Iterate loop in php associative array using Jquery

Question

I have PHP associative array and I use JQuery AJAX to get the result array but my problem is when that result is pass to jquery and use looping to extract each Sequence,Percent and Date then that extracted data will store to new Jquery Array for data manipulation purposes. Please see my sample code so far.

sample code PHP ARRAY:

$Sequence=array(
    array("Seq1","20%"),
    array("Seq2","40%"),
    array("Seq3","60%"),
    array("Seq4","80%"),
    array("Seq5","100%")
);

****For loop here****

$ResultArray[$arrayIndex]=array(
    'Sequence' => $Sequence[$arrayIndex][0],
    'Percent' => $Sequence[$arrayIndex][1],
    'Date' => $row['exactDate']
);


echo json_encode($ResultArray); // then pass result array to jquery

JQUERY :

$(document).ready(function(){

    var ListOfSequence = []
    var ListOfPercentage = [];
    var ListOfDates = [];

    $("#button").click(function(){

        var _WOID = $('#txtWOID').val();

        $.ajax({
            url:'getStatus.php',
            type:'POST',
            data:{id:_WOID},
            dataType:'json',
            success:function(output){
                //here is where the problem begin
                for (var key in output) {
                    if (output.hasOwnProperty(key)) {
                        //here where extracted data will store to designated array
                        ListOfSequence.push(key);//<---store extracted Sequence
                        ListOfPercentage.push(key);//<---store percentage
                        ListOfDates.push(output[key]);//<---store dates                
                    }
                }

                ListOfPercentage.reverse();

                console.log(ListOfPercentage);
                console.log(ListOfDates);
                console.log(ListofSequence);

            }

        });

    });

});

and here's the console.log:

enter image description here

Thank you in advance

DarkMukke · Accepted Answer · 2014-10-16 09:03:04Z

1

Since you are already using jQuery you could use $.each() :

$(document).ready(function(){

  var ListOfSequence = []
  var ListOfPercentage = [];
  var ListOfDates = [];

  $("#button").click(function(){

    var _WOID = $('#txtWOID').val();

    $.ajax({
      url:'getStatus.php',
      type:'POST',
      data:{id:_WOID},
      dataType:'json',
      success:function(json){
         $.each(json, function(index, object){
             ListOfSequence.push(object.Sequence);
             ListOfPercentage.push(object.Percent);
             ListOfDates.push(object.Date);
         });

      }

    });

  });

});

answered Oct 16, 2014 at 9:03

DarkMukke

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Comments

mocanuga · Accepted Answer · 2014-10-16 09:01:45Z

1

You should set the json response header before sending the content to the browser like so:

header('Content-type: application/json'); die(json_encode($ResultArray);)

answered Oct 16, 2014 at 9:01

mocanuga

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9 Comments

Waelhi Over a year ago

ok sir... if you don't mind, what will happen if I don't set json response header?

DarkMukke Over a year ago

@mocanuga never ever tell anyone to use eval() its really bad practice unless you know what you are doing (and OP clearly is not), a better aswer would have been var json = JSON.parse(output);

DarkMukke Over a year ago

@mocanuga since OP is using dataType : json it does not need to be eval'ed or w/e, jquery will take care of that

mocanuga Over a year ago

You're right. It's just that I always set the header just to be safe. Also, I didn't recommend anyone to use eval. I just told him that, without the json header, he might need to. JSON.parse might not be available in older browsers.

DarkMukke Over a year ago

and that's why I didn't down rate your answer itself because imo that is an improvement and good practice, just the comments aren't (and sorry I explode every time I see someone say eval()), and you are right about older browsers but that must be something like IE6/7 : stackoverflow.com/questions/4908875/… if MS stops supporting IE6/7 why should we still support them ?

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