Javascript: accessing multidimensional arrays

Question

I'm trying to create a new set of arrays from an existing array, using the code:

var stock = new Array()
    for (i = 0; i < seOutput.length; i++) {
      stock[i] = new Array(seOutput[i][0], seOutput[i][1], seOutput[i][2], seOutput[i][3], seOutput[i][4])      
}

If I console.log(seOutput); I see this:

Array
(
  [0] => Array
      (
          [0] => Blue
          [1] => Yes
          [2] => No
          [3] => Yes
          [4] => Yes
      )

  [1] => Array
      (
          [0] => Red
          [1] => No
          [2] => No
          [3] => No
          [4] => Yes
      )

This worked logically in my head, where the code seOutput[0][0] would equal Blue, but that's not what's given back in my console.log(seOutput[0][0] statement. For the statements console.log(seOutput[0]) and console.log(seOutput[0][0]) I get A (the first letter in the word Array ).

What is the proper way to access this array?

To give a bit more detail, seOutput is generated from my ajax success call to a .php page where the final bit of code is:

foreach ($json['Result']['Data']['Type'] as $i) {

  $y = "{$i['Color']},{$i['Object']},{$i['Crs']},{$i['CrsX']},{$i['CrsB']}"; 
  $x = explode("," , $y);
  array_push($output, $x);
}

then

print_r($output);

and my ajax code:

$.ajax({
              url: "seSearch.php", 
              type: "post",                  
              data: {partNumber: q , c:c},
              success: function(data) {

                  var seOutput = data;
  }
});

What's generated from that is the "array" I pasted above. Why ajax has converted this into a string I don't know....

stock is built from seOutput, and the for loop works, it just doesn't populates with bogus data because I'm not referencing seOutput correctly. — Brian Powell
– Brian Powell, Commented Feb 18, 2015 at 15:27

farincz · Accepted Answer · 2015-02-18 15:29:48Z

1

I think you assigned string to seOutput (containing array string representation) instead of real array in. So you access it properly but you have no array. Try print typeof(seOutput) to check it.

answered Feb 18, 2015 at 15:29

farincz

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5 Comments

Brian Powell Over a year ago

string - that's weird. So I don't even have an array here. Is there a simple way to convert this into an array? (Also, aside from that, was my logic on how to access this if it WERE an array correct?)

atmd Over a year ago

Yes, the code is correct IF it was an array. how are you building the array?

farincz Over a year ago

I don't know where you get this string. But there must exists original array before conversion to string. Maybe you accidentally make conversion somewhere.

farincz Over a year ago

you need php.net/manual/en/function.json-encode.php instead of print_r(). Also set header Content-Type: application/json to proper parsing in $.ajax

Brian Powell Over a year ago

how silly of me... yeah, doing json for both solved it. Thanks @farincz

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