Working with arrays in jQuery

Question

I am stuck on a arrays in jQuery. I am returning back from php an array In php file I am adding array like this in a loop:

$table_data[]= array("id"=>mysql_result($query,$i,"id"),"name"=>trim(mysql_result($query,$i,"name")));

at the end of php file :

echo json_encode($table_data);

On my jquery :

 $.ajax({
   type: "POST",
   url: "phpfilename.php",
   data: ({
     newtask: "grab"
   }),
   dataType: "json",
   success: function(data){
     alert("value - "+data.length);
   }
 });

(this returns correct record counts)

$.each(data, function(key, value) {
 alert( "The key is '" + key + "' and the value is '" + value + "'" );
});

(the above loop returns me : 0,1,2 as a key , Object as a value )

I need help with understanidn how to pass arrays from php to jquery and how to handle them . Maybe I am totally on wrong way.

What's the problem? You're getting the array and looping through it. What are you asking how to do? — aquinas
– aquinas, Commented May 22, 2012 at 12:05
You should read the MDN JavaScript Guide to learn the basics about arrays and objects in JavaScript. alert is the worse debug tool you could have chosen. The output is correct though, since you have an array of objects and the default string representation of an object is [object Object]. — Felix Kling
– Felix Kling, Commented May 22, 2012 at 12:05

thecodeparadox · Accepted Answer · 2012-05-22 12:20:21Z

1

To access any property of an object, the process is:

ObjectName.property

or

ObjectName[''+ property +''];

read here

$.each(data, function(key, value) {

  // try

  console.log(value.id);
  console.log(value.name);

});

edited May 22, 2012 at 12:20

answered May 22, 2012 at 12:05

thecodeparadox

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3 Comments

Samir Memmedov Over a year ago

that is it, great. thank you , I was searching it for hours. Thank you very much

Roko C. Buljan Over a year ago

you mean ObjectName["property"];

Samir Memmedov Over a year ago

$.each(data, function(key, value) { // try console.log(value.id); console.log(value.name); }); this one was helpfull

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