Question

$.post("/diabetes/ropimages/getcount.php",{pid:$("#patient_id").val()} ,function(data1){
  //alert(data1);
  var count = data1;
  var pid = $("#patient_id").val();
  var rid;

  for( var i = 1 ; i <= count ; i++) {
    var link ='<img src="/diabetes/ropimages/thumbpicdisplay.php?pid=+pid+&rid=1" />';
    $("#content1").empty().html(link);
  }
});

I am trying to pass pid value in url ..but its taking directly as +pid+ as value ..how do i give it the value of pid.

And how do i print 3 images in a div? like the one in code

Answer 1

You simply need to terminate the string after ?pid= and then use the concatenation operator (+) to “insert” the pid variable in the appropriate location:

'<img src="/diabetes/ropimages/thumbpicdisplay.php?pid=' + pid + '&rid=1" />'

As for attaching the 3 images to the div, you might have more luck doing the following:

var link = '';

for(var i = 1; i <= count; i++) {
   link += '<img src="...thumbpicdisplay.php?pid=' + pid + '&rid=1" />';
}

$("#content1").empty().html(link);