($dataArray[$r]); yields $fields = 1, but as you can see in the image below, ECHO json_encode($dataArray); demonstrates that $fields = 9 should be observed.

$dataArray = $_POST['array'];

$validRows = array();
$invalidRows = array();

echo sizeof($dataArray); // correct output

for($r = 0; $r < sizeof($dataArray); $r++){

  $fields = sizeof($dataArray[$r]); // $fields is not what I expect

  ECHO json_encode($dataArray); // see image below (sent to javaScript alert)
  ECHO sizeof($fields); // returns 1

  ...
}

ECHO json_encode($dataArray); Seen in a javaScript alert

What's my issue here?

Answer 1

Based on PHP Documentation, sizeOf (alias of Count) returns 1 if the array is not countable .

http://php.net/manual/en/function.count.php

Returns the number of elements in array_or_countable. If the parameter is not an array or not an object with implemented Countable interface, 1 will be returned. There is one exception, if array_or_countable is NULL, 0 will be returned.

But appart from that, on your second ECHO, you are displaying the sizeOff ot the result of a sizeoff , that would be a single number pseudo: a = count("a","b","c") a should be 3 ECHO count(a) ECHO Display 1, as (a) is a single element